# Sequence eliminating function (SEF) and the interval formulas of prime
numbers

Ahmed Hamdy A. Diab

###### Abstract

In this paper I will derive a function that eliminates any sequence of
equidistant numbers from the integer numbers, then I will derive its inverse,
Then I will use this function to eliminate the multiples of the prime numbers
from the integer numbers to produce formulas that generate all prime numbers
in some interval.

###### Contents

  1. 1 Introduction
  2. 2 Elimination function
    1. 2.1 Derivation of the elimination function
    2. 2.2 Derivation of the inverse of SEF
  3. 3 Derivation of prime numbers formulas
    1. 3.1 Elimination of the multiples of prime number 2
    2. 3.2 Elimination of the multiples of prime number 3
    3. 3.3 Elimination of the multiples of prime number 5
      1. 3.3.1 Eliminating the first sequence
      2. 3.3.2 Eliminating the second sequence
    4. 3.4 Elimination of the multiples of prime number 7
    5. 3.5 The interval of prime numbers produced
    6. 3.6 Generalization of my prime numbers formulas
  4. 4 conclusion

##  1 Introduction

Formulas that generate prime numbers are one of the most important prime
numbers problems, which have drawn the interest of mathematicians over the
centuries [2, 6, 1, 3], Much work has been done on this subject like, Euler’s
prime generating polynomial [5, 1], Fermat numbers [4, 1], Mersenne primes [1]
and many other works, Each of them made their own contribution to the prime
numbers, which helped to understand the prime numbers more broadly than
before, but the topic remains shrouded in mystery and many unanswered
questions[2, 3]. In this paper I will focus on producing a prime generating
formulas, I will explain how to derive an infinite number of these formulas,
which generates all of the prime numbers in some interval see 3.5, we will
start with deriving an equation that eliminates a sequence of an equidistant
numbers see 2.1, then we will derive its inverse see 2.2, then we will use
this equation to eliminate the multiples of prime numbers one by one from the
integer numbers in a certain method see 3, this will produce the desired
formulas, at the end of this paper I will make a generalized formula to
generally describe my prime numbers formulas see 3.6.

##  2 Elimination function

###  2.1 Derivation of the elimination function

The elimination function is a function that eliminates a sequence of
equidistant numbers from the integer numbers.

These are the steps to derive the elimination function:

  1. 1.

Let $n=\ \ 1,\ \ 2,\ \ 3,\ \ldots.\ $ , $y\in\mathbb{Z}$
then let

$\frac{n}{y}=p$ (1)

$p$ is an integer number when $n$ equals $y$ or a multiple of $y$, the
difference between every two adjacent integer values of $p$ is the same, and
equals $y$, we will use these properties to obtain the elimination of a
sequence of equidistant numbers.

  2. 2.

Adding constant $\frac{c}{y}\ $ to the left-hand side of eq(1) where
$c\in\mathbb{Z}$

$\frac{n}{y}+\frac{c}{y}=p$

$\therefore\ \frac{n+c}{y}=p$ (2)

This constant will shift our results $p$ corresponding to $n$ depending on the
value of $c$.

  3. 3.

Taking the floor to the left-hand side of eq(2)

$\therefore\ \left\lfloor\frac{n+c}{y}\right\rfloor=p$ (3)

The results will be integers, This function’s mechanism is adding one to the
results after $y$ numbers.

  4. 4.

Adding $n$ to the left-hand side of eq(3)

$\therefore\ n+\left\lfloor\frac{n+c}{y}\right\rfloor=p$ (4)

substituting by $n$ in eq(4) the result is the integer numbers without a
specific sequence of numbers where the number of numbers between each two
successive numbers in the sequence equals $y$ and the difference between these
two adjacent numbers is $y+1$ .
Let $x_{1}$: Is the first number in the eliminated sequence.
$x_{2}$: Is the second number in the eliminated sequence.

$\therefore\ x_{2}-x_{1}=y+1$ (5)

Simplifying eq(4)

$n+\left\lfloor\frac{n+c}{y}\right\rfloor=p$

$\left\lfloor\frac{ny+n+c}{y}\right\rfloor=p$

$\left\lfloor\frac{(y+1)n+c}{y}\right\rfloor=p$

From eq(5)

$\left\lfloor\frac{(x_{2}-x_{1})n+c}{(x_{2}-x_{1})-1}\right\rfloor=p$ (6)

We will call eq(6) by SEF for (Sequence Eliminating Function).

To avoid some problems while using the equation we will set the following
condition

$x_{1\ }\leq\ y+1$

Example (1): Eliminate the next sequence of numbers from the integers where
the difference between the adjacent numbers is 5.

The sequence | 3 | 8 | 13 | 18 | 23 | …
---|---|---|---|---|---|---

sol.
$x_{2}=8$, $x_{1}=3$
Sub. in eq(6)

$\left\lfloor\frac{5n+c}{4}\right\rfloor=p$ (7)

The method of calculating constant c:
Substituting by $n$ equals the first number in the sequence the result will be
the next number $(n+1)$, and the value of the equation inside the floor in
this case is an integer number so we can neglect the floor.

Applying this to the example:
Substituting by $n=3$, $p=4$ in eq(7)

$\frac{15+c}{4}=4$

$15+c=16$

$c=1$

Sub. by $c$ in eq(7)

$\left\lfloor\frac{5n+1}{4}\right\rfloor=p$

Sub. by n=1, 2, 3, ….

n | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | ….
---|---|---|---|---|---|---|---|---|---|---|---
p | 1 | 2 | 4 | 5 | 6 | 7 | 9 | 10 | 11 | 12 | ….

The sequence is eliminated from the results.

###  2.2 Derivation of the inverse of SEF

  1. 1.

Let $l=1,\ 2,\ 3,\ ....$ , $m\in\mathbb{Z}$ Then let

$\frac{l}{m}=d\ $ (8)

The result is an integer numbers when $l$ equals $m$ or a multiple of it.

  2. 2.

adding $\frac{g}{m}$ to the left-hand side of eq(8) where $g$ is an integer
constant

$\frac{l}{m}+\frac{g}{m}=d$

$\therefore\ \frac{l+g}{m}=d$ (9)

This will shift the results depending to the value of $g$.

  3. 3.

Subtracting the left-hand side of eq(9) from $l$

$l-\frac{l+g}{m}=d$ (10)

  4. 4.

Taking the ceil to the left-hand side of eq(10)

$\left\lceil l-\frac{l+g}{m}\right\rceil=d$ (11)

The results will be just integers.
There is a sequence of numbers from the inputs has the same differences
between each two adjacent numbers in the sequence and this difference equals
$m$, the results $\left\\{d\right\\}$ of this sequence is the same as the
results of the next input after each number in the sequence like, the result
of one equals one and the result of two equals one also, this is the same for
all the numbers in the sequence, this will stand to infinity as we see from
the derivation.The results of the inputs without this sequence is the numbers
$1,2,3,4,…$, so if we took this inputs without this specific sequence we will
get the inverse effect that SEF makes. we will use this by adjusting our
inputs and the other constants to get the inverse of SEF, first thing is to
take the inputs $l=p$ where $p$ is the output of SEF and it is the numbers
$(1,2,3,4,…)$ without a specific sequence of numbers with the same differences
between each two adjacent numbers, this difference equals $y+1$, from this $m$
must equal $y+1$ and this is the second substitution, the last step is
substituting by $c=g$ because we want to get the same shift at the results
that $c$ makes, by these adjustments the results $d$ will equal $n=1,\ 2,\ 3,\
....$ .
Before substituting we will simplify eq(11).

$\left\lceil l-\frac{l+g}{m}\right\rceil=d$

$\left\lceil\frac{ml-l\ -\ g}{m}\right\rceil=d$

$\left\lceil\frac{(m-1)l-g}{m}\right\rceil=d$ (12)

  5. 5.

Applying the next substitutions to eq(12)

     * •

$l=p$

     * •

$m=y+1$

     * •

$c=g$

     * •

$d=n=1,\ 2,\ 3,\ ....$

So eq (12) will be

$\left\lceil\frac{(y+1-1)p-c}{y+1}\right\rceil=d$

$\left\lceil\frac{yp-c}{y+1}\right\rceil=n$

$\left\lceil\frac{(x_{2}-x_{1}-1)p-c}{(x_{2}-x_{1})}\right\rceil=n$ (13)

Eq(13) is the inverse of SEF and we can call it ISEF.

Example (2): solve:

$\left\lfloor\frac{5n+1}{4}\right\rfloor=p$

sol.

$\left\lceil\frac{4p-1}{5}\right\rceil=n$

##  3 Derivation of prime numbers formulas

We will use eq(6) to make the prime numbers formulas by eliminating the
multiples of prime numbers one by one, The elimination process includes the
prime number itself.

###  3.1 Elimination of the multiples of prime number 2

The first prime number whose multiples we want to eliminate is 2.

This is the produced formula after eliminating the multiples of 2 by using
eq(6)

$2n-1=p$ (14)

where $n=\ 1,\ \ 2,\ \ 3,\ \ldots.\ $

The first number in the results is one, To shift our results and get rid of
number one which is not a prime number, we will replace $n$ by $n+1$ in
eq(14).

$2(n+1)-1=p$

$2n+1=p$ (15)

where $n=\ 1,\ \ 2,\ \ 3,\ \ldots.\ $

This formula - eq(15) - gives us all the prime numbers in the interval
$\left]2,3^{2}\right[$.

###  3.2 Elimination of the multiples of prime number 3

The second prime whose multiples we want to eliminate is 3, In the results of
eq(15) we don’t have all the multiples of 3 this because there are common
multiples between 2 and 3 that have already been eliminated, In this case this
will not create any problem (I will explain the possible problems and how to
solve them while eliminating the multiples of 5) because multiples of three in
the results of eq(15) are a sequence of equidistant numbers, So we can easily
eliminate them by SEF.

We will eliminate the multiples of 3 in $p$ from the results of eq(15) by
eliminating its corresponding $n$ which is $(n=1,4,7,10,....)$ from $n$,

Using SEF to eliminate the sequence $(n=1,4,7,10,....)$ from $n$,

$\therefore\left\lfloor\frac{3n_{1}+1}{2}\right\rfloor=n$ (16)

Sub. by $n$ from eq(16) in eq(15)

$2\left\lfloor\frac{3n_{1}+1}{2}\right\rfloor+1=p$ (17)

where $n_{1}=1,\ 2,\ 3,\ ....$

This is the formula that eliminates the multiples of 2 and 3 from the
integers, And gives us all the prime numbers in the interval
$\left]3,5^{2}\right[$.

###  3.3 Elimination of the multiples of prime number 5

The third prime whose multiples we want to eliminate from eq(17) is 5, we can
easily determine it from our last formula.

In the results of eq(17) We don’t have all the multiples of 5 because of the
common multiples between 2 and 3 with 5 that are already eliminated, In this
case the multiples remaining are not a sequence of equidistant numbers, So we
cannot eliminate them directly.

There are common multiples between 2, 3 and 5 that you can get by multiplying
the multiples of 6 by 5, Since the multiples of 6 are common multiples between
2 and 3.

The common multiples between 2, 3 and 5 are a sequence of equidistant numbers,
The distribution of the multiples of 5,3 and 2 is the same after each number
in this sequence, and the uncommon multiples of 5 with 2 and 3 too, From this
we will divide the multiples of 5 into a number of sequences equal to the
number of uncommon multiples of 5 with 2 and 3 in a region bounded by two
common multiples between 2, 3 and 5, We can use the region between 0 and the
first common multiple between 2, 3 and 5, The resulting sequences are a
sequences of equidistant numbers, All sequences have the same difference
between its adjacent numbers, So that we can eliminate its corresponding
inputs from our last formula - eq(17) in our example - by SEF, for 5 there
will be two sequences because the number of the uncommon multiples of 5 with 2
and 3 from 0 to the first common multiple between 2, 3 and 5 is 2.

####  3.3.1 Eliminating the first sequence

The first number in the first sequence is 5, Its corresponding $n_{1}$ is 1,
The second number in the first sequence is 35, its corresponding $n_{1}$ is
11, The corresponding number can be determined using the inverse of eq(17).

using SEF

$\left\lfloor\frac{10n_{2}+8}{9}\right\rfloor=n_{1}$ (18)

Sub. by $n_{1}$ in eq(17)

$2\left\lfloor\frac{3\left\lfloor\frac{10n_{2}+8}{9}\right\rfloor+1}{2}\right\rfloor+1=p$
(19)

Where $n_{2}=1,\ 2,\ 3,\ ....$

####  3.3.2 Eliminating the second sequence

The second sequence had the same differences between its adjacent numbers as
the first, but the difference is reduced by one, because we will eliminate it
from the last formula - eq(19) -, where the first sequence is eliminated this
causes a missing number between each two adjacent numbers in the second
sequence,so the difference between their corresponding $n_{2}$ is 9.

The first number in the second sequence is 25, its corresponding $n_{2}$ is 7.

Using SEF

$\left\lfloor\frac{9n_{3}+1}{8}\right\rfloor=n_{2}$ (20)

Sub. by $n_{2}$ in eq(19)

$2\left\lfloor\frac{3\left\lfloor\frac{10\left\lfloor\frac{9n_{3}+1}{8}\right\rfloor+8}{9}\right\rfloor+1}{2}\right\rfloor+1=p$
(21)

Where $n_{3}=1,\ 2,\ 3,\ ....$

Eq(21) gives us all the prime numbers in the interval $\left]5,7^{2}\right[$.

