# General convolution sums involving Fibonacci $m$-step numbers

Robert Frontczak1 and Karol Gryszka2

###### Abstract

In this paper, using a generating function approach, we derive several new
convolution sum identities involving Fibonacci $m$-step numbers. As special
instances of the results derived herein, we will get many new and known
results involving Fibonacci, Tribonacci, Tetranacci and Pentanacci numbers. In
addition, we establish some general results providing insights into the inner
structure of such convolutions. Finally, some mixed convolutions involving
Fibonacci $m$-step numbers and Jacobsthal and Pell numbers will be stated.

1Independent Researcher
Reutlingen, Germany
2Corresponding author
Institute of Mathematics
University of the National Education Commission, Krakow
Podchora̧żych 2, 30-084 Kraków, Poland

E-mail addresses<EMAIL_ADDRESS>
<EMAIL_ADDRESS>
Keywords: Convolution, Fibonacci $m$-step number, generating function.

MSC 2020: 05A15, 11B37, 11B39.

##  1 Motivation and Preliminaries

The motivation for writing this paper comes from three recently published
papers by Dresden and his collaborators: Dresden and Wang [3, 5] and Dresden
and Tulskikh [4]. In these papers, convolutions involving important number
sequences like Fibonacci numbers $F_{n}$, Lucas numbers $L_{n}$, Pell numbers
$\mathcal{P}_{n}$, Jaconsthal numbers $J_{n}$, Tribonacci numbers $T_{n}$, and
others are studied. Dresden and Wang close their article [5] with a short
discussion of the two seemingly unrelated convolutions

$\sum_{j=0}^{n}J_{j}F_{n-j}=J_{n+1}-F_{n+1}\quad(\mbox{see}\,\,\cite[cite]{[\@@bibref{}{Koshy}{}{}]})$
(1.1)

and

$\sum_{j=0}^{n}T_{j}F_{n-j}=T_{n+2}-F_{n+2}\quad(\mbox{see}\,\,\cite[cite]{[\@@bibref{}{BenBook,Frontczak2018,Frontczak2019}{}{}]}).$
(1.2)

A third example of this nature is the following convolution

$\sum_{j=0}^{n}\mathcal{P}_{j}F_{n-j}=\mathcal{P}_{n}-F_{n}\quad(\mbox{see}\,\,\cite[cite]{[\@@bibref{}{Greubel,Seiffert}{}{}]}).$
(1.3)

A hidden link between (1.2) and (1.3) will be revealed later. At this point we
can clearly see that identities (1.1)–(1.3) look suspiciously similar. Dresden
and Wang [5] ask ”… if there are other general convolution formulas waiting to
be discovered.”? The answer to this questions is ”Definitely Yes!”. Building
on a generating function approach (see also [8, 13, 14, 15, 17]) for Fibonacci
$m$-step numbers we will prove many new convolution identities, recovering
known results as special cases, including the identities presented above.

We note that the generating function approach is not a novel discovery.
However, the referenced articles show that this approach has received a lot of
attention in the recent years. In this context, we mention the latest article
by Gessel and Kar [9] who provide an extensive analysis of (binomial)
convolutions of sequences having rational generating functions.

As for the Fibonacci $m$-step sequences, there seems to be a lack of a general
approach to some of the convolution identities involving these sequences and
other recurrence relations, mentioned at the beginning of the article and many
more. Our response to this is a step forward towards a better understanding of
the general inner structure of these convolutions. To keep things coherent and
focused on a single topic, this article is devoted exclusively to Fibonacci
$m$-step numbers. Thus, identities related to Lucas ($m$-step) sequences (see
for example [3, 4]) will be the subject of another study.

We start with a definition. The $m$-step Fibonacci numbers are defined for all
$m\geq 1$ by

$F_{1}^{(m)}=1,\quad\text{and}\quad F_{n}^{(m)}=0\quad\mbox{for all}\quad
n=-(m-2),\ldots,0$

($F_{0}^{(1)}=0$) and for all $n\geq 2$,

$F_{n}^{(m)}=\sum_{j=1}^{m}F_{n-j}^{(m)}.$

Their arithmetic structure was studied in a recent article by the second
author [12]. The ordinary generating function for $m$-step Fibonacci numbers
is given by

$F^{(m)}(x)=\frac{x}{1-x-x^{2}-\cdots-x^{m}}.$

We introduce the following notation for a selection of particular Fibonacci
$m$-step numbers. Let $F_{n}^{(2)}=F_{n}$, $F_{n}^{(3)}=T_{n}$,
$F_{n}^{(4)}=Q_{n}$, and $F_{n}^{(5)}=P_{n}$ denote the Fibonacci, Tribonacci,
Tetranacci, and Pentanacci numbers, respectively. So, the following can be
stated.