###  3.4 Elimination of the multiples of prime number 7

The fourth prime is 7, using the same steps we used with 5 we can eliminate
the multiples of 7 and all the next primes, The difference will be the number
of sequences into which multiples (not eliminated multiples) of this prime
will be divided, for 7 it will be 8 sequences, and the formula that eliminates
the multiples of 7 from eq(21) is:

$\left\lfloor{\frac{\mathbf{50}\ast\left\lfloor{\frac{\mathbf{49}\ast\mathbf{n_{4}}+\mathbf{1}}{\mathbf{48}}}\right\rfloor+\mathbf{13}}{\mathbf{49}}}\right\rfloor=\mathbf{k}$

$\left\lfloor{\frac{\mathbf{54}\ast\left\lfloor{\frac{\mathbf{53}\ast\left\lfloor{\frac{\mathbf{52}\ast\left\lfloor{\frac{\mathbf{51}\ast\mathbf{k}+\mathbf{20}}{\mathbf{50}}}\right\rfloor+\mathbf{24}}{\mathbf{51}}}\right\rfloor+\mathbf{31}}{\mathbf{52}}}\right\rfloor+\mathbf{35}}{\mathbf{53}}}\right\rfloor=\mathbf{b}$

$\mathbf{2}\ast\left\lfloor{\frac{\mathbf{3}\ast\left\lfloor{\frac{\mathbf{10}\ast\left\lfloor{\frac{\mathbf{9}\ast\left\lfloor{\frac{\mathbf{56}\ast\left\lfloor{\frac{\mathbf{55}\ast\mathbf{b}+\mathbf{42}}{\mathbf{54}}}\right\rfloor+\mathbf{54}}{\mathbf{55}}}\right\rfloor+\mathbf{1}}{\mathbf{8}}}\right\rfloor+\mathbf{8}}{\mathbf{9}}}\right\rfloor+\mathbf{1}}{\mathbf{2}}}\right\rfloor+\mathbf{1}=\mathbf{p}$
(22)

Where $n_{4}=1,\ 2,\ 3,\ ....$

This formula - eq(22) - gives us all the prime numbers in the interval
$\left]7,11^{2}\right[$, I divided it into 3 parts because the length of the
floor brackets will be too long for one part.

###  3.5 The interval of prime numbers produced

All the previous prime numbers formulas give us all the prime numbers in an
open interval.

  1. 1.

The first term in this open interval is the last prime number whose multiples
we have eliminated in the formula, because it has already been eliminated and
also all the prime numbers smaller than it.

  2. 2.

The last term in the interval is the square of the first prime larger than the
last prime eliminated, as it is the first uncommon multiple of this prime with
the primes eliminated.

###  3.6 Generalization of my prime numbers formulas

This is a generalized formula that describes generally my prime numbers
formulas, to be able to refer to any of these formulas easily.

${A(P_{j})}^{M}=P$ (23)

The inverse is

${A^{-1}(P_{j})}^{M}=P$ (24)

Where
$P_{j}$: The last prime number whose multiples we have eliminated.
$M$: The number of sequences eliminated.

\- The generalized formula for:

  1. 1.

Eq(15):

${A(2)}^{1}=P$

  2. 2.

Eq(17):

${A(3)}^{2}=P$

  3. 3.

Eq(19):

${A(5)}^{3}=P$

  4. 4.

Eq(21):

${A(5)}^{4}=P$

##  4 conclusion

In this paper we have derived the sequence eliminating function (SEF) step by
step as well as its inverse, and illustrated how to use it in making the
formulas of prime numbers, in a process of eliminating the multiples of prime
numbers one by one from the integer numbers, we started by eliminating the
multiples of 2 then the multiples of 3 then the multiples of 5, in the process
of eliminating the multiples of 5 we had some problems then we showed how to
solve these problems, after solving these problems we will not have any
problem with eliminating the multiples of any prime greater than 5, then we
showed the formula that eliminates the multiples of 7 and all prime numbers
smaller than 7 which generates all prime numbers in the interval
$\left]7,11^{2}\right[$, and we showed how to determine these intervals for
all my prime numbers formulas, and finally I made a generalized formula for my
prime numbers formulas.

## References

  * [1] Richard Crandall and Carl B Pomerance.  Prime numbers: a computational perspective, volume 182.  Springer Science & Business Media, 2006.
  * [2] Richard Guy.  Unsolved problems in number theory, volume 1.  Springer Science & Business Media, 2004.
  * [3] Godfrey Harold Hardy, Edward Maitland Wright, et al.  An introduction to the theory of numbers.  Oxford university press, 1979.
  * [4] Michal Krizek, Florian Luca, and Lawrence Somer.  17 Lectures on Fermat Numbers: From Number Theory to Geometry.  Springer Science & Business Media, 2002.
  * [5] Paulo Ribenboim.  Euler’s famous prime generating polynomial and the class number of imaginary quadratic fields.  My Numbers, My Friends: Popular Lectures on Number Theory, pages 91–111, 2000.
  * [6] Eric W Weisstein.  Prime-generating polynomial.  https://mathworld. wolfram. com/, 2005.


#  $N$-tuple sum analogues for Ramanujan-type congruences

Mohamed El Bachraoui<EMAIL_ADDRESS>
###### Abstract.

In this paper we prove supercongruence relations for truncated $N$-tuple sums
of basic hypergeometric series. As an application, we give double, triple, and
quadruple sum analogues of some Ramanujan-type supercongruences.

###### Key words and phrases:

$q$-analogue; super(congruence); cyclotomic polynomial; truncated sum

###### 1991 Mathematics Subject Classification:

11B65; 11F33; 33C20

##  1\. Introduction

Throughout the paper $n$ is a nonnegative integer, $p$ is a prime number, and
$x$ and $q$ are complex numbers such that $|q|<1$. We will use the following
notation from the theory of basic hypergeometric series, see Gasper-Rahman
[3]. The $q$-shifted factorials are given by

$(x;q)_{0}=1,\ \quad(x;q)_{n}=\prod_{j=0}^{n-1}(1-xq^{j}),\ \text{and\
}(x;q)_{\infty}=\prod_{j=0}^{\infty}(1-xq^{j}).$

The $q$-integer is given for any nonnegative integer $n$ by

$[n]_{q}=[n]=\frac{1-q^{n}}{1-q}=1+q+\cdots+q^{n-1}$

and it is easy to check that $\lim_{q\to 1}[n]=n$. Let $\mathbb{Z}$,
$\mathbb{N}_{0}$, and $\mathbb{N}$ denote the sets of integers, nonnegative
integers, and positive integers respectively. Let $\mathbb{Z}[q]$ denote the
set of polynomials in $q$ with integer coefficients and let $\mathbb{Z}(q)$
denote the set of rational functions in $q$ with integer coefficients. The
$n$-th cyclotomic polynomial is the polynomial in $\mathbb{Z}[q]$ given by

$\Phi_{n}(q)=\prod_{\begin{subarray}{c}j=1\\\
\gcd(j,n)=1\end{subarray}}^{n}(q-\zeta^{j}),$

where $\zeta=e^{2\pi i/n}$ is the $n$-th root of unity. Given polynomials
$A_{1}(q),A_{2}(q),P(q)\in\mathbb{Z}[q]$, the congruence
$\frac{A_{1}(q)}{A_{2}(q)}\equiv 0\pmod{P(q)}$ means that $P(q)$ divides
$A_{1}(q)$ and that $\gcd\big{(}P(q),A_{2}(q)\big{)}=1$; and in general for
rational functions $A(q),B(q)\in\mathbb{Z}(q)$, the congruence $A(q)\equiv
B(q)\pmod{P(q)}$ means that $A(q)-B(q)\equiv 0\pmod{P(q)}$. Ramanujan-type
supercongruences for truncated sums of basic hypergeometric series have
received much interest in recent years, see for instance [4, 5, 6, 7, 8, 9,
11, 12, 13, 15, 17, 20]. Most of the publications dealing with supercongruence
relations were motivated by Ramanujan’s formulas for $\frac{1}{\pi}$ as in
Ramanujan [16] and Berndt [1] and the later work by Van Hamme [19] where he
stated a variety of conjectures on the $p$-adic analogues of Ramanujan-type
formulas for $1/\pi$. Among Van Hamme’s $p$-adic conjectures one finds

$\begin{split}\text{(E2)\quad}&\sum_{k=0}^{p-1}(-1)^{k}\frac{\big{(}\frac{1}{3}\big{)}_{k}^{3}}{k!^{3}}(6k+1)\equiv
p\pmod{p^{3}}\quad\text{for\ }p\equiv 1\pmod{3}\\\
\text{(F2)\quad}&\sum_{k=0}^{p-1}(-1)^{k}\frac{\big{(}\frac{1}{4}\big{)}_{k}^{3}}{k!^{3}}(8k+1)\equiv
p\Big{(}\frac{-2}{p}\Big{)}\pmod{p^{3}}\quad\text{for\ }p\equiv 1\pmod{4}\\\
\text{(G2)\quad}&\sum_{k=0}^{p-1}\frac{\big{(}\frac{1}{4}\big{)}_{k}^{4}}{k!^{4}}(8k+1)\equiv
p\frac{\big{(}\frac{1}{2}\big{)}_{(p-1)/4}}{\big{(}1\big{)}_{(p-1)/4}}\pmod{p^{3}}\quad\text{for\
}p\equiv 1\pmod{4}\end{split}$

where $(x)_{n}=x(x+1)\cdots(x+n-1)$ is the Pochhammer function and
$\big{(}\frac{\cdot}{p}\big{)}$ is the Legendre symbol modulo $p$. All three
conjectures (E2), (F2), and (G2) have been settled by Swisher [18]. Guo and
Zudilin [11] recently developed the _$q$ -microscope_ approach and proved a
large number of supercongruences for truncated sums of basic hypergeometric
series along with a $q$-analogue for the following Ramanujan-type congruence

$\text{(GZ)\quad}\sum_{k=0}^{p-1}\frac{{4k\choose 2k}{2k\choose
k}^{2}}{2^{8k}3^{2k}}(8k+1)\equiv
p\Big{(}\frac{-3}{p}\Big{)}\pmod{p^{3}}\quad\text{for\ }p>3.$

Another example of this type of supercongruences which is important for our
work is

$\text{(LW)\quad}\sum_{k=0}^{p-1}\frac{\big{(}\frac{1}{4}\big{)}_{k}^{4}}{k!^{4}}(8k+1)\equiv-p\frac{\big{(}\frac{1}{2}\big{)}_{(p-1)/4}}{\big{(}1\big{)}_{(p-1)/4}}\pmod{p^{3}}\quad\text{for\
}p\equiv 1\pmod{4}$

which is recently obtained by Liu and Wang’s [14, p. 3] using the
$q$-microscope method. It is natural to raise the question about $N$-tuple sum
analogues of the Ramanujan-type congruences. In [2] we dealt with the case
$N=2$ as we established double sum versions of some Ramanujan-type congruences
including the following double sum analogue of (GZ)

$\sum_{k=0}^{p-1}\sum_{j=0}^{k}\frac{{4j\choose 2j}{2j\choose
j}^{2}{4(k-j)\choose 2(k-j)}{2(k-j)\choose
k-j}^{2}}{2^{8k}3^{2k}}(8j+1)\big{(}8(k-j)+1\big{)}\equiv p^{2}\pmod{p^{3}}.$

To achieve this, we combined the $q$-microscope approach with a result on
double sums of series satisfying certain properties. Recently, Li [13] and
Song and Wang [17] gave other supercongruences for truncated sums of squares
of $q$-series. In this note we consider the general case where $N\geq 2$. As
an application we deduce supercongruences for truncated $N$-tuple sums of some
(basic) hypergeometric series including double, triple, and quadruple sum
analogues of the relations (E2), (F2), (G2), (GZ), and (LW). Proofs of all our
main results rely on the $q$-microscope technique and the following lemma.

######  Lemma 1.

Let $N\geq 2$ be a positive integer, let
$\\{z_{1}(k)\\}_{k=0}^{\infty},\ldots,\\{z_{N}(k)\\}_{k=0}^{\infty}$ be
sequences of complex numbers and let

$t(k)=\sum_{i_{1}=0}^{k}\sum_{i_{2}=0}^{k-i_{1}}\cdots\sum_{i_{N-1}=0}^{k-i_{1}-\cdots-
i_{N-1}}z_{1}(i_{1})\cdots z_{N-1}(i_{N-1})z_{N}(k-i_{1}-\cdots-i_{N-1}).$

(a) If there is $d\in\mathbb{N}$ such that for all $j=1,\ldots,N$

(1) $z_{j}(k)=0\ \text{for all $k\in\mathbb{N}_{0}$\ such that\
}\frac{d-1}{N}<k<d,$

then

$\sum_{i=1}^{N}\sum_{k=0}^{d-1}z_{i}(k)=\sum_{k=0}^{d-1}t(k).$

(b) If moreover,

(2) $\begin{split}z_{j}(ld+k)&=z_{j}(ld)z_{j}(k)\ \text{for all
$k,l\in\mathbb{N}_{0}$ with $0\leq k<d$},\end{split}$

then for any $k,l\in\mathbb{N}_{0}$ such that $0\leq k<d$, there holds

$t(ld+k)=t(k)\sum_{i_{1}=0}^{l}\sum_{i_{2}=0}^{l-i_{1}}\cdots\sum_{i_{N}=0}^{l-i_{1}-\cdots-
i_{N-1}}z_{1}(i_{1}d)\cdots z_{N-1}(i_{N-1}d)z_{N}\big{(}(l-i_{1}-\cdots-
i_{N-1})d\big{)}.$

(c) If for all $j=1,\ldots,N$, the sequence $\\{z_{j}(k)\\}_{k=0}^{\infty}$
satisfies both (1)-(2) for some $d$ and if $n$ is a positive multiple of $d$,
then

$\sum_{k=0}^{n-1}t(k)=\Big{(}\sum_{l=0}^{\frac{n}{d}-1}\sum_{i_{1}=0}^{l}\sum_{i_{2}=0}^{l-i_{1}}\cdots\sum_{i_{N}=0}^{l-i_{1}-\cdots-
i_{N-1}}z_{1}(i_{1}d)\cdots z_{N-1}(i_{N-1}d)z_{N}\big{(}(l-i_{1}-\cdots-
i_{N-1})d\big{)}\Big{)}$ $\times\sum_{k=0}^{d-1}t(k).$

The paper is organized as follows. In Section 2 we state our main theorems
along with their consequences. Proofs for these theorems are given in Sections
3-6. Section 7 is devoted to the proof of Lemma 1. In Section 8 we close with
some comments and suggestions for further research.