$\displaystyle F_{n}$ $\displaystyle=F_{n-1}+F_{n-2},$ $\displaystyle
F_{0}=0,F_{1}=1,$ $\displaystyle T_{n}$
$\displaystyle=T_{n-1}+T_{n-2}+T_{n-3},$ $\displaystyle
T_{0}=0,T_{1}=T_{2}=1,$ $\displaystyle Q_{n}$
$\displaystyle=Q_{n-1}+Q_{n-2}+Q_{n-3}+Q_{n-4},$ $\displaystyle
Q_{0}=0,Q_{1}=Q_{2}=1,Q_{3}=2,$ $\displaystyle P_{n}$
$\displaystyle=P_{n-1}+P_{n-2}+P_{n-3}+P_{n-4}+P_{n-5},$ $\displaystyle
P_{0}=0,P_{1}=P_{2}=1,P_{3}=2,P_{4}=4,$

Let us denote by $F(x)$, $T(x)$, $Q(x)$, and $P(x)$ the ordinary generating
functions of these sequences, respectively. Hence, we have

$F(x)=F^{(2)}(x),\quad T(x)=F^{(3)}(x),\quad
Q(x)=F^{(4)}(x)\quad\text{and}\quad P(x)=F^{(5)}(x).$

These special candidates will be used later to highlight particular cases of
the results obtain in this paper.

##  2 General convolution identities with Fibonacci $m$-step numbers

This section is for general formulas for Fibonacci $m$-step numbers. We
consider several forms of formulas. First, we investigate convolutions of
$F_{n}^{(m)}$ with $F_{n}^{(m^{\prime})}$ for $m\neq m^{\prime}$. Then we show
a few somewhat curious identities with convolutions exhibiting what we call a
”switch effect”. One such convolution is

$\sum_{j=0}^{n}F_{j}\big{(}T_{n-j}-Q_{n-j}\big{)}=\sum_{j=0}^{n-1}\big{(}F_{j}-T_{j}\big{)}Q_{n-1-j},$

where ”switching” is referred to ”switch the parentheses”. Finally, we
consider convolutions those steps differ by $2$.

###  2.1 Mixed convolution

######  Theorem 2.1.

For all $n\geq m$ we have

$\sum_{j=0}^{n-m}F_{j}^{(m)}F_{n-m-j}^{(m+1)}=F_{n}^{(m+1)}-F_{n}^{(m)}.$
(2.1)

###### Proof.

Notice that

$1-x-x^{2}-\cdots-x^{m}=\frac{x}{F^{(m)}(x)}.$

Subtracting $x^{m+1}$ and rearranging we obtain

$1-x-x^{2}-\cdots-x^{m}-x^{m+1}=\frac{x-x^{m+1}F^{(m)}(x)}{F^{(m)}(x)},$

and thus

$\frac{F^{(m)}(x)}{x-x^{m+1}F^{(m)}(x)}=\frac{F^{(m+1)}(x)}{x}.$

This gives

$F^{(m+1)}(x)-F^{(m)}(x)=x^{m}F^{(m)}(x)F^{(m+1)}(x)$ (2.2)

or

$\displaystyle\sum_{n=0}^{\infty}(F_{n}^{(m+1)}-F_{n}^{(m)})x^{n}$
$\displaystyle=x^{m}\Big{(}\sum_{n=0}^{\infty}F_{n}^{(m)}x^{n}\Big{)}\Big{(}\sum_{n=0}^{\infty}F_{n}^{(m+1)}x^{n}\Big{)}$
$\displaystyle=\sum_{n=0}^{\infty}\sum_{j=0}^{n}F_{j}^{(m)}F_{n-j}^{(m+1)}x^{n+m}$
$\displaystyle=\sum_{n=m}^{\infty}\sum_{j=0}^{n-m}F_{j}^{(m)}F_{n-m-j}^{(m+1)}x^{n}.$

This proves the theorem as for all $n<m$ we have

$\sum_{j=0}^{n-m}F_{j}^{m}F_{n-m-j}^{m+1}=0\quad\mbox{and}\quad
F_{n}^{m+1}-F_{n}^{m}=0.$

∎

As a corollary, we obtain Theorem 2.1 from [6] (identity (2.3) below) and many
more.

######  Corollary 2.2.

Let $n\geq 0$ be an integer. Then

$\displaystyle\sum\limits_{j=0}^{n}F_{j}T_{n-j}$
$\displaystyle=T_{n+2}-F_{n+2},$ (2.3)
$\displaystyle\sum\limits_{j=0}^{n}T_{j}Q_{n-j}$
$\displaystyle=Q_{n+3}-T_{n+3},$ (2.4)
$\displaystyle\sum\limits_{j=0}^{n}Q_{j}P_{n-j}$
$\displaystyle=P_{n+4}-Q_{n+4}.$ (2.5)

###### Proof.

Set $m=2$, $m=3$ and $m=4$, respectively, in Theorem 2.1. ∎

It is interesting that Theorem 2.1 can be generalized further. Let $p\geq 0$
be an integer. Then we have the following result.

######  Theorem 2.3.

Let $p\geq 0$ be an integer. For all $n\geq m$ we have the identity

$\sum_{k=0}^{p-1}\sum_{j=0}^{n-m-k}F_{j}^{(m)}F_{n-m-
k-j}^{(m+p)}=F_{n}^{(m+p)}-F_{n}^{(m)}.$ (2.6)

###### Proof.

This follows from the relation

$\frac{x}{F^{(m+p)}(x)}=\frac{x-\big{(}\sum_{k=1}^{p}x^{m+k}\big{)}F^{(m)}(x)}{F^{(m)}(x)}$

or equivalently

$F^{(m+p)}(x)-F^{(m)}(x)=\Big{(}\sum_{k=1}^{p}x^{m+k-1}\Big{)}F^{(m)}(x)F^{(m+p)}(x).$

The remaining part of the proof is similar to the proof of Theorem 2.1. ∎

When $p=1$ then we get Theorem 2.1. When $p=2$ then we get the following.