##  2\. Statement of results

######  Theorem 1.

Let $s,N\in\mathbb{N}$ such that $2\leq N\leq s$, let

$\alpha_{q}(k)=(-1)^{k}q^{s{k+1\choose
2}-k}[2sk+1]\frac{(xq;q^{s})_{k}(q/x;q^{s})_{k}(q;q^{s})_{k}}{(xq^{s};q^{s})_{k}(q^{s}/x;q^{s})_{k}(q^{s};q^{s})_{k}},$

and let

$t_{q}(k)=\sum_{i_{1}=0}^{k}\sum_{i_{2}=0}^{k-i_{1}}\cdots\sum_{i_{N}=0}^{k-i_{1}-\cdots
i_{N-1}}\alpha_{q}(i_{1})\cdots\alpha_{q}(i_{N-1})\alpha_{q}(k-i_{1}-\cdots-
i_{N-1}).$

Then for any positive integer $n$ whose prime factors are all congruent to $1$
modulo $s$, there holds

$\sum_{k=0}^{n-1}t_{q}(k)\equiv
q^{N(n-1)(n-s+1)/(2s)}[n]^{N}(-1)^{N(n-1)/s}\pmod{[n](1-aq^{n})(a-q^{n})}.$

In our applications of Theorem 1 we restrict ourselves to $s=3$ and $s=4$ as
they are related to Van Hamme’s relations (E2) and (F2). So, letting in
Theorem 1, $s=3$, $x=1$ and $q\to 1$, we obtain the following double and
triple sum analogues (E2).

######  Corollary 1.

If $p\equiv 1\pmod{3}$, then

$\sum_{k=0}^{p-1}\sum_{j=0}^{k}(-1)^{k}\frac{\big{(}\frac{1}{3}\big{)}_{j}^{3}}{j!^{3}}\frac{\big{(}\frac{1}{3}\big{)}_{k-j}^{3}}{(k-j)!^{3}}(6j+1)\big{(}6(k-j)+1\big{)}\equiv
p^{2}\pmod{p^{3}}$

and

$\sum_{k=0}^{p-1}\sum_{j=0}^{k}\sum_{i=0}^{k-j}(-1)^{k}\frac{\big{(}\frac{1}{3}\big{)}_{j}^{3}}{j!^{3}}\frac{\big{(}\frac{1}{3}\big{)}_{i}^{3}}{i!^{3}}\frac{\big{(}\frac{1}{3}\big{)}_{k-j-i}^{3}}{(k-j-i)!^{3}}(6i+1)(6j+1)\big{(}6(k-j-i)+1\big{)}\equiv
0\pmod{p^{3}}.$

Letting in Theorem 1, $s=4$, $x=1$ and $q\to 1$, we deduce the following
double, triple, and quadruple sum analogues of (F2).

######  Corollary 2.

If $p\equiv 1\pmod{4}$, then

$\sum_{k=0}^{p-1}\sum_{j=0}^{k}(-1)^{k}\frac{\big{(}\frac{1}{4}\big{)}_{j}^{3}}{j!^{3}}\frac{\big{(}\frac{1}{4}\big{)}_{k-j}^{3}}{(k-j)!^{3}}(8j+1)\big{(}8(k-j)+1\big{)}\equiv
p^{2}\pmod{p^{3}}$

and

$\sum_{k=0}^{p-1}\sum_{j=0}^{k}\sum_{i=0}^{k-j}(-1)^{k}\frac{\big{(}\frac{1}{4}\big{)}_{j}^{3}}{j!^{3}}\frac{\big{(}\frac{1}{4}\big{)}_{i}^{3}}{i!^{3}}\frac{\big{(}\frac{1}{4}\big{)}_{k-j-i}^{3}}{(k-j-i)!^{3}}(8i+1)(8j+1)\big{(}8(k-j-i)+1\big{)}$
$\equiv\sum_{k=0}^{p-1}\sum_{j=0}^{k}\sum_{i=0}^{k-j}\sum_{h=0}^{k-j-i}(-1)^{k}\frac{\big{(}\frac{1}{4}\big{)}_{j}^{3}}{j!^{3}}\frac{\big{(}\frac{1}{4}\big{)}_{i}^{3}}{i!^{3}}\frac{\big{(}\frac{1}{4}\big{)}_{h}^{3}}{h!^{3}}\frac{\big{(}\frac{1}{4}\big{)}_{k-j-
i-h}^{3}}{(k-j-i-h)!^{3}}(8i+1)(8j+1)(8h+1)$ $\qquad\times\big{(}8(k-j-
i-h)+1\big{)}\equiv 0\pmod{p^{3}}.$

Theorem 1 can be generalized as follows.

######  Theorem 2.

Let $(r,s)\in\mathbb{Z}\times\mathbb{N}$ such that $\gcd(r,s)=1$, let

$\alpha_{q}(k)=(-1)^{k}q^{s{k+1\choose
2}-rk}[2sk+1]\frac{(xq^{r};q^{s})_{k}(q^{r}/x;q^{s})_{k}(q^{r};q^{s})_{k}}{(xq^{s};q^{s})_{k}(q^{s}/x;q^{s})_{k}(q^{s};q^{s})_{k}},$

and let for a positive integer $N\geq 2$

$t_{q}(k)=\sum_{i_{1}=0}^{k}\sum_{i_{2}=0}^{k-i_{1}}\cdots\sum_{i_{N}=0}^{k-i_{1}-\cdots
i_{N-1}}\alpha_{q}(i_{1})\cdots\alpha_{q}(i_{N-1})\alpha_{q}(k-i_{1}-\cdots-
i_{N-1}).$

Then for any prime $p\equiv r\pmod{s}$ such that $r\leq p$ and
$(p-r)/s\leq(p-1)/N$, there holds

(3) $\sum_{k=0}^{p-1}t_{q}(k)\equiv
q^{N(p-r)(p-s+r)/(2s)}[p]^{N}(-1)^{N(p-r)/s}\pmod{[p](1-aq^{p})(a-q^{p})}.$

Theorem 2 obviously has a wide range of applications, but we restrict
ourselves to the following two cases. Letting in Theorem 2, $(r,s)=(2,3)$,
$x=1$ and $q\to 1$, we get.

######  Corollary 3.

If $p\equiv 2\pmod{3}$, then

$\sum_{k=0}^{p-1}\sum_{j=0}^{k}(-1)^{k}\frac{\big{(}\frac{2}{3}\big{)}_{j}^{3}}{j!^{3}}\frac{\big{(}\frac{2}{3}\big{)}_{k-j}^{3}}{(k-j)!^{3}}(6j+1)\big{(}6(k-j)+1\big{)}\equiv
p^{2}\pmod{p^{3}}$

and

$\sum_{k=0}^{p-1}\sum_{j=0}^{k}\sum_{i=0}^{k-j}(-1)^{k}\frac{\big{(}\frac{2}{3}\big{)}_{j}^{3}}{j!^{3}}\frac{\big{(}\frac{2}{3}\big{)}_{i}^{3}}{i!^{3}}\frac{\big{(}\frac{2}{3}\big{)}_{k-j-i}^{3}}{(k-j-i)!^{3}}(6i+1)(6j+1)\big{(}6(k-j-i)+1\big{)}\equiv
0\pmod{p^{3}}.$

Upon letting in Theorem 2, $(r,s)=(3,4)$, $x=1$ and $q\to 1$, we derive.

######  Corollary 4.

If $p\equiv 3\pmod{4}$, then

$\sum_{k=0}^{p-1}\sum_{j=0}^{k}(-1)^{k}\frac{\big{(}\frac{3}{4}\big{)}_{j}^{3}}{j!^{3}}\frac{\big{(}\frac{3}{4}\big{)}_{k-j}^{3}}{(k-j)!^{3}}(8j+1)\big{(}8(k-j)+1\big{)}\equiv
p^{2}\pmod{p^{3}}$

and

$\sum_{k=0}^{p-1}\sum_{j=0}^{k}\sum_{i=0}^{k-j}(-1)^{k}\frac{\big{(}\frac{3}{4}\big{)}_{j}^{3}}{j!^{3}}\frac{\big{(}\frac{3}{4}\big{)}_{i}^{3}}{i!^{3}}\frac{\big{(}\frac{3}{4}\big{)}_{k-j-i}^{3}}{(k-j-i)!^{3}}(8i+1)(8j+1)\big{(}8(k-j-i)+1\big{)}$
$\equiv\sum_{k=0}^{p-1}\sum_{j=0}^{k}\sum_{i=0}^{k-j}\sum_{h=0}^{k-j-i}(-1)^{k}\frac{\big{(}\frac{3}{4}\big{)}_{j}^{3}}{j!^{3}}\frac{\big{(}\frac{3}{4}\big{)}_{i}^{3}}{i!^{3}}\frac{\big{(}\frac{3}{4}\big{)}_{h}^{3}}{h!^{3}}\frac{\big{(}\frac{3}{4}\big{)}_{k-j-
i-h}^{3}}{(k-j-i-h)!^{3}}(8i+1)(8j+1)(8h+1)$ $\qquad\times\big{(}8(k-j-
i-h)+1\big{)}\equiv 0\pmod{p^{3}}.$

######  Theorem 3.

Let

$\alpha_{q}(k)=q^{2k}[8k+1]\frac{(q;q^{4})_{k}^{2}(xq;q^{4})_{k}(q/x;q^{4})_{k}}{(q^{4};q^{4})_{k}^{2}(xq^{4};q^{4})_{k}(q^{4}/x;q^{4})_{k}},$
$s_{q}(k)=\sum_{j=0}^{k}\alpha_{q}(j)\alpha_{q}(k-j),\quad
t_{q}(k)=\sum_{j=0}^{k}\sum_{i=0}^{k-j}\alpha_{q}(j)\alpha_{q}(i)\alpha_{q}(k-j-i),$

and

$u_{q}(k)=\sum_{j=0}^{k}\sum_{i=0}^{k-j}\sum_{h=0}^{k-j-i}\alpha_{q}(j)\alpha_{q}(i)\alpha_{q}(h)\alpha_{q}(k-j-
i-h).$

Then for any positive integer $n$ whose prime factors are all congruent to $1$
modulo $4$, there holds

$\sum_{k=0}^{n-1}s_{q}(k)\equiv[n]^{2}\frac{(q^{2};q^{4})_{(n-1)/4}^{2}}{(q^{4};q^{4})_{(n-1)/4}^{2}}q^{(1-n)/2}\pmod{[n](1-xq^{n})(x-q^{n})},$
$\sum_{k=0}^{n-1}t_{q}(k)\equiv[n]^{3}\frac{(q^{2};q^{4})_{(n-1)/4}^{3}}{(q^{4};q^{4})_{(n-1)/4}^{3}}q^{3(1-n)/4}\pmod{[n](1-xq^{n})(x-q^{n})},$

and

$\sum_{k=0}^{n-1}u_{q}(k)\equiv[n]^{4}\frac{(q^{2};q^{4})_{(n-1)/4}^{4}}{(q^{4};q^{4})_{(n-1)/4}^{4}}q^{(1-n)}\pmod{[n](1-xq^{n})(x-q^{n})}.$

Upon letting in Theorem 3, $x=1$ and $q\to 1$, we get the following double,
triple, and quadruple sum versions of Van Hamme’s congruence (G2).