######  Corollary 2.4.

For all $n\geq m$ we have the following identity

$\sum_{j=0}^{n-m-1}F_{j}^{(m)}(F_{n-m-j}^{(m+2)}+F_{n-m-j-1}^{(m+2)})=F_{n}^{(m+2)}-F_{n}^{(m)}.$

Corollary 2.4 allows to obtain many different identities involving
convolutions of two sequences. We show the following two examples.

######  Example 2.5.

Set $m=1$ in Corollary 2.4 to get

$T_{n}-1=\sum_{j=0}^{n-2}T_{j}+\sum_{j=0}^{n-3}T_{j}=2\sum_{j=0}^{n-3}T_{j}+T_{n-2}.$

This gives (after replacing $n$ by $n+3$)

$\sum_{j=0}^{n}T_{j}=\frac{1}{2}(T_{n+3}-T_{n+1}-1)=\frac{1}{2}(T_{n+2}+T_{n}-1),$
(2.7)

which is a well-known partial sum formula (see for example [1, 7, 11]).

######  Example 2.6.

For any $n\geq 0$ we have

$\sum_{j=0}^{n}F_{j+2}Q_{n-j}=Q_{n+3}-F_{n+3}.$ (2.8)

This follows from setting $m=2$ in Corollary 2.4 and calculating

$\displaystyle Q_{n}-F_{n}$
$\displaystyle=\sum_{j=0}^{n-3}F_{j}(Q_{n-2-j}+Q_{n-3-j})$
$\displaystyle=\sum_{j=1}^{n-3}F_{j}Q_{n-2-j}+\sum_{j=0}^{n-3}F_{j}Q_{n-3-j}$
$\displaystyle=\sum_{j=0}^{n-4}F_{j+1}Q_{n-3-j}+\sum_{j=0}^{n-3}F_{j}Q_{n-3-j}$
$\displaystyle=\sum_{j=0}^{n-4}F_{j+2}Q_{n-3-j}$
$\displaystyle=\sum_{j=0}^{n-3}F_{j+2}Q_{n-3-j}.$

Finally, we point out that when $m=1$ in Theorem 2.3 then we get the partial
sum formula for Fibonacci $m$-step numbers.

######  Corollary 2.7.

For any $n\geq p$,

$F^{(p+1)}_{n}-1=\sum_{k=0}^{p-1}\sum\limits_{j=0}^{n-2-k}F_{j}^{(p+1)}.$
(2.9)

In fact, rewriting (2.9) to a slightly more convenient way we can get the
following general identity.

######  Theorem 2.8 (Partial sum formula for Fibonacci $m$-step numbers).

For any $m\geq 2$ we have

$\sum\limits_{k=0}^{n}F_{k}^{(m)}=\frac{1}{m-1}\left(F_{n+m}^{(m)}-\sum_{k=1}^{m-2}kF_{n+k}^{(m)}-1\right).$
(2.10)

###### Proof.

Substitute $p+1\to m$ in (2.9) to get

$\displaystyle F_{n}^{(m)}-1$
$\displaystyle=\sum\limits_{k=0}^{m-2}\sum\limits_{j=0}^{n-2-k}F_{k}^{(m)}$
$\displaystyle=(m-1)\sum\limits_{k=0}^{n-m}F_{k}^{(m)}+\sum\limits_{k=1}^{m-2}kF_{n-m+k}^{(m)}.$

Substituting $n\to n+m$ and rearranging the terms leads to the desired
formula. ∎

We note that Theorem 2.8 is not new. For instance, in [18] the partial sum
formula is proved using induction but the proof is two pages long. A shorter
proof of an equivalent version of the partial sum formula is given by Dresden
and Wang in [3]. Second author has recently shown a proof without words of the
identity in [11]. Here, we obtained it as a corollary of a more general
result.

We have seen from (2.8) that it is possible to derive a formula for the
convolution of Fibonacci and Tetranacci numbers. In the following, we gather
all closed formulas for mixed convolutions of two Fibonacci $m$-step numbers
with $2\leq m\leq 5$, that are not present in Corollary 2.2.

######  Corollary 2.9.

Let $n\geq 0$ be an integer. Then

$\displaystyle\sum\limits_{j=0}^{n}F_{j}Q_{n-j}$
$\displaystyle=Q_{n+1}+Q_{n-1}-F_{n+1},$ (2.11)
$\displaystyle\sum\limits_{j=0}^{n}F_{j}P_{n-j}$
$\displaystyle=\frac{1}{2}\big{(}P_{n+2}+P_{n-1}-F_{n+2}\big{)},$ (2.12)
$\displaystyle\sum\limits_{j=0}^{n}P_{j}T_{n-j}$
$\displaystyle=\frac{1}{2}\big{(}P_{n+3}+P_{n+1}+P_{n-1}-T_{n+3}-T_{n+1}\big{)}.$
(2.13)

###### Proof.

Identity (2.11) follows from (2.8) after fixing the summation range. To show
(2.12) we use Theorem 2.3 with $m=2$ and $p=3$. Then

$\displaystyle P_{n}-F_{n}$
$\displaystyle=\sum\limits_{j=0}^{n-4}P_{j}F_{n-4-j}+\sum\limits_{j=0}^{n-3}P_{j}F_{n-3-j}+\sum\limits_{j=0}^{n-2}P_{j}F_{n-2-j}$
$\displaystyle=2\sum\limits_{j=0}^{n-4}P_{j}F_{n-2-j}+P_{n-3}.$

It follows that

$\frac{1}{2}\big{(}P_{n+1}-P_{n-2}-F_{n+1}\big{)}=\sum\limits_{j=0}^{n-3}P_{j}F_{n-1-j}=\sum\limits_{j=0}^{n-1}P_{j}F_{n-1-j}-P_{n-2}$

and this implies the identity.