######  Corollary 5.

If $p\equiv 1\pmod{4}$ is a prime number, then we have

$\sum_{k=0}^{p-1}\sum_{j=0}^{k}\frac{\big{(}\frac{1}{4}\big{)}_{j}^{4}}{j!^{4}}\frac{\big{(}\frac{1}{4}\big{)}_{k-j}^{4}}{(k-j)!^{4}}(8j+1)(8k-8j+1)\equiv
p^{2}\frac{\big{(}\frac{1}{2}\big{)}_{(p-1)/4}^{2}}{\big{(}1\big{)}_{(p-1)/4}^{2}}\pmod{p^{3}}$

and

$\sum_{k=0}^{p-1}\sum_{j=0}^{k}\sum_{i=0}^{k-j}\frac{\big{(}\frac{1}{4}\big{)}_{j}^{4}}{j!^{4}}\frac{\big{(}\frac{1}{4}\big{)}_{i}^{4}}{i!^{4}}\frac{\big{(}\frac{1}{4}\big{)}_{k-j-i}^{4}}{(k-j-i)!^{4}}(8i+1)(8j+1)(8k-8j-8i+1)$
$\equiv\sum_{k=0}^{p-1}\sum_{j=0}^{k}\sum_{i=0}^{k-j}\sum_{h=0}^{k-j-i}\frac{\big{(}\frac{1}{4}\big{)}_{j}^{4}}{j!^{4}}\frac{\big{(}\frac{1}{4}\big{)}_{i}^{4}}{i!^{4}}\frac{\big{(}\frac{1}{4}\big{)}_{h}^{4}}{h!^{4}}\frac{\big{(}\frac{1}{4}\big{)}_{k-j-
i-h}^{4}}{(k-j-i-h)!^{4}}(8i+1)(8j+1)(8h+1)$ $\times\big{(}8(k-j-
i-h)+1\big{)}\equiv 0\pmod{p^{3}}.$

######  Theorem 4.

Let

$\alpha_{q}(k)=q^{-4k}[8k+1]^{2}[8k+1]_{q^{2}}\frac{(q^{2};q^{8})_{k}^{2}(xq^{2};q^{8})_{k}(q^{2}/x;q^{8})_{k}}{(q^{8};q^{8})_{k}^{2}(xq^{8};q^{8})_{k}(q^{8}/x;q^{8})_{k}},$
$s_{q}(k)=\sum_{j=0}^{k}\alpha_{q}(j)\alpha_{q}(k-j),\quad
t_{q}(k)=\sum_{j=0}^{k}\sum_{i=0}^{k-j}\alpha_{q}(j)\alpha_{q}(i)\alpha_{q}(k-j-i),$

and

$u_{q}(k)=\sum_{j=0}^{k}\sum_{i=0}^{k-j}\sum_{h=0}^{k-j-i}\alpha_{q}(j)\alpha_{q}(i)\alpha_{q}(h)\alpha_{q}(k-j-
i-h).$

Then for any positive integer $n$ whose prime factors are all congruent to $1$
modulo $4$, modulo $[n]_{q^{2}}(1-xq^{2n})(x-q^{2n})$ there holds

$\sum_{k=0}^{n-1}s_{q}(k)\equiv[n]_{q^{2}}^{2}\frac{(q^{4};q^{8})_{(n-1)/4}^{2}}{(q^{8};q^{8})_{(n-1)/4}^{2}}q^{(1-n)}\Big{(}1-\frac{(1-xq^{2})(1-q^{2}/x)}{(1-q)^{2}(1+q^{2})}\Big{)}^{2},$
$\sum_{k=0}^{n-1}t_{q}(k)\equiv[n]_{q^{2}}^{3}\frac{(q^{4};q^{8})_{(n-1)/4}^{3}}{(q^{8};q^{8})_{(n-1)/4}^{3}}q^{3(1-n)/2}\Big{(}1-\frac{(1-xq^{2})(1-q^{2}/x)}{(1-q)^{2}(1+q^{2})}\Big{)}^{3},$

and

$\sum_{k=0}^{p-1}u_{q}(k)\equiv[n]_{q^{2}}^{4}\frac{(q^{4};q^{8})_{(n-1)/4}^{4}}{(q^{8};q^{8})_{(n-1)/4}^{4}}q^{2(1-n)}\Big{(}1-\frac{(1-xq^{2})(1-q^{2}/x)}{(1-q)^{2}(1+q^{2})}\Big{)}^{4}.$

Letting in Theorem 4, $x=1$ and $q\to 1$, we derive the following double,
triple, and quadruple versions of (LW).

######  Corollary 6.

If $p\equiv 1\pmod{4}$ is a prime number, then we have

$\sum_{k=0}^{p-1}\sum_{j=0}^{k}\frac{\big{(}\frac{1}{4}\big{)}_{j}^{4}}{j!^{4}}\frac{\big{(}\frac{1}{4}\big{)}_{k-j}^{4}}{(k-j)!^{4}}(8j+1)^{3}(8k-8j+1)^{3}\equiv
p^{2}\frac{\big{(}\frac{1}{2}\big{)}_{(p-1)/4}^{2}}{\big{(}1\big{)}_{(p-1)/4}^{2}}\pmod{p^{3}}$

and

$\sum_{k=0}^{p-1}\sum_{j=0}^{k}\sum_{i=0}^{k-j}\frac{\big{(}\frac{1}{4}\big{)}_{j}^{4}}{j!^{4}}\frac{\big{(}\frac{1}{4}\big{)}_{i}^{4}}{i!^{4}}\frac{\big{(}\frac{1}{4}\big{)}_{k-j-i}^{4}}{(k-j-i)!^{4}}(8i+1)^{3}(8j+1)^{3}(8k-8j-8i+1)^{3}$
$\equiv\sum_{k=0}^{p-1}\sum_{j=0}^{k}\sum_{i=0}^{k-j}\sum_{h=0}^{k-j-i}\frac{\big{(}\frac{1}{4}\big{)}_{j}^{4}}{j!^{4}}\frac{\big{(}\frac{1}{4}\big{)}_{i}^{4}}{i!^{4}}\frac{\big{(}\frac{1}{4}\big{)}_{h}^{4}}{h!^{4}}\frac{\big{(}\frac{1}{4}\big{)}_{k-j-
i-h}^{4}}{(k-j-i-h)!^{4}}(8i+1)^{3}(8j+1)^{3}(8h+1)^{3}$
$\qquad\qquad\qquad\times\big{(}8(k-j-i-h)+1\big{)}^{3}\equiv 0\pmod{p^{3}}.$

Our last application is related to the congruence (GZ).

######  Theorem 5.

Let

$\alpha_{q}(k)=\frac{(xq;q^{2})_{k}(q/x;q^{2})_{k}(q;q^{2})_{2k}}{(xq^{6};q^{6})_{k}(q^{6}/x;q^{6})_{k}(q^{2};q^{2})_{2k}},[8k+1]q^{2k^{2}},$

let

$s_{q}(k)=\sum_{j=0}^{k}\alpha_{q}(j)\alpha_{q}(k-j),\quad
t_{q}(k)=\sum_{j=0}^{k}\sum_{i=0}^{k-j}\alpha_{q}(j)\alpha_{q}(i)\alpha_{q}(k-j-i),$

and

$u_{q}(k)=\sum_{j=0}^{k}\sum_{i=0}^{k-j}\sum_{h=0}^{k-j-i}\alpha_{q}(j)\alpha_{q}(i)\alpha_{q}(i)\alpha_{q}(k-j-
i-h).$

Then for any positive integer $n$ which is coprime to $6$ there holds

$\sum_{k=0}^{n-1}s_{q}(k)\equiv q^{1-n}[n]^{2}\pmod{[n](1-xq^{n})(x-q^{n})}$

and

$\sum_{k=0}^{n-1}t_{q}(k)\equiv\sum_{k=0}^{n-1}u_{q}(k)\equiv 0\pmod{[n]}.$

Note that the supercongruence in Theorem 5 for the double sum
$\sum_{k=0}^{n-1}s_{q}(k)$ already appeared in [2, Theorem 3] and we include
it here for completeness. Upon letting $x=1$ and then $q\to 1$ in Theorem 5 we
obtain the following triple and quadruple sum versions of (GZ).

######  Corollary 7.

If $p>3$ be a prime number, then we have

$\sum_{k=0}^{p-1}\sum_{j=0}^{k}\sum_{i=0}^{k-j}\frac{{2j\choose
j}^{2}{4j\choose 2j}{2i\choose i}^{2}{4i\choose 2i}{2(k-j-i)\choose
k-j-i}^{2}{4(k-j-i)\choose
2(k-j-i)}}{2^{8k}3^{2k}}(8i+1)(8j+1)\big{(}8(k-j-i)+1\big{)}$
$\equiv\sum_{k=0}^{p-1}\sum_{j=0}^{k}\sum_{i=0}^{k-j}\sum_{h=0}^{k-j-i}\frac{{2j\choose
j}^{2}{4j\choose 2j}{2i\choose i}^{2}{4i\choose 2i}{2h\choose h}^{2}{4h\choose
2h}{2(k-j-i-h)\choose k-j-i}^{2}{4(k-j-i-h)\choose 2(k-j-i-h)}}{2^{8k}3^{2k}}$
$\times(8i+1)(8j+1)(8h+1)\big{(}8(k-j-i-h)+1\big{)}\equiv 0\pmod{p}.$

##  3\. Proof of Theorem 2

Let $p$ be a prime number such that $p\equiv r\pmod{s}$ and let $\zeta$ be a
$p$-th root of unity. From [11, p. 349] we have

(4) $\sum_{k=0}^{p-1}\alpha_{\zeta}(k)=0.$

Furthermore, note that from the assumptions $p\equiv r\pmod{s}$, $r\leq p$,
and $(p-r)/s\leq(p-1)/N$ it follows that

$\alpha_{\zeta}(k)=0\ \text{for\ }\frac{p-1}{N}\leq k\leq p-1,$

implying that the sequence $\\{\alpha_{\zeta}(k)\\}_{k=0}^{\infty}$ satisfies
condition (1). Then by an appeal to Lemma 1(a) with
$z_{j}(k)=\alpha_{\zeta}(k)$ for $j=1,\ldots,N$, we find

$\sum_{k=0}^{p-1}t_{\zeta}(k)=\Big{(}\sum_{k=0}^{p-1}\alpha_{\zeta}(k)\Big{)}^{N}$

and so by (4),

(5) $\sum_{k=0}^{p-1}t_{q}(k)\equiv 0\equiv
q^{N(p-r)(p-s+r)/(2s)}[p]^{N}(-1)^{N(p-r)/s}\pmod{[p]}.$

Furthermore, from [11, p. 349] by letting $x=q^{-p}$ or $x=q^{p}$ in
$\alpha_{q}(k)$, we have

(6) $\sum_{k=0}^{p-1}\alpha_{q}(k)=q^{(p-r)(p-s+r)/(2s)}[p](-1)^{(p-r)/s}.$

Noting that with these choices of $x$, the sequence
$\\{\alpha_{q}(k)\\}_{k=0}^{\infty}$ fulfills property (1), we deduce from
Lemma 1(a) that

(7)
$\sum_{k=0}^{p-1}t_{q}(k)=\Big{(}\sum_{k=0}^{p-1}\alpha_{q}(k)\Big{)}^{N}.$

As the polynomials $[p]$, $(1-xq^{p})$, and $(x-q^{p})$ are relatively prime,
we achieve the desired congruence by combining (5)-(7).

##  4\. Proof of Theorem 1

Let $d\mid n$ and let $\zeta$ be a $d$-th primitive root of unity. From the
assumptions that $d\equiv 1\pmod{s}$ and that $N\leq s$ we get

$\alpha_{\zeta}(k)=0\ \text{for\ }\frac{d-1}{N}\leq k\leq d-1$

and by [11] that

$\sum_{k=0}^{d-1}\alpha_{\zeta}(k)=0$

and so, by Lemma 1(a)

$\sum_{k=0}^{d-1}t_{\zeta}(k)=\Big{(}\sum_{k=0}^{d-1}\alpha_{\zeta}(k)\Big{)}^{N}=0.$

Then the forgoing identity and Lemma 1(c) give

$\sum_{k=0}^{n-1}t_{\zeta}(k)=\sum_{k=0}^{d-1}t_{\zeta}(k)\Big{(}\sum_{l=0}^{n/d-1}\sum_{m=0}^{l}\sum_{i=0}^{l-m}\alpha_{\zeta}(md)\alpha_{\zeta}(id)\alpha_{\zeta}\big{(}(l-m-i)d\big{)}\Big{)}=0,$

yielding that

(8) $\sum_{k=0}^{n-1}t_{q}(k)\equiv 0\equiv
q^{N(n-1)(n-s+1)/(2s)}[n]^{N}(-1)^{N(n-1)/s}\pmod{[n]}.$

In addition, from [11] by letting $x=q^{-n}$ or $x=q^{n}$ in $\alpha_{q}(k)$,
we have

(9) $\sum_{k=0}^{n-1}\alpha_{q}(k)=q^{(n-1)(n-s+1)/(2s)}[n](-1)^{(n-1)/s}$

and it easily verified that the sequence $\\{\alpha_{q}(k)\\}_{k=0}^{\infty}$
fulfills property (1) as $x=q^{-n}$ or $x=q^{n}$. Then we deduce by Lemma 1(a)
that

(10)
$\sum_{k=0}^{n-1}t_{q}(k)=\Big{(}\sum_{k=0}^{n-1}\alpha_{q}(k)\Big{)}^{N}.$

Now use (8)-(10) and the fact the polynomials $[n]$, $(1-xq^{n})$, and
$(x-q^{n})$ are relatively prime to derive the desired congruence.