To show (2.13) we use Theorem 2.3 with $p=2$ and $m=3$ and we get

$P_{n}-T_{n}=\sum\limits_{j=0}^{n-4}P_{j}(T_{n-3-j}+T_{n-4-j})$ (2.14)

Then, using (2.14) twice, second time with $n+2$ in place of $n$, and adding
up we get

$P_{n+2}+P_{n}-T_{n+2}-T_{n}=2\sum\limits_{j=0}^{n-4}P_{j}T_{n-1-j}+2P_{n-3}+P_{n-2}.$

or, equivalently

$P_{n+2}+P_{n}-P_{n-2}-2P_{n-3}-T_{n+2}-T_{n}=2\sum\limits_{j=0}^{n-4}P_{j}T_{n-1-j}.$
(2.15)

Notice that

$\sum\limits_{j=0}^{n-4}P_{j}T_{n-1-j}=\sum\limits_{j=0}^{n-1}P_{j}T_{n-1-j}-P_{n-2}-P_{n-3},$

thus substituting that to (2.15) and simplifying we get

$\sum\limits_{j=0}^{n-1}P_{j}T_{n-1-j}=\frac{1}{2}\big{(}P_{n+2}+P_{n}+P_{n-2}-T_{n+2}-T_{n}\big{)}.$

This implies (2.13). ∎

The method used to obtain identities (2.12) and (2.13) will be explored in the
general case in Section 4.

###  2.2 Mixed convolutions with the ”switch effect”

In the following theorem we show a convolution-type formula where by switching
a place of parentheses and the minus sign with minor adjustment of indices, we
obtain the equality. It is also important to mention that the formula has an
interesting connection to convolution sums of three sequences. The connection
will be established in the next two sections.

######  Theorem 2.10.

For all $n\geq 1$ and $m\geq 3$ we have

$\sum_{j=0}^{n}F_{j}^{(m-2)}\big{(}F_{n-j}^{(m)}-F_{n-j}^{(m-1)}\big{)}=\sum_{j=0}^{n-1}F_{j}^{(m)}\big{(}F_{n-1-j}^{(m-1)}-F_{n-1-j}^{(m-2)}\big{)}.$

###### Proof.

Since

$\displaystyle 1-x-x^{2}-\cdots-x^{m}$
$\displaystyle=1-x-x^{2}-\cdots-x^{m-1}-x(1-x-x^{2}-\cdots-x^{m-2})$
$\displaystyle\qquad+x(1-x-x^{2}-\cdots-x^{m-1})$

it follows that

$\frac{1}{F^{(m)}(x)}=\frac{1}{F^{(m-1)}(x)}-\frac{x}{F^{(m-2)}(x)}+\frac{x}{F^{(m-1)}(x)}$

or equivalently

$F^{(m-2)}(x)\Big{(}F^{(m)}(x)-F^{(m-1)}(x)\Big{)}=xF^{(m)}(x)\Big{(}F^{(m-1)}(x)-F^{(m-2)}(x)\Big{)}.$

Passing to the power series and comparing coefficients of $x^{n}$ we obtain
the identity. ∎

The next corollary is again a rediscovery of Theorem 2.1 in [6].

######  Corollary 2.11.

For all $n\geq 1$ we have

$T_{n+1}-F_{n+1}=\sum_{j=0}^{n-1}T_{j}F_{n-1-j}.$

###### Proof.

Set $m=3$ in Theorem 2.10 and simplify using

$F_{n}^{(1)}=\begin{cases}0,&\text{\rm if $n=0$};\\\ 1,&\text{\rm if $n\geq
1$};\end{cases}$

equation (2.7) and

$\sum_{j=0}^{n}F_{j}=F_{n+2}-1.$

∎

######  Corollary 2.12.

For all $n\geq 1$ we have

$\sum_{j=0}^{n}F_{j}\big{(}T_{n-j}-Q_{n-j}\big{)}=\sum_{j=0}^{n-1}\big{(}F_{j}-T_{j}\big{)}Q_{n-1-j}$

and

$\sum\limits_{j=0}^{n}T_{j}\big{(}Q_{n-j}-P_{n-j}\big{)}=\sum\limits_{j=0}^{n-1}\big{(}T_{j}-Q_{j}\big{)}P_{n-1-j}.$
(2.16)

###### Proof.

Set $m=4$ and $m=5$, respectively, in Theorem 2.10 and simplify. ∎

It is also possible to obtain the ”switch” effect in the following sense.

######  Theorem 2.13.

For all $m,n\geq 2$ we have

$\sum_{j=0}^{n}F_{j}^{(m-1)}F_{n-j}^{(m)}=\sum_{j=0}^{n-2}2^{j}\big{(}F_{n-1-j}^{(m-1)}-F_{n-2-j}^{(m)}\big{)}.$

###### Proof.

The relation

$1-x-x^{2}-\cdots-x^{m}=1-2x+x(1-x-x^{2}-\cdots-x^{m-1})$

translates to

$\frac{1}{F^{(m)}(x)}=\frac{1}{P_{2}(x)}+\frac{x}{F^{(m-1)}(x)}$

with

$P_{2}(x)=\frac{x}{1-2x}=x\sum_{n=0}^{\infty}2^{n}x^{n}.$

This gives

$P_{2}(x)F^{(m-1)}(x)=F^{(m)}(x)F^{(m-1)}(x)+xP_{2}(x)F^{(m)}(x).$

Passing to the power series and comparing coefficients of $x^{n}$ we obtain
the identity. ∎

######  Corollary 2.14.

For all $n\geq 0$

$\sum_{j=0}^{n}2^{j}F_{n-j}=2^{n+1}-F_{n+3}.$ (2.17)

###### Proof.

Set $m=2$ in Theorem 2.13, simplify, and replace $n$ by $n+2$. ∎

######  Corollary 2.15.

For all $n\geq 0$

$\sum_{j=0}^{n}2^{j}\big{(}F_{n+1-j}-T_{n-j}\big{)}=T_{n+4}-F_{n+4}.$

###### Proof.

Set $m=3$ in Theorem 2.13, use Corollary 2.11, and replace $n$ by $n+2$. ∎

######  Corollary 2.16.

For all $n\geq 0$

$\sum_{j=0}^{n}2^{j}T_{n-j}=2^{n+2}-T_{n+4}.$ (2.18)

###### Proof.

Combine Corollary 2.14 with Corollary 2.15. ∎

######  Corollary 2.17.

For all $n\geq 0$

$\sum_{j=0}^{n+2}T_{j}Q_{n-j}=\sum_{j=0}^{n}2^{j}\big{(}T_{n+1-j}-Q_{n-j}\big{)}.$

###### Proof.

Set $m=4$ in Theorem 2.13, and replace $n$ by $n+2$. ∎

This is a general formula of (2.17) and (2.18) for Fibonacci $m$-step numbers.

######  Theorem 2.18.

For all $n\geq 0$ and $m\geq 1$

$\sum_{j=0}^{n}2^{j}F_{n-j}^{(m)}=2^{n-1+m}-F_{n+1+m}^{(m)}.$

###### Proof.

We use induction on $m$. The statement is true for $m=1$ and $m=2$. Assume the
identity is true for a fixed $m-1>2$ (and all $n$). Replacing $n$ by $n+2$ in
Theorem 2.13 yields

$\sum_{j=0}^{n}2^{j}F_{n-j}^{(m)}=\sum_{j=0}^{n}2^{j}F_{n+1-j}^{(m-1)}-\sum_{j=0}^{n+2}F_{j}^{(m-1)}F_{n+2-j}^{(m)}.$

But from Theorem 2.1 upon making the replacements $n\mapsto n+2+m$ and
$m\mapsto m-1$ we get

$\sum_{j=0}^{n+2}F_{j}^{(m-1)}F_{n+2-j}^{(m)}=F_{n+1+m}^{(m)}-F_{n+1+m}^{(m-1)}.$

From here using the inductive hypothesis we can calculate

$\displaystyle\sum_{j=0}^{n}2^{j}F_{n-j}^{(m)}$
$\displaystyle=\sum_{j=0}^{n}2^{j}F_{n+1-j}^{(m-1)}-F_{n+1+m}^{(m)}+F_{n+1+m}^{(m-1)}$
$\displaystyle=\sum_{j=0}^{n+1}2^{j}F_{n+1-j}^{(m-1)}-F_{n+1+m}^{(m)}+F_{n+1+m}^{(m-1)}$
$\displaystyle=2^{n+1-1+(m-1)}-F_{n+1+1+(m-1)}^{(m-1)}-F_{n+1+m}^{(m)}+F_{n+1+m}^{(m-1)}$
$\displaystyle=2^{n-1+m}-F_{n+1+m}^{(m)}.$

∎

###  2.3 Three other general identities

The next convolution identities involve an alternating sum. They all share the
same structure, we convolve two sequences with indices $m$ that differ by $2$.

######  Theorem 2.19.

For any $n\geq 2m-1$ we have

$\sum\limits_{j=0}^{n-1}(-1)^{j}\big{(}F_{j}^{(2m)}-F_{j}^{(2m-2)}\big{)}=(-1)^{n+1}\sum\limits_{j=0}^{n-2m+1}F_{j}^{(2m)}F_{n-2m+1-j}^{(2m-2)}.$
(2.19)

###### Proof.

As

$1+x-x^{2}+x^{3}\mp\cdots-x^{2(m-1)}+x^{2m-1}-x^{2m}=1+x-x^{2}+x^{3}\mp\cdots-x^{2(m-1)}+x^{2m-1}(1-x),$

we obtain the functional equation

$\frac{1}{F^{(2m)}(-x)}=\frac{1}{F^{(2m-2)}(-x)}-x^{2m-1}\frac{1}{F^{(1)}(x)}$

or equivalently

$\Big{(}F^{(2m)}(-x)-F^{(2m-2)}(-x)\Big{)}F^{(1)}(x)=x^{2m-1}F^{(2m)}(-x)F^{(2m-2)}(-x).$

When simplifying we again use

$F_{n}^{(1)}=\begin{cases}0,&\text{\rm if $n=0$};\\\ 1,&\text{\rm if $n\geq
1$}.\end{cases}$

∎

As a corollary, we can show a different proof of identity (2.11).