##  5\. Proof of Theorem 3 and Theorem 4

_Proof of Theorem 3._ Let $n$ be a positive integer whose prime factors are
all congruent to $1$ modulo $4$, Let $d\mid n$, and let $\zeta$ be a $d$-th
root of unity. Clearly $n\equiv d\equiv 1\pmod{4}$. Liu-Wang [14, p. 5] proved
that

(11) $\sum_{k=0}^{d-1}\alpha_{\zeta}(k)=0$

and for $x=q^{n}$ or $x=q^{-n}$,

(12)
$\sum_{k=0}^{n-1}\alpha_{q}(k)=[n]\frac{(q^{2};q^{4})_{(n-1)/4}}{(q^{4};q^{4})_{(n-1)/4}}q^{(1-n)/4}.$

It easy to see that the sequence $\\{\alpha_{\zeta}(k)\\}_{k=0}^{\infty}$
satisfies both conditions (1) and (2) for $N=1,2,3$. Then by Lemma 1(a) and
(11), we find

$\sum_{k=0}^{d-1}s_{\zeta}(k)=\Big{(}\sum_{k=0}^{d-1}\alpha_{\zeta}(k)\Big{)}^{2}\sum_{k=0}^{d-1}t_{\zeta}(k)=\Big{(}\sum_{k=0}^{d-1}\alpha_{\zeta}(k)\Big{)}^{3}=\sum_{k=0}^{d-1}u_{\zeta}(k)=\Big{(}\sum_{k=0}^{d-1}\alpha_{\zeta}(k)\Big{)}^{4}=0.$

Thus with the help of Lemma 1(c)

$\sum_{k=0}^{n-1}s_{\zeta}(k)\sum_{k=0}^{n-1}t_{\zeta}(k)=\sum_{k=0}^{n-1}u_{\zeta}(k)=0,$

implying that

$\sum_{k=0}^{n-1}s_{q}(k)\equiv\sum_{k=0}^{n-1}t_{q}(k)\equiv\sum_{k=0}^{n-1}u_{q}(k)\equiv
0\pmod{[n]}.$

Besides, noticing that for $x=q^{n}$ or $x=q^{-n}$, the sequence
$\\{\alpha_{q}(k)\\}_{k=0}^{\infty}$ meets the property (1), we deduce from
Lemma 1(a) that

$\sum_{k=0}^{n-1}s_{q}(k)=\Big{(}\sum_{k=0}^{n-1}\alpha_{q}(k)\Big{)}^{2},\
\sum_{k=0}^{n-1}t_{q}(k)=\Big{(}\sum_{k=0}^{n-1}\alpha_{q}(k)\Big{)}^{3},\
\text{and\
}\sum_{k=0}^{n-1}u_{q}(k)=\Big{(}\sum_{k=0}^{n-1}\alpha_{q}(k)\Big{)}^{4}.$

This together with (12) yield that modulo $(1-xq^{n})(x-q^{n})$,

$\begin{split}\sum_{k=0}^{n-1}s_{q}(k)&\equiv[n]^{2}\frac{(q^{2};q^{4})_{(n-1)/4}^{2}}{(q^{4};q^{4})_{(n-1)/4}^{2}}q^{(1-n)/2}\\\
\sum_{k=0}^{n-1}t_{q}(k)&\equiv[n]^{3}\frac{(q^{2};q^{4})_{(n-1)/4}^{3}}{(q^{4};q^{4})_{(n-1)/4}^{3}}q^{3(1-n)/4}\\\
\sum_{k=0}^{n-1}u_{q}(k)&\equiv[n]^{4}\frac{(q^{2};q^{4})_{(n-1)/4}^{4}}{(q^{4};q^{4})_{(n-1)/4}^{4}}q^{1-n}.\end{split}$

Now combine the above congruences to achieve the desired result.

_Proof of Theorem 4._ Follows similarly by a combination of Lemma 1 with the
proof of [14, Theorem 4].

##  6\. Proof of Theorem 5

The result is clear for $n=1$. Let $n>1$ with $\gcd(n,6)=1$, let $d\mid n$,
and let $\zeta$ be a $d$-th primitive root of unity. From [11, p. 339] we have

(13) $\sum_{k=0}^{d-1}\alpha_{\zeta}(k)=0.$

As $(q;q^{2})_{2k}=(q;q^{4})_{k}(q^{3};q^{4})_{k}$ appears in the numerator of
$\alpha_{q}(k)$ we see that the sequence
$\\{\alpha_{\zeta}(k)\\}_{k=0}^{\infty}$ satisfies condition (1) for
$N\in\\{3,4\\}$. Then by (13) and Lemma 1(a) applied to
$z_{j}(k)=\alpha_{q}(k)$ for $j=1,2,3,4$, we obtain

$\sum_{k=0}^{d-1}t_{\zeta}(k)=\Big{(}\sum_{k=0}^{d-1}\alpha_{\zeta}(k)\Big{)}^{3}=0\
\text{and\
}\sum_{k=0}^{d-1}u_{\zeta}(k)=\Big{(}\sum_{k=0}^{d-1}\alpha_{\zeta}(k)\Big{)}^{4}=0.$

Moreover, it is easy to check by the basic properties of roots of unity that
$\\{\alpha_{\zeta}(k)\\}_{k=0}^{\infty}$ fulfills the property (2) as well and
that $\alpha_{\zeta}(0)=1$. It follows by virtue of Lemma 1(c) with $N=3$ that

$\sum_{k=0}^{n-1}t_{\zeta}(k)=\sum_{k=0}^{d-1}t_{\zeta}(k)\Big{(}\sum_{l=0}^{n/d-1}\sum_{m=0}^{l}\sum_{i=0}^{l-m}\alpha_{\zeta}(md)\alpha_{\zeta}(id)\alpha_{\zeta}\big{(}(l-m-i)d\big{)}\Big{)}=0,$

implying that $\sum_{k=0}^{n-1}t_{q}(k)$ is divisible by the cyclotomic
polynomial $\Phi_{d}(q)$ for any divisor $d>1$ of $n$. Hence

(14) $\sum_{k=0}^{n-1}t_{q}(k)\equiv 0\pmod{[n]}.$

We similarly show by an appeal to Lemma 1(b, c) with $N=4$ that
$\sum_{k=0}^{n-1}u_{q}(k)\equiv 0\pmod{[n]}$. This completes the proof.

##  7\. Proof of Lemma 1

Part (a) is a direct consequence of (1). Part (c) follows from part (b) and
the relation

$\sum_{k=0}^{n-1}t(k)=\sum_{l=0}^{\frac{n}{d}-1}\sum_{k=0}^{d-1}t(ld+k).$

As for part (b), we use induction on $N$. Let $N=2$ and for simplicity of
notation, let $a(k)=z_{1}(k)$, $b(k)=z_{2}(k)$, and
$c(k)=\sum_{j=0}^{k}a(j)b(k-j)$. Then we have

(15) $c(ld+k)=\sum_{j=0}^{ld+k}a(j)b(ld+k-j)=\sum_{j=0}^{d-1}a(j)b(ld+k-j)$
$+\sum_{j=d}^{2d-1}a(j)b(ld+k-j)+\cdots+\sum_{j=(l-1)d}^{ld-1}a(j)b(ld+k-j)+\sum_{j=ld}^{ld+k}a(j)b(ld+k-j).$

We now handle the individual terms in the foregoing identity. Let $r_{j}$ be
the remainder of the division of $j$ by $d$ for $j=0,1,\ldots,ld+k$. Then for
$m=0,1,\ldots,l-1$ and $md\leq j\leq(m+1)d-1$, we have $j=md+r_{j}$ with
$0\leq r_{j}<d$. Then

(16) $\sum_{j=md}^{(m+1)d-1}a(j)b(ld+k-j)$ $=\sum_{j:r_{j}\leq
k}b(md+r_{j})b\big{(}(l-m)d+k-r_{j}\big{)}+\sum_{j:r_{j}>k}a(md+r_{j})b\big{(}(l-m)d+k-r_{j}\big{)}$
$=\sum_{j=0}^{k}a(md+j)b\big{(}(l-m)d+k-j\big{)}+\sum_{j:r_{j}>k}a(r_{j})b\big{(}(l-m)d+k-r_{j}\big{)}$
$=a(md)b\big{(}(l-m)d\big{)}\sum_{j=0}^{k}a(j)b(k-j)+a(md)b\big{(}(l-m)d\big{)}\sum_{j:r_{j}>k}a(r_{j})\frac{b\big{(}(l-m)d+k-r_{j}\big{)}}{b\big{(}(l-m)d\big{)}}$

We now claim that

(17)
$\sum_{j:r_{j}>k}a(r_{j})\frac{b\big{(}(l-m)d+k-r_{j}\big{)}}{b\big{(}(l-m)d\big{)}}=\sum_{r_{j}=k+1}^{d-1}a(r_{j})\frac{b\big{(}(l-m)d+k-r_{j}\big{)}}{b\big{(}(l-m)d\big{)}}=0.$

Note first that the claim is clear if $k>\frac{d-1}{2}$ since $a(r_{j})=0$ for
$\frac{d-1}{2}<r_{j}\leq d-1$ by assumption. By the same assumption, it easy
to see that the terms in the foregoing sum for which $r_{j}>\frac{d-1}{2}\geq
k$ vanish. Now suppose that $k<r_{j}\leq\frac{d-1}{2}$. It follows that
$r_{j}-k<\frac{d-1}{2}$ and so, $d+k-r_{j}>\frac{d-1}{2}$. Thus we get

$a(r_{j})\frac{b\big{(}(l-m)d+k-r_{j}\big{)}}{b\big{(}(l-m)d\big{)}}=a(r_{j})\frac{b\big{(}(l-m-1)d\big{)}}{b\big{(}(l-m)d\big{)}}\frac{b\big{(}(l-m-1)d+d+k-j\big{)}}{b\big{(}(l-m-1)d\big{)}}$
$=a(r_{j})\frac{b\big{(}(l-m-1)d\big{)}}{b\big{(}(l-m)d\big{)}}b\big{(}d+k-j\big{)}=0.$

This proves the claim. Similarly, we have

(18) $\sum_{j=ld}^{ld+k}a(j)b(ld+k-j)=a(ld)\sum_{j=0}^{k}a(j)b(k-j).$

Now combine the relations (15)–(18) to deduce that

$c(ld+k)=c(k)\sum_{m=0}^{l-1}a(md)b\big{(}(l-m)d\big{)}+c(k)a(ld)=c(k)\sum_{m=0}^{l}a(md)b\big{(}(l-m)d\big{)}$

which gives the desired formula for the case $N=2$. Suppose now that the
statement is true for $N-1$ and let

$s(k)=\sum_{i_{2}=0}^{k}\sum_{i_{3}=0}^{k-i_{2}}\cdots\sum_{i_{N-1}=0}^{k-i_{2}-\cdots-
i_{N-1}}z_{2}(i_{2})\cdots z_{N-1}(i_{N-1})z_{N}(k-i_{2}-\cdots-i_{N-1}).$

We have

(19)
$t(ld+k)=\sum_{i_{1}=0}^{ld+k}\sum_{i_{2}=0}^{ld+k-i_{1}}\cdots\sum_{i_{N}=0}^{ld+k-i_{1}-\cdots-
i_{N-1}}z_{1}(i_{1})\cdots z_{N-1}(i_{N-1})z_{N}(ld+k-i_{1}-\cdots-i_{N-1}).$
$=\sum_{i_{1}=0}^{d-1}\sum_{i_{2}=0}^{ld+k-i_{1}}\cdots\sum_{i_{N}=0}^{ld+k-i_{1}-\cdots-
i_{N-1}}z_{1}(i_{1})\cdots z_{N-1}(i_{N-1})z_{N}(ld+k-i_{1}-\cdots-i_{N-1})$
$+\sum_{i_{1}=d}^{2d-1}\sum_{i_{2}=0}^{ld+k-i_{1}}\cdots\sum_{i_{N}=0}^{ld+k-i_{1}-\cdots-
i_{N-1}}z_{1}(i_{1})\cdots z_{N-1}(i_{N-1})z_{N}(ld+k-i_{1}-\cdots-i_{N-1})$
$+\cdots+\sum_{i_{1}=(l-1)d}^{ld-1}\sum_{i_{2}=0}^{ld+k-i_{1}}\cdots\sum_{i_{N}=0}^{ld+k-i_{1}-\cdots-
i_{N-1}}z_{1}(i_{1})\cdots z_{N-1}(i_{N-1})z_{N}(ld+k-i_{1}-\cdots-i_{N-1})$
$+\sum_{i_{1}=ld}^{ld+k}\sum_{i_{2}=0}^{ld+k-i_{1}}\cdots\sum_{i_{N}=0}^{ld+k-i_{1}-\cdots-
i_{N-1}}z_{1}(i_{1})\cdots z_{N-1}(i_{N-1})z_{N}(ld+k-i_{1}-\cdots-i_{N-1}).$

Let $r$ be the remainder of $i_{1}$ divided by $d$ for $i_{1}=0,1\ldots,ld+k$.
Clearly for $m=0,1,\ldots,l-1$ and $md\leq j\leq(m+1)d-1$ we have $i_{1}=md+r$
with $0\leq r<d$. Then for $m=0,1,\ldots,l-1$ we have

(20)
$\sum_{i_{1}=md}^{(m+1)d-1}\sum_{i_{2}=0}^{ld+k-i_{1}}\cdots\sum_{i_{N}=0}^{ld+k-i_{1}-\cdots-
i_{N-1}}z_{1}(i_{1})\cdots z_{N-1}(i_{N-1})z_{N}(ld+k-i_{1}-\cdots-i_{N-1})$
$=\sum_{i_{1}=md}^{(m+1)d-1}z_{1}(md+r)\sum_{i_{2}=0}^{ld+k-i_{1}}\cdots\sum_{i_{N}=0}^{ld+k-i_{1}-\cdots-
i_{N-1}}z_{2}(i_{2})\cdots z_{N-1}(i_{N-1})$ $\times z_{N}\big{(}(l-m)d+k-r-
i_{2}-\cdots-i_{N-1})\big{)}$
$=z_{1}(md)\sum_{j=md}^{(m+1)d-1}z_{1}(r)\sum_{i_{2}=0}^{ld+k-i_{1}}\cdots\sum_{i_{N}=0}^{ld+k-i_{1}-\cdots-
i_{N-1}}z_{2}(i_{2})\cdots z_{N-1}(i_{N-1})$ $\times z_{N}\big{(}(l-m)d+k-r-
i_{2}-\cdots-i_{N-1})\big{)}$ $=z_{1}(md)\sum_{\begin{subarray}{c}j=md\\\
r\leq
k\end{subarray}}^{(m+1)d-1}z_{1}(r)\sum_{i_{2}=0}^{ld+k-i_{1}}\cdots\sum_{i_{N}=0}^{ld+k-i_{1}-\cdots-
i_{N-1}}z_{2}(i_{2})\cdots z_{N-1}(i_{N-1})$ $\times z_{N}\big{(}(l-m)d+k-r-
i_{2}-\cdots-i_{N-1})\big{)}$ $+z_{1}(md)\sum_{\begin{subarray}{c}j=md\\\
r>k\end{subarray}}^{(m+1)d-1}z_{1}(r)\sum_{i_{2}=0}^{ld+k-i_{1}}\cdots\sum_{i_{N}=0}^{ld+k-i_{1}-\cdots-
i_{N-1}}z_{2}(i_{2})\cdots z_{N-1}(i_{N-1})$ $\times z_{N}\big{(}(l-m)d+k-r-
i_{2}-\cdots-i_{N-1})\big{)}.$

Let $A$ and $B$ respectively be the first and the second term on the right
hand-side of the foregoing identity. Then by the induction hypothesis

$A=z_{1}(md)\sum_{i_{1}=0}^{k}z_{1}(i_{1})\sum_{i_{2}=0}^{(l-m)d+k-i_{1}}\cdots\sum_{i_{N}=0}^{(l-m)d+k-i_{1}-\cdots-
i_{N-1}}z_{2}(i_{2})\cdots z_{N-1}(i_{N-1})$ (21)
$\qquad\qquad\qquad\qquad\qquad\qquad\times
z_{N}\big{(}(l-m)d+k-i_{1}-i_{2}-\cdots-i_{N-1})\big{)}$
$=z_{1}(md)\sum_{i_{1}=0}^{k}z_{1}(i_{1})s\big{(}(l-m)d+k-i_{1}\big{)}$
$=z_{1}(md)\sum_{i_{1}=0}^{k}z_{1}(i_{1})s(k-i_{1})\sum_{i_{2}=0}^{l-m}\sum_{i_{3}=0}^{l-m-
i_{2}}\cdots\sum_{i_{N}=0}^{(l-m)d+k-i_{2}-\cdots-i_{N-1}}z_{2}(i_{2}d)\cdots
z_{N-1}(i_{N-1}d)$ $\qquad\qquad\qquad\qquad\qquad\qquad\times
z_{N}\big{(}(l-m-i_{2}-\cdots-i_{N-1})d\big{)}$
$=z_{1}(md)t(k)\sum_{i_{2}=0}^{l-m}\sum_{i_{3}=0}^{l-m-
i_{2}}\cdots\sum_{i_{N}=0}^{(l-m)d+k-i_{2}-\cdots-i_{N-1}}z_{2}(i_{2}d)\cdots
z_{N-1}(i_{N-1}d)$

As for $B$, we find by the induction hypothesis and (1)

$\begin{split}B&=z_{1}(md)\sum_{\begin{subarray}{c}i_{1}=md\\\
r>k\end{subarray}}^{(m+1)d-1}z_{1}(r)\sum_{i_{2}=0}^{ld+k-i_{1}}\cdots\sum_{i_{N}=0}^{ld+k-i_{1}-\cdots-
i_{N-1}}z_{2}(i_{2})\cdots z_{N-1}(i_{N-1})\\\
&\qquad\qquad\qquad\qquad\qquad\qquad\times z_{N}\big{(}(l-m)d+k-r-
i_{2}-\cdots-i_{N-1})\big{)}\\\
&=z_{1}(md)\sum_{i_{1}=k+1}^{d-1}z_{1}(i_{1})\sum_{i_{2}=0}^{(l-m)d+k-i_{1}}\cdots\sum_{i_{N}=0}^{(l-m)d+k-i_{1}-\cdots-
i_{N-1}}z_{2}(i_{2})\cdots z_{N-1}(i_{N-1})\\\
&\qquad\qquad\qquad\qquad\qquad\qquad\times
z_{N}\big{(}(l-m)d+k-i_{1}-i_{2}-\cdots-i_{N-1})\big{)}\\\
&=z_{1}(md)\sum_{i_{1}=k+1}^{d-1}z_{1}(i_{1})\sum_{i_{2}=0}^{(l-m-1)d+d+k-i_{1}}\cdots\sum_{i_{N}=0}^{(l-m-1)d+d+k-i_{1}-\cdots-
i_{N-1}}z_{2}(i_{2})\cdots z_{N-1}(i_{N-1})\\\
&\qquad\qquad\qquad\qquad\qquad\qquad\times
z_{N}\big{(}(l-m-1)d+d+k-i_{1}-i_{2}-\cdots-i_{N-1})\big{)}\\\
&=z_{1}(md)\sum_{i_{1}=k+1}^{d-1}z_{1}(i_{1})s\big{(}(l-m-1)d+d+k-i\big{)}\\\
&=z_{1}(md)\sum_{i_{1}=k+1}^{\frac{d-1}{N}}z_{1}(i_{1})\sum_{i_{2}=0}^{l-m-
i_{1}}\cdots\sum_{i_{N}=0}^{l-m-i_{1}-\cdots-i_{N-1}}z_{1}(i_{2})\cdots
z_{N-1}(i_{N-1})z_{N}(d+k-i_{1}-\cdots-i_{N-1})\\\
&\qquad\qquad\qquad\qquad\qquad\qquad\times\sum_{i_{1}=k+1}^{d-1}z_{1}(i_{1})s(d+k-i_{1})\\\
&=z_{1}(md)\sum_{i_{1}=k+1}^{\frac{d-1}{N}}z_{1}(i_{1})\sum_{i_{2}=0}^{\frac{d-1}{N}}\cdots\sum_{i_{N}=0}^{\frac{d-1}{N}}z_{1}(i_{2})\cdots
z_{N-1}(i_{N-1})z_{N}(d+k-i_{1}-\cdots-i_{N-1})\\\
&\qquad\qquad\qquad\qquad\qquad\qquad\times\sum_{i_{1}=k+1}^{d-1}z_{1}(i_{1})v(d+k-i_{1})\\\
\end{split}$

As $k-i_{1},i_{2},\ldots,i_{N-1}\leq(d-1)/N$, we deduce that $d-1\geq
d+k-i_{1}-\cdots-i_{N-1}>(d-1)/N$ and therefore $z_{N}(d+k-i_{1}-\cdots-
i_{N-1})=0$. It follows that

$s(d+k-i_{1})=\sum_{i_{2}=0}^{\frac{d-1}{n}}\cdots\sum_{i_{N}=0}^{\frac{d-1}{N}}z_{1}(i_{2})\cdots
z_{N-1}(i_{N-1})z_{N}(d+k-i_{1}-\cdots-i_{N-1})=0$

and hence $B=0$. Regarding the last term of (19), we similarly find

$\sum_{i_{1}=ld}^{ld+k}\sum_{i_{2}=0}^{ld+k-i_{1}}\cdots\sum_{i_{N}=0}^{ld+k-i_{1}-\cdots-
i_{N-1}}z_{1}(i_{1})\cdots z_{N-1}(i_{N-1})z_{N}(ld+k-i_{1}-\cdots-i_{N-1})$
(22) $=z_{1}(ld)t(k).$

Then by (19)-(22), we deduce after replacing $m$ with $i_{1}$ that

$t(ld+k)=t(k)\sum_{i_{1}=0}^{l}\sum_{i_{2}=0}^{l-i_{1}}\cdots\sum_{i_{N}=0}^{l-i_{1}-\cdots-
i_{N-1}}z_{1}(i_{1}d)\cdots z_{N-1}(i_{N-1}d)z_{N}\big{(}(l-i_{1}-\cdots-
i_{N-1})d\big{)}.$

This completes the proof.

##  8\. Final comments

Note that in Theorem 5 while the sequence $\\{\alpha_{q}(k)\\}_{k=0}^{\infty}$
with the choices $x=q^{-n}$ or $x=q^{n}$ satisfies (1) for $N=2$, it does not
satisfy this condition for $N>2$. This justifies the fact that while for any
prime $p>1$ we have

$\sum_{k=0}^{p-1}\sum_{j=0}^{k}\frac{{2j\choose j}^{2}{4j\choose
2j}{2(k-j)\choose k-j}^{2}{4(k-j)\choose
2(k-j)}}{2^{8k}3^{2k}}(8j+1)\big{(}8(k-j)+1\big{)}\equiv p^{2}\pmod{p^{3}},$

it is in general not true that

$\sum_{k=0}^{p-1}\sum_{j=0}^{k}\sum_{i=0}^{k-j}\frac{{2j\choose
j}^{2}{4j\choose 2j}{2i\choose i}^{2}{4i\choose 2i}{2(k-j-i)\choose
k-j-i}^{2}{4(k-j-i)\choose
2(k-j-i)}}{2^{8k}3^{2k}}(8i+1)(8j+1)\big{(}8(k-j-i)+1\big{)}$ $\equiv
p^{3}\equiv 0\pmod{p^{3}}.$

Related to this we have the following conjecture which is based on
computational evidence.

######  Conjecture.

If $p>3$ is a prime number, then

$\sum_{k=0}^{p-1}\sum_{j=0}^{k}\sum_{i=0}^{k-j}\frac{{2j\choose
j}^{2}{4j\choose 2j}{2i\choose i}^{2}{4i\choose 2i}{2(k-j-i)\choose
k-j-i}^{2}{4(k-j-i)\choose
2(k-j-i)}}{2^{8k}3^{2k}}(8i+1)(8j+1)\big{(}8(k-j-i)+1\big{)}\equiv
0\pmod{p^{2}}.$

Data Availability Statement. The authors confirm that the manuscript is self-
contained and that the data supporting the findings of this study are
available within the article.

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# Study of charm hadronization and in-medium modification at the Electron-ion
Collider in China

Senjie Zhu<EMAIL_ADDRESS>University of Science and Technology of
China, Hefei, Anhui Province 230026, China  Xiao Huang  University of Science
and Technology of China, Hefei, Anhui Province 230026, China  Lei Xia
University of Science and Technology of China, Hefei, Anhui Province 230026,
China  Aiqiang Guo  Institute of Modern Physics, Chinese Academy of Sciences,
Lanzhou, Gansu Province 730000, China  University of Chinese Academy of
Sciences, Beijing 100049, China  Yutie Liang  Institute of Modern Physics,
Chinese Academy of Sciences, Lanzhou, Gansu Province 730000, China  University
of Chinese Academy of Sciences, Beijing 100049, China  Guangdong Provincial
Key Laboratory of Nuclear Science,
Institute of Quantum Matter, South China Normal University, Guangzhou 510006,
China  Yifei Zhang<EMAIL_ADDRESS>University of Science and Technology of
China, Hefei, Anhui Province 230026, China  Yuxiang Zhao  Institute of Modern
Physics, Chinese Academy of Sciences, Lanzhou, Gansu Province 730000, China
Southern Center for Nuclear-Science Theory (SCNT), Institute of Modern
Physics, Chinese Academy of Sciences, Huizhou, Guangdong Province 516000,
China  University of Chinese Academy of Sciences, Beijing 100049, China  Key
Laboratory of Quark and Lepton Physics (MOE) and Institute of Particle
Physics,
Central China Normal University, Wuhan 430079, China

###### Abstract

Charm quark production and its hadronization in $ep$ and $eA$ collisions at
the future Electron-Ion Collider in China (EicC) will help us understand the
quark/gluon fragmentation processes and the hadronization mechanisms in the
nuclear medium, especially within a poorly constrained kinematic region
($x<0.1$). In this paper, we report a study on the production of charmed
hadrons, $D^{0}$ and $\Lambda_{c}^{+}$, reconstructed with a dedicated geant4
simulation of vertex$\,\&\,$tracking detectors designed for EicC. The
$\Lambda_{c}^{+}$/$D^{0}$ ratios as functions of multiplicity and $p_{T}$, as
well as the $D^{0}$ double ratio are presented with projected statistical
precision.

Suggested keywords

††preprint: APS/123-QED

##  I Introduction

Exploring the most fundamental building blocks of the universe is the
paramount objective of modern physics. Over the past five decades, we’ve come
to understand that matter, in its essence, is composed of nuclei and nucleons.
These, in turn, are made up of even more basic components known as quarks,
which are confined in a colorless bound state through the exchange of gluons.
This interaction among quarks and gluons is governed by the theory of strong
force, known as Quantum Chromodynamics (QCD). However, up to now, the
constituent interactions and the dynamical distributions of quarks and gluons
inside nucleons, in particular in the kinematic regime dominated by sea quarks
and gluons, are still unclear or poorly constrained by existing experiments
[1, 2]. The next-generation experimental facilities with electron-ion
collision have been proposed for the quantitative study of matter in this new
regime, such as Electron-Ion Collider at BNL (EIC-BNL) [3, 4] and Electron-ion
collider in China (EicC) [5].

EicC is proposed to operate at a center-of-mass (c.m.) energy ranging from 15
GeV to 20 GeV, with a luminosity of approximately $2.0~{}\times~{}10^{33}$
cm${}^{-2}\cdot$s-1 [5]. This would enable EicC to explore the kinematic
region dominated by sea quarks, effectively bridging the gap between the EIC-
BNL and JLab 12 GeV experiments. The primary objective of EicC is to delve
into the partonic structure and three-dimensional tomography of nucleons and
nuclei, the cold nuclear matter effects, as well as the origin of proton mass.
This ambitious endeavor aims to significantly advance our understanding of
these fundamental aspects of nuclear physics.