######  Corollary 2.20.

For any $n\geq 0$ we have

$\sum_{j=0}^{n}F_{j}Q_{n-j}=Q_{n+1}+Q_{n-1}-F_{n+1}.$ (2.20)

###### Proof.

Set $m=2$ in Theorem 2.19 and use

$\sum_{j=0}^{n}(-1)^{j}F_{j}=(-1)^{n}F_{n-1}-1$

as well as [20]

$\sum_{j=0}^{n}(-1)^{j}Q_{j}=(-1)^{n}(Q_{n+3}-2Q_{n+2}+Q_{n+1}-Q_{n})-1.$

These results produce

$\sum_{j=0}^{n-3}F_{j}Q_{n-3-j}=Q_{n+2}-2Q_{n+1}+Q_{n}-Q_{n-1}-F_{n-2},$

and the statement follows upon replacing $n$ by $n+3$ and simplifying. ∎

The companion result for Theorem 2.19 for odd $m$ is stated next.

######  Theorem 2.21.

For any $n\geq 2m$ we have

$\sum\limits_{j=0}^{n-1}(-1)^{j+1}\big{(}F_{j}^{(2m+1)}-F_{j}^{(2m-1)}\big{)}=(-1)^{n}\sum\limits_{j=0}^{n-2m}F_{j}^{(2m+1)}F_{n-2m-j}^{(2m-1)}.$
(2.21)

###### Proof.

This result follows from the functional equation

$\frac{1}{F^{(2m+1)}(-x)}=\frac{1}{F^{(2m-1)}(-x)}+x^{2m}\frac{1}{F^{(1)}(x)}$

or equivalently

$\Big{(}F^{(2m-1)}(-x)-F^{(2m+1)}(-x)\Big{)}F^{(1)}(x)=x^{2m}F^{(2m-1)}(-x)F^{(2m+1)}(-x).$

∎

We proceed with two corollaries. The first identity is known (see for instance
Equation (18) in [7]), the second identity is a rediscovery of (2.13) with a
slightly different (but equivalent) right hand side.

######  Corollary 2.22.

For any $n\geq 0$ we have

$\sum_{j=0}^{n}(-1)^{j}T_{j}=\frac{1}{2}((-1)^{n}(T_{n+1}-T_{n-1})-1).$

###### Proof.

Set $m=1$ in Theorem 2.21 and simplify. ∎

######  Corollary 2.23.

For any $n\geq 0$ we have

$\sum_{j=0}^{n}P_{j}T_{n-j}=\frac{1}{2}(-P_{n+7}+2P_{n+6}-P_{n+5}+2P_{n+4}+P_{n+3}-T_{n+4}+T_{n+2}).$

###### Proof.

Set $m=2$ in Theorem 2.21 and simplify while making use of (see [21])

$\sum_{j=0}^{n}(-1)^{j}P_{j}=\frac{1}{2}((-1)^{n}(-P_{n+4}+2P_{n+3}-P_{n+2}+2P_{n+1}+P_{n})-1).$

∎

We conclude this section with the remark that identities involving Fibonacci
$m$-step numbers and other important number sequences can be obtained fairly
easily using the generating function approach. We give an example involving
Jacobsthal numbers, which is related to identity (1.1).

######  Theorem 2.24.

Let $J_{n}$ be the Jacobsthal numbers, i.e., $J_{0}=0,J_{1}=1$, and
$J_{n+2}=J_{n+1}+2J_{n}$. Then, for $m\geq 3$ and any $n\geq 2$ we have

$\sum\limits_{j=0}^{n}F_{j}^{(m)}F_{n-j}^{(m-2)}=J_{n-1}+\sum\limits_{j=0}^{n-2}J_{j}\big{(}F_{n-j}^{(m-2)}-F_{n-2-j}^{(m)}\big{)}.$
(2.22)

###### Proof.

We have

$1-x-x^{2}-x^{3}-\cdots-x^{m}=1-x-2x^{2}+x^{2}(1-x-x^{2}-\cdots-x^{m-2}),$

which implies

$\frac{x}{F^{(m)}(x)}=\frac{x}{J(x)}+\frac{x^{3}}{F^{(m-2)}(x)},$

where $J(x)=\frac{x}{1-x-2x^{2}}$ is the ordinary generating function for
Jacobsthal numbers. ∎

######  Corollary 2.25.

For any $n\geq 0$ we have

$\sum_{j=0}^{n}J_{j}T_{n-j}=J_{n+1}+\frac{1}{2}(J_{n+2}-T_{n+3}-T_{n+1}).$

###### Proof.

Use the previous theorem with $m=3$ in conjunction with (2.7) and

$\sum_{j=0}^{n}J_{j}=\frac{1}{2}(J_{n+2}-1).$

∎

##  3 Convolutions of multiple sequences

The main goal of this section is to derive convolution identities of three and
more Fibonacci $m$-step sequences. For the convenience, we introduce the
following notation:

$K(\ell,b)=\\{(k_{1},\ldots,k_{\ell})\in\mathbb{Z}^{\ell}_{\geq 0}\colon
k_{1}+\cdots+k_{\ell}=b\\}.$

Before providing an analysis of the $m$-step sequences, we present a Pell
sequence result. Then we consider a general convolution of three Fibonacci
$m$-step sequences and derive all mixed convolutions with $2\leq m\leq 5$.
Finally, we delve with the convolution of four sequences.