Understanding the interactions of partons with nuclear matter, as well as
their fragmentation [6] or combination [7] to form hadrons—a process known as
hadronization—is a fundamental question that future EicC experiments aim to
answer. Heavy quarks, due to their large masses, are predominantly produced in
the initial hard scatterings of collisions and undergo the complete evolution
of the nuclear medium system. This makes them an ideal probe for measuring
nuclear medium effects and heavy quark hadronizations in high-energy heavy-ion
collisions [8, 9, 10].

For example, the enhancement of $\Lambda_{c}^{+}/D^{0}$ ratios in heavy-ion
collisions compared with elementary particle collisions has been measured by
STAR [9] and ALICE [11, 12], revealing the hadronization mechanism of charm
quark with light quarks in hot [7, 13] and dense medium [14]. However, in
these collisions, it is a complex task to separate the effect of the cold
nuclear medium (CNM) from the dominant medium response caused by the hot
quark-gluon plasma (QGP). This is where electron-ion collisions come into
play, offering an ideal platform to study the CNM effects [15] in a more
transparent system where it is believed no QGP is formed. Moreover, two
theories with different time scales - parton energy loss and the hadron
absorption model [16] \- can successfully describe the suppression of light
hadrons in electron-ion collisions. The measurement of heavy quark production
presents a novel opportunity to reveal the mechanisms of hadronization. To
address these inquiries, it is essential to investigate charm hadronization
and the production of open charmed hadrons in future EicC experiments.

In this paper, we perform a comprehensive simulation study with a geant4 [17]
detector configuration for open charm hadron production. The report is
organized as follows. Section II introduces the simulation setup and process.
Section III presents the simulation results and a summary is given in Section
IV.

##  II Simulation Setup

This study employs the pythia event generator, as referenced in [18]. Two
distinct versions of this generator, namely pythiaeRHIC (pythia 6.4) and
pythia 8.3, are utilized in our analysis. The configuration for pythia 6.4 is
detailed extensively in Ref. [19]. The physical processes including vector-
meson diffractive and resolved processes, semihard QCD $2\rightarrow 2$
scattering, neutral boson scattering off heavy quarks within the proton, and
photon-gluon fusion, are turned on. Several alternative hadronization models
in pythia 8.3 are used for charm hadronization studies, e.g., implemented
color reconnection models. The kinematic variables used are listed in Table 1.

Table 1: Kinematic variable definition.  $k$ and $k^{\prime}$ |  The four-momentum of the incoming electron and the scattered electron
---|---
$p$ and $p_{h}$ |  The four-momentum of the incoming proton and the produced hadron
$q=k-k^{\prime}$ |  The four-momentum of emitted virtual photon
$Q^{2}=-q^{2}$ |  The negative square invariant momentum transfer of emitted virtual photon
$x_{B}=Q^{2}/(2p\cdot q)$ |  Bjorken scaling variable
$y=p\cdot q/(p\cdot k)$ |  The fraction of the incoming electron’s energy transferred to the hadronic system
$z=p\cdot p_{h}/(p\cdot q)$ |  The momentum fraction of the virtual photon to be carried by the produced hadron
$W=\sqrt{(q+p)^{2}}$ |  The center-of-mass energy of $\gamma^{*}$-Nucleon system
$\nu=q\cdot p/m_{N}$ |  The energy of $\gamma^{*}$ in nucleon rest frame ($m_{N}$ is the mass of the nucleon)
Figure 1:  (Top) $\rm{log}_{10}$$(x_{B})-\rm{log}_{10}$$(Q^{2}/GeV^{2})$
distribution of charmed events per $1\,\rm{fb}^{-1}$. (Bottom)
$\rm{log}_{10}$$(x_{B})$ distribution of inclusive DIS and charm per
$1\,\rm{fb}^{-1}$ ($\rm Q^{2}>1\,GeV^{2}$ ). Figure 2:  The
momentum-$\rm\theta$ distributions of the particles produced by pythia. The
particle types are labeled respectively in the panel for $D^{0}/\bar{D^{0}}$,
$\Lambda_{c}^{\pm}$ and pions from their decay.

The kinematic coverage of the collision with electron (3.5 GeV) and proton (20
GeV) is shown in Fig.1 (top panel), which presents the
$\rm{log}_{10}$$(x_{B})-\rm{log_{10}}(Q^{2})$ distribution of charmed events
per $1\,\rm{fb}^{-1}$ integral luminosity. As shown in the $x_{B}-Q^{2}$
distribution, one step along $Q^{2}$ distribution exists at
$\rm{log}_{10}($$Q^{2}$$)=0.4$. The reason is that the subprocess cross-
section ($\gamma^{*}q\rightarrow q$) in pythia 6 is deliberately set to zero
when the photon’s virtuality $Q^{2}$ approaches zero. This is done to prevent
double-counting with real-photon physics processes[18]. In the above beam
setting and with DIS requirement $Q^{2}>1\,$GeV2, the $x_{B}$ distributions
are shown in the bottom panel of Fig.1, including inclusive DIS events and
charmed events. Charmed events account for $1.4\%$ of inclusive DIS events.
The momentum-$\theta$ phase space 2-dimension distribution of $D^{0}$ is shown
in Fig.2 (the first panel). Due to the higher momentum of the beam protons
compared to the beam electron, the $D^{0}/\bar{D^{0}}$ distribution is skewed
towards the direction of the proton beam. Consequently, as illustrated in the
second panel of Fig.2, the pions from $D^{0}\rightarrow\pi K$ decay are given
a boost, causing them to be produced more in the forward direction of the
proton beam. As illustrated in Fig.2 for $\Lambda_{c}^{\pm}$, the
distributions of $\Lambda_{c}^{\pm}$ and the pions from its decay are even
more skewed towards the direction of the proton beam. There is a region near
$\theta=0$ where the number of $\Lambda_{c}^{\pm}$ is significantly larger
than in other areas, and the $\Lambda_{c}^{\pm}$ in this region has much
higher energy. Furthermore, $\Lambda_{c}^{+}$ is much more abundant than
$\Lambda_{c}^{-}$ in this region. This phenomenon is caused by the formation
of beam remnants [20], a charm quark produced via the hard scattering and $u$,
$d$ quarks from the beam proton form a $\Lambda_{c}^{+}$ together. Because the
$u$, $d$ quarks from protons carry a large fraction of the proton momentum,
the beam remnant $\Lambda_{c}^{+}$ is very energetic.

The specific design and layout for EicC detectors have been detailed in Ref.
[21], which include a tracking and vertex detector system. This report
primarily focuses on the tracking and vertex detectors. The detector
acceptance is $-3<\eta<3$ ($5.7^{\circ}\sim 174.3^{\circ}$). In Fig.2 (panel
(b) and (c)), the acceptance region is marked by two black lines. In Ref.
[21], the performances have been obtained through geant4 simulation for
different particle species ($e,\mu,\pi,K,p$) under a $1.5\,$T magnetic field
configuration.

Figure 3:  The detector’s performance for $\pi$: (a) the efficiency of track
reconstruction, (b) the resolution of the reconstructed track momentum, (c)
the resolution of the reconstructed track $DCA$ in $z$ direction, (d) the
resolution of the reconstructed track $DCA$ in the $r\phi$ plane. The
different marker styles and line colors of the lines represent the different
$\eta$ regions of $\pi$, which are shown in the legend of the panel (a).

The performances for $\pi$ are shown as an example in Fig.3. In this study,
the Distance of the Closest Approach ($DCA$) are defined as the closest
distance between the reconstructed track and the primary vertex in $r\phi$
plane ($DCA_{r\phi}$) and $z$-axis ($DCA_{z}$). The tracking efficiency, the
momentum resolution, the $DCA_{z}$ and $DCA_{r\phi}$ resolutions are shown in
Panel (a), (b), (c), and (d), respectively. In the absence of particle
identification (PID) detector in the simulation, a $3\sigma$ separation power
for PID is assumed in the prospected PID detector acceptance coverage listed
in Table 2. It is applied as a momentum hard cut-off as a function of $\eta$.

Table 2: PID acceptance for hadrons at different $\eta$ regions.  $\eta$ | $[-3,-1)$ | $[-1,1)$ | $(1,3]$
---|---|---|---
$p_{max}[\rm{GeV}/$$c]$ | 4 | 6 | 15
Figure 4:  Topological cuts applied to reconstruct $D^{0}$ (a) and
$\Lambda_{c}^{+}$ (b). Figure 5:  Normalized $D^{0}$ Topological cut
distributions, where the blue lines are for signal and the red lines are for
background.

All the performance parameters mentioned above have been used to conduct a
fast simulation for the physical projection. In this simulation, the track
information generated by pythia is adjusted, or “smeared,” according to a
Gaussian function to mimic the limitations of the detection capability. The
number of the tracks in the acceptance ($-3<\eta<3$) and passing the tracking
efficiency filters is used to determine the primary vertex resolution, which
is also smeared in the pythia events. After smearing, different topological
requirements are used to separate the signal and background.

$D^{0}$ is reconstructed from channel $D^{0}\rightarrow\pi^{+}K^{-}$ with a
branch ratio ($\mathcal{B}$) of 3.83$\%$ and $\Lambda_{c}^{+}$ is
reconstructed from channel $\Lambda_{c}^{+}\rightarrow\pi^{+}K^{-}p$
($\mathcal{B}$=2.96% in pythia 6.4 and $\mathcal{B}$=3.4% in pythia 8.3). The
$\mathcal{B}$ difference of $\Lambda_{c}^{+}$ between pythia 6.4 and pythia
8.3 is caused by a missing channel in pythia 6.4
($\Lambda_{c}^{+}\rightarrow\Lambda\pi^{+},\Lambda\rightarrow pK^{-}$). In the
following part of this report, we use $D^{0}$ to represent both $D^{0}$ and
$\bar{D^{0}}$ and use $\Lambda_{c}^{+}$ to represent both $\Lambda_{c}^{+}$
and $\Lambda_{c}^{-}$. Figure. 4, we show topological cut diagrams for $D^{0}$
(a) and for $\Lambda^{+}_{c}$ (b). For $D^{0}$, three topological requirements
are applied, including $DCA_{\pi K}$, $DecayL_{XY}^{D^{0}}$ and
$\cos\theta_{XY}$, which are defined as:

  * $\bullet$

$DCA_{\pi K}$ is the closest distance between two daughter tracks.

  * $\bullet$

$DecayL_{XY}^{D^{0}}$ is the distance between the decay vertex and the
reconstructed primary vertex (in the $xy$ plane). Once $DCA_{\pi K}$ is
defined, two points that symbolize $DCA_{\pi K}$ on each of the two tracks can
be found. Then, the decay vertex of $D^{0}$ is defined as the middle of the
two points.

  * $\bullet$

$\cos\theta_{XY}$ is the cosine value of the angle between the $D^{0}$ decay
length and the $D^{0}$ momentum (in the $xy$ plane).

The reason for defining both $\cos\theta_{XY}$ and $DecayL_{XY}^{D^{0}}$ in
the $xy$ plane is that the spatial resolution in the $xy$ plane is much better
than that in the $z$ dimension, especially in the endcap region. However, if
the projections of two non-parallel lines in $xy$ plane cross, $DCA_{\pi K}$
will always be zero in the $xy$ plane. Therefore, $DCA_{\pi K}$ is defined in
3 dimensions instead of $xy$ plane. For $\Lambda_{c}^{+}$, the $DCA_{\pi K}$
is replaced by $DCA_{\rm daughters}^{\rm max}$, the maximum of all three
$DCA_{\rm daughters}$, where $DCA_{\rm daughters}$ is the DCA between two of
the three daughters. The decay vertex is the center of all three decay
vertexes defined by each daughter pair. All three topological cut
distributions are shown in Fig.5 for the signal (blue dashed line) and
background (red solid line).

Figure 6:  Topological requirements are optimized to ensure the highest
significance of $D^{0}$. The previous best results are applied to the next
round of optimization.

The $\pi K$ invariant mass distributions in the $D^{0}$ mass region are fitted
with a Gaussian function for signal and a linear function for background to
obtain the $D^{0}$ signal yield. The significance is defined as
$S/\sqrt{S+B}$, where $S$ and $B$ stand for the counts of $D^{0}$ signal and
background, respectively. All the criteria are optimized by maximizing the
significance iteratively. Figure. 6 shows $D^{0}$ significance as a function
of all three topological requirements. The optimal topological criteria are:
(i) $DCA_{\pi K}<110\,\mu m$, (ii) $\cos\theta_{XY}>-0.75$. As seen in Fig.2
(the first panel), the $p_{T}$ of most $D^{0}$ is lower than 1GeV/$c$ and
$D^{0}$ with lower $p_{T}$ tends to have a smaller $DecayL^{D^{0}}_{XY}$, so
the $DecayL^{D^{0}}_{XY}$ distribution of signal is similar to that of
background. Therefore, the significance is insensitive to
$DecayL^{D^{0}}_{XY}$ as shown in Fig.6.

The $\pi K$ invariant mass distributions with different detector performance
configurations are shown in Fig.7. There are three configurations included: no
PID, with PID and PID+Vertex. In the first two configurations, the vertex
detectors in the geometry for geant4 simulation are removed to determine the
significance of the vertex detectors in $D^{0}$ reconstruction. For no PID,
the PID acceptance in Table 2 isn’t applied, and all possible combinations
formed by particles with opposite charges are considered. As shown in Fig.7,
the signal peak of the green line (no PID) is almost invisible. Compared to
the green line (no PID), the background of the blue line (with PID) is greatly
suppressed when the PID acceptance is applied. Similarly, compared to the
green line (no PID), the background of the red line (PID+Vertex) is also
suppressed, and the mass peak is narrowed. This is because the momentum
resolution becomes better when the vertex detectors are installed. After
adding the vertex detectors, the significance improves from $13.9$ to $21.8$.