###  3.1 A Pell-Fibonacci relation

######  Theorem 3.1.

Let $\mathcal{P}_{n}$ be the Pell numbers, i.e.,
$\mathcal{P}_{0}=0,\mathcal{P}_{1}=1$, and
$\mathcal{P}_{n+2}=2\mathcal{P}_{n+1}+\mathcal{P}_{n}$. Then, for all $n\geq
1$ and $m\geq 2$ we have the following Pell-Fibonacci-$m$-step-relation:

$\sum_{K(3,n-1)}\mathcal{P}_{k_{1}}F_{k_{2}}^{(m-1)}F_{k_{3}}^{(m)}=\sum_{j=0}^{n}F_{j}^{(m-1)}\left(\mathcal{P}_{n-j}-F_{n-j}^{(m)}\right)-\sum_{j=0}^{n-1}\mathcal{P}_{j}F_{n-1-j}^{(m)}.$

###### Proof.

Let $\mathcal{P}(x)=\frac{x}{1-2x-x^{2}}$ denote the generating function for
the Pell numbers. From

$1-x-x^{2}-x^{3}-\cdots-x^{m}=1-2x-x^{2}+x(1-x-x^{2}-\cdots-x^{m-1})+x^{2},$

we get

$\frac{1}{F^{(m)}(x)}=\frac{1}{\mathcal{P}(x)}+x\frac{F^{(m-1)}(x)+1}{F^{(m-1)}(x)},$

or

$x\mathcal{P}(x)F^{(m-1)}(x)F^{(m)}(x)=F^{(m-1)}(x)\left(\mathcal{P}(x)-F^{(m)}(x)\right)-x\mathcal{P}(x)F^{(m)}(x).$

The result follows upon passing to power series and comparing the coefficients
of $x^{n}$. ∎

As a first corollary, we rediscover identity (1.3).

######  Corollary 3.2.

For any $n\geq 0$ we have

$\sum_{j=0}^{n}\mathcal{P}_{j}F_{n-j}=\mathcal{P}_{n}-F_{n}.$

###### Proof.

Set $m=2$ in Theorem 3.1 and simplify. ∎

######  Corollary 3.3.

For any $n\geq 0$ we have

$\sum_{K(3,n)}\mathcal{P}_{k_{1}}F_{k_{2}}T_{k_{3}}=\mathcal{P}_{n+1}+F_{n+2}-T_{n+3}-\sum_{j=0}^{n}\mathcal{P}_{j}T_{n-j}.$

or

$\sum_{K(3,n)}\mathcal{P}_{k_{1}}F_{k_{2}}T_{k_{3}}=\frac{1}{2}\left(\mathcal{P}_{n+1}-T_{n+3}-T_{n+2}\right)+F_{n+2}.$
(3.1)

###### Proof.

Set $m=3$ in Theorem 3.1, use (1.2) and (1.3), and simplify. To get (3.1), let
$R(x)=\frac{x}{1-x^{2}}$ and notice that

$\frac{1}{T(x)}=\frac{1}{\mathcal{P}(x)}+x\cdot\frac{1}{R(x)},$

which is equivalent to

$x\mathcal{P}(x)T(x)=R(x)(\mathcal{P}(x)-T(x)).$

Using

$R(x)=\frac{x}{2}\left(\frac{1}{1-x}+\frac{1}{1+x}\right)$

and passing to power series, we get

$\sum_{j=0}^{n}\mathcal{P}_{j}T_{n-j}=\frac{1}{2}\sum_{j=0}^{n}\mathcal{P}_{j}+\frac{1}{2}\sum_{j=0}^{n}(-1)^{n-j}\mathcal{P}_{j}-\frac{1}{2}\sum_{j=0}^{n}T_{j}-\frac{1}{2}\sum_{j=0}^{n}(-1)^{n-j}T_{j}.$

Recall the well-known identities for Pell numbers (see for example [2]):

$\sum_{j=0}^{n}\mathcal{P}_{j}=\frac{1}{2}\left(\mathcal{P}_{n+1}+\mathcal{P}_{n}-1\right)\quad\mbox{and}\quad\sum_{j=0}^{n}(-1)^{j}\mathcal{P}_{j}=\frac{1}{2}\left((-1)^{n}(\mathcal{P}_{n+1}-\mathcal{P}_{n})-1\right).$

These relations show that

$\sum_{j=0}^{n}\mathcal{P}_{j}T_{n-j}=\frac{1}{2}\left(\mathcal{P}_{n+1}-T_{n+1}-T_{n}\right)$

and (3.1) follows. ∎

From

$\mathcal{P}(x)F(x)=\mathcal{P}(x)-F(x),$

we get (by induction) for all $r\geq 1$

$\mathcal{P}(x)F^{r}(x)=\mathcal{P}(x)-\sum_{s=1}^{r}F^{s}(x),$

or equivalently

$\sum_{K(r+1,n)}\mathcal{P}_{k_{1}}F_{k_{2}}F_{k_{3}}\cdots
F_{k_{r+1}}=\mathcal{P}_{n}-\sum_{s=1}^{r}\sum_{K(s,n)}{F}_{k_{1}}\cdots
F_{k_{s}}.$