Figure 7:  The $\pi K$ invariant mass distributions with different detector
performance configurations:(no PID) without either PID system or vertex
detectors, (with PID) with the PID system but without vertex detectors,
(PID+Vertex) with the PID system and the vertex detectors. The significance is
achieved at an integrated luminosity 0.04fb-1. The significances of with PID
and PID+Vertex are 13.9 and 21.8, respectively. The $S/B$ of with PID and
PID+Vertex are 0.06 and 0.215, respectively. Figure 8:  $\Lambda_{c}^{+}$
topological cut distributions. The blue dashed and red solid lines represent
the distributions for the background and signal, respectively. Because the
differences of $decayL^{\Lambda^{+}_{c}}_{XY}$ and $\cos{\theta_{XY}}$
distribution between signal and background are small, $DCA^{max}_{\rm
Daughters}$ is the only topological requirement taken into account.  Figure 9:
The results of $\Lambda_{c}^{+}$ reconstruction.(a) the significance scanning
for $DCA^{max}_{\rm Daughters}$. (b) the $\pi Kp$ invariant mass
distributions. The read solid and blue dashed lines represent the
distributions without and with requirement application, respectively.

The reconstruction procedure is also carried out for $\Lambda_{c}^{+}$. The
difference is that the shape of the $\Lambda_{c}^{+}$ peak is non-Gaussian, as
shown in the bottom panel of Fig.9. Therefore, we use the line shape of
$\Lambda_{c}^{+}$ after smearing the daughter tracks from pure
$\Lambda_{c}^{+}$ and a linear function in the fit. The distributions of
$\Lambda_{c}^{+}$ topological cuts are shown in Fig.8, where we can find that
$DCA_{\rm daughters}^{\rm max}$ is the only useful criterion. Therefore, in
Fig.9, only the criterion optimization for $DCA_{\rm daughters}^{\rm max}$ is
performed and the optimal result is $DCA_{\rm daughters}^{\rm max}<1000\,\mu
m$. The significance is improved from 34 to 38 with this requirement.

##  III Results

###  III.1 Baryon-to-meson ratios

Figure 10:  The values of the red dashed and red solid lines are calculated
from pythia 8.3 with QCD-CR and MPI-CR. The red solid triangles and circles
are the centers between the curves of the two models at both mid- and forward-
rapidity. The red error bars represent the projection of
$\Lambda_{c}^{+}/D^{0}$ ratio statistical uncertainties at an integrated
luminosity $100\,\rm{fb}^{-1}$. The black points are the result of ALICE
measurement where the error bars are statistical uncertainties and the grey
boxes are the systematic uncertainties [11]. Figure 11:  The projection of the
statistical uncertainty of $\Lambda_{c}^{+}/D^{0}$ ratio as a function of
multiplicity. The description of both lines and points is similar to what has
been discussed in Fig.10. The red and blue points are for EicC results and
ATHENA results [22].

The fragmentation ratios of the charm quark obtained from different
experiments were thought to be universal. Both HERA $ep$ collision
measurements [23, 24] and the previous $pp$ collision measurements at LHC [25]
are consistent with $e^{+}e^{-}$ collision measurements [26]. However, an
enhancement of $\Lambda_{c}^{+}/D^{0}$ ratio has been observed in the recent
measurement at ALICE [11, 27]. The deviation of the $\Lambda_{c}^{+}/D^{0}$
ratio from $pp$ collision to $e^{+}e^{-}$ collision experiment indicates that
even a small hadronic system can affect charm hadronization. Therefore, the
$\Lambda_{c}^{+}/D^{0}$ ratio should be re-examined in the $ep$ collision
system, specifically at EicC, which provides unique kinematic coverage and
high luminosity.

Two projections of $\Lambda_{c}^{+}/D^{0}$ ratios are provided in this
subsection. $ep$ collision data with beam energies 3.5 GeV $\times$ 20 GeV are
generated by pythia 8.3 with QCD color reconnection (QCD-CR) tuning [28] and
multiple-parton-interaction color reconncetion (MPI-CR) tuning [20]. In
Fig.10, the statistical uncertainty projection as a function of $p_{T}$ is
shown. The curves are from pythia 8.3, in which the solid and dashed curves
are from the QCD-CR and MPI-CR settings. The values of the red points are the
average of the two models at the corresponding rapidity regions. The
estimation of statistical uncertainties is the root mean square of the sum of
squares of errors propagated from all sources. For $\Lambda_{c}^{+}/D^{0}$
ratio, the expression is

$\sigma(\frac{\Lambda_{c}^{+}}{D^{0}})=\sqrt{(\frac{\sigma(\Lambda_{c}^{+})}{N_{D^{0}}})^{2}+(\frac{N_{\Lambda_{c}^{+}}}{N^{2}_{D^{0}}}\sigma(D^{0}))^{2}}$

, where the statistical uncertainties of $D^{0}$ and
$\Lambda_{c}^{+}$($\sigma(D^{0})$ and $\sigma(\Lambda_{c}^{+})$) are
calculated by the signal yields divided by the corresponding significance. The
statistical uncertainty is scaled to an integral luminosity
$100\,\rm{fb}^{-1}$. The results of $\Lambda_{c}^{+}/D^{0}$ ratios are
presented in two rapidity regions, the mid-rapidity region ($|y|<1$) and
forward-rapidity region ($1<y<3$), shown as solid circles and triangles
respectively. The measurement in the $pp$ collision from ALICE [11] is also
presented, and shown as black solid circles in Fig.10. The black points
represent the center values, the black line is the statistical uncertainty,
and the grey box is the systematic uncertainty. Compared to ALICE, the
statistical precision is significantly improved in EicC due to high luminosity
and much less hadronic combinatorial background. EicC results show that the
uncertainties are smaller than the difference between the two models at both
mid- and forward-rapidity. This suggests that EicC has sufficient power to
distinguish between different hadronization models. In addition, the
$\Lambda_{c}^{+}/D^{0}$ ratios have different behaviors in mid- and forward-
rapidity regions. A wide rapidity coverage, benefiting from large acceptance
of the detector design, can be achieved at EicC. As mentioned above, there is
an enhancement of $\Lambda_{c}^{+}$ production caused by beam remnant
formation in the very forward rapidity region. With wider rapidity coverage of
EicC, the interaction between beam remnant and charm quark can also be studied
in greater detail.

Fig.11 shows the statistical uncertainty projections of
$\Lambda_{c}^{+}/D^{0}$ as a function of charged particle multiplicity by red
marks. The high multiplicity results from EIC-BNL simulation [22] are shown as
the open blue marks. Only the tracks in $|\eta|<3$, $p_{T}>0.2\,\rm{GeV}/$$c$
and not from $D^{0}$ or $\Lambda_{c}^{+}$ decay are counted in the number of
charge tracks. The dashed and solid lines are from the calculation of MPI-CR
and QCD-CR pythia 8.3 models, respectively. The red and blue curves are the
results for EicC and EIC-BNL, respectively. EicC can provide a measurement of
$\Lambda_{c}^{+}/D^{0}$ ratio at low multiplicity as a complement to EIC-BNL
measurements. The measurement of the charm baryon-to-meson ratio can provide
insight into the interplay between the hard and soft processes that produce
particles. In $pp$ or $AA$ collisions, a higher charged particle multiplicity
corresponds to higher collision energy and lower impact parameters. The events
with different collision energies and impact parameters will have different
underlying event structures. Thus, a wider coverage of multiplicity is
necessary to study the charm hadronization at $\gamma-Nucleon$ and $\gamma-
Nuclei$ collision systems.

###  III.2 $D^{0}$ double ratio

Figure 12: The $D^{0}$ double ratio: The statistical uncertainty projections
are corresponding to an integral luminosity
$\mathcal{L}_{int}=10\,\rm{fb}^{-1}$. The open and solid circles represent the
results of EicC total kinematic coverage (EicC Total:
$2\,\rm{GeV}^{2}<Q^{2}<200\,\rm{GeV^{2}},\nu<96\,\rm{GeV}$) and a smaller
specific kinematic coverage (EicC Specific:
$2\,\rm{GeV}^{2}<Q^{2}<8\,\rm{GeV}^{2}$, $48\,\rm{GeV}<\nu<77\,\rm{GeV}$) .

Hadron production is affected by initial and final state interactions. The
primary initial effect is EMC effect, which is not of interest in this
particular study. We focus on the final state interaction and the effect of
the surrounding nuclear medium on hadronization. To minimize the initial state
effect, the observable double ratio is defined as

$R^{h}_{eA}(Q^{2},x_{B},z)=\frac{N^{h}_{A}(Q^{2},x_{B},z)/N^{DIS}_{A}(Q^{2},x_{B})}{N^{h}_{p}(Q^{2},x_{B},z)/N^{DIS}_{p}(Q^{2},x_{B})}$

where $N^{DIS}$ and $N^{h}$ are the number of DIS events and the number of the
SIDIS events in which $h$ hadrons are produced, respectively.

DIS requirements, including (i) $Q^{2}>2\,$GeV2, (ii) $\nu<96\,\rm{GeV}$, and
(iii) $W^{2}>4\,$GeV2 are applied. The statistical uncertainty projection of
$D^{0}$ double ratio as a function of $z$ is shown in Fig.12. Here, the
electron-nucleus collision is the $eAu$ collision with electron momentum
$3.5\,\rm{GeV}/c$ and $Au$ momentum $12.93\,\rm{GeV}/$$c$$/u$. The center
values of all the points are set to 1 because there should be no medium
modification in pythia. The statistical uncertainty is scaled to the integral
luminosity $10\,\rm{fb}^{-1}/u$ of electron-nucleus and electron-proton
collision data. The open red circles are EicC results for the whole kinematic
region. The solid red circles are EicC results for additional DIS criteria:
$Q^{2}<8$ GeV2 and $48$ GeV $<\nu<77$ GeV. The EicC specific kinematic
coverage in Fig.12 is chosen to be a region where $\nu$ is rather low and the
$D^{0}$ are produced abundantly.

There are precise measurements of the light hadron double ratio
($\pi^{+/-},K^{+/-},p$ and $\bar{p}$) versus multiple variables at HERMES.
However, as mentioned earlier, two theories with different time scales can
adequately describe the HERMES data [29, 16]. To provide an additional
powerful constraint to these theories, $D^{0}$ double ratio is necessary
because that $D^{0}$ double ratio is very different from that of light
hadrons. For light hadrons, the values of the double ratio at all $z$ bins are
below unity for nuclear attenuation or parton energy loss. However, for
$D^{0}$, there is a peak in $c\rightarrow D^{0}$ fragmentation function at low
$z$($z<0.1$). So $D^{0}$ double ratio is higher than unity at low $z$ bins and
lower than unity at high $z$ bins for the same reason as light hadron
suppression. In addition, a lower c.m. energy can produce a larger cold
nuclear effect [30]. Therefore, EicC, which has a lower c.m. energy than EIC-
BNL but can abundantly produce $D^{0}$, can be an ideal facility to study how
the nuclear medium affects the hadronization process. Additionally, it can
complement EIC-BNL by covering a wider kinematic region.

##  IV Summary

In summary, open charm reconstruction is studied with fast simulation, where
the detector performance parameterizations are derived from the geant4
simulation with the developing detector geometry designed for EicC. The
reconstruction of $D^{0}(\rightarrow K^{-}\pi^{+})$ and
$\Lambda_{c}^{+}(\rightarrow p\pi^{+}K^{-})$ in the $\sqrt{s}=16.7\,$GeV$\,ep$
collision demonstrates the importance of vertex detectors in reducing
background and improving momentum resolutions. This leads to higher signal
significance and lower statistical uncertainties, providing sufficient
precision for future charm hadronization studies in EicC.

We have projected the statistical uncertainty of $\Lambda_{c}^{+}/D^{0}$
ratios as functions of $p_{T}$ and charged-particle multiplicity. Different
hadronization models (QCD-CR and MPI-CR) can be distinguished with the
projected statistical uncertainties at an integrated luminosity of
$\mathcal{L}_{int}=100\,\rm{fb}^{-1}$. A broader rapidity coverage than ALICE
can provide abundant information about hadronization in $\gamma-Nucleon$ and
$\gamma-Nuclei$ collision systems.

The statistical uncertainties of the double ratio of $D^{0}$, which behave
differently from light hadron double ratios, are projected as a function of z
at an integrated luminosity of $\mathcal{L}_{int}=10\,\rm{fb}^{-1}/u$. Precise
measurements at EicC can provide an excellent opportunity to understand charm
hadronization mechanisms and how charm interacts with the nuclear medium.

###### Acknowledgements.

This work is supported in part by the Strategic Priority Research Program of
the Chinese Academy of Sciences under grant number XDB34000000, the Guangdong
Major Project of Basic and Applied Basic Research No. 2020B0301030008, the
Guangdong Provincial Key Laboratory of Nuclear Science with No.
2019B121203010, the National Natural Science Foundation of China with Grant
No. 11890712, 12061141008 and the National Key R&D Program of China with Grant
No. 2018YFE0104700 and 2018YFE0205200. The authors acknowledge the computing
resources available at the Southern Nuclear Science Computing Center.

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