Special cases of this convolution include (see Zhang’s paper [22] for the
Fibonacci convolutions)

$\sum_{K(3,n)}\mathcal{P}_{k_{1}}F_{k_{2}}F_{k_{3}}=\mathcal{P}_{n}-F_{n}-\frac{1}{5}\left((n-1)F_{n}+2nF_{n-1}\right)\quad(n\geq
1)$

as well as

$\displaystyle\sum_{K(4,n)}\mathcal{P}_{k_{1}}F_{k_{2}}F_{k_{3}}F_{k_{4}}$
$\displaystyle=\mathcal{P}_{n}-F_{n}-\frac{1}{5}\left((n-1)F_{n}+2nF_{n-1}\right)$
$\displaystyle\quad-\frac{1}{50}\left((5n^{2}-9n-2)F_{n-1}+(5n^{2}-3n-2)F_{n-2}\right)\quad(n\geq
2).$ (3.2)

###  3.2 Convolution of three Fibonacci $m$-step sequences

The next theorem is the main observation and will be used as a basis for some
observations what will follow.

######  Theorem 3.4.

Let $m,p,q\geq 1$ be integers. Then the following functional identity holds
true:

$\displaystyle x^{2m+p}(1-x^{p})(1-x^{q})F^{(m)}(x)F^{(m+p)}(x)F^{(m+p+q)}(x)$
$\displaystyle=x^{m}(1-x)(1-x^{p})F^{(m)}(x)F^{(m+p+q)}(x)-(1-x)^{2}(F^{(m+p)}(x)-F^{(m)}(x)).$
(3.3)

In particular,

$x^{2m+1}F^{(m)}(x)F^{(m+1)}(x)F^{(m+2)}(x)=x^{m}F^{(m)}(x)F^{(m+2)}(x)-F^{(m+1)}(x)+F^{(m)}(x).$
(3.4)

###### Proof.

We utilize the proof of Theorem 2.3 and work with the identity:

$F^{(a+b)}(x)-F^{(a)}(x)=\left(\sum\limits_{k=0}^{b-1}x^{a+k}\right)F^{(a)}(x)F^{(a+b)}(x).$
(3.5)

Applying it twice we get:

$\displaystyle F^{(m+p)}(x)$
$\displaystyle=\frac{F^{(m)}(x)}{1-\left(\sum\limits_{k=0}^{p-1}x^{m+k}\right)F^{(m)}(x)},$
$\displaystyle F^{(m+p+q)}(x)-F^{(m+p)}(x)$
$\displaystyle=\left(\sum\limits_{k=0}^{q-1}x^{m+p+k}\right)F^{(m+p)}(x)F^{(m+p+q)}(x).$
(3.6)

Hence, we obtain

$F^{(m+p+q)}(x)-\frac{F^{(m)}(x)}{1-\left(\sum\limits_{k=0}^{p-1}x^{m+k}\right)F^{(m)}(x)}=\left(\sum\limits_{k=0}^{q-1}x^{m+p+k}\right)F^{(m+p)}(x)F^{(m+p+q)}(x),$

or, after rearranging,

$\displaystyle\left(\sum\limits_{k=0}^{p-1}x^{m+k}\right)\left(\sum\limits_{k=0}^{q-1}x^{m+p+k}\right)F^{(m)}(x)F^{(m+p)}(x)F^{(m+p+q)}(x)$
$\displaystyle\qquad=\left(\sum\limits_{k=0}^{q-1}x^{m+p+k}\right)F^{(m+p)}(x)F^{(m+p+q)}(x)$
$\displaystyle\qquad\qquad+\left(\sum\limits_{k=0}^{p-1}x^{m+k}\right)F^{(m)}(x)F^{(m+p+q)}(x)-F^{(m+p+q)}(x)+F^{(m)}(x).$

Simplifying using (3.6) yields

$\displaystyle\left(\sum\limits_{k=0}^{p-1}x^{m+k}\right)\left(\sum\limits_{k=0}^{q-1}x^{m+p+k}\right)F^{(m)}(x)F^{(m+p)}(x)F^{(m+p+q)}(x)$
$\displaystyle\qquad=\left(\sum\limits_{k=0}^{p-1}x^{m+k}\right)F^{(m)}(x)F^{(m+p+q)}(x)-F^{(m+p)}(x)+F^{(m)}(x).$

Finally, from the geometric series

$\left(\sum_{k=0}^{p-1}x^{m+k}\right)\left(\sum_{k=0}^{q-1}x^{m+p+k}\right)=x^{2m+p}\frac{(1-x^{p})(1-x^{q})}{(1-x)^{2}},\quad\sum_{k=0}^{p-1}x^{m+k}=x^{m}\frac{1-x^{p}}{1-x}$

and the functional equation follows. The particular case belongs to $p=q=1$. ∎

The special case of Theorem 3.4 leads to yet another identity involving triple
convolutions. Namely, we have the following.

######  Theorem 3.5.

For any $p\geq 1$ and any $m\geq 1$ we have

$x^{m}F^{(m)}(x)F^{(m+1)}(x)F^{(p)}(x)=F^{(p)}(x)F^{(m+1)}(x)-F^{(p)}(x)F^{(m)}(x).$

In particular,

$\sum\limits_{K(3,n-m)}F^{(m)}_{k_{1}}F^{(m+1)}_{k_{2}}F^{(p)}_{k_{3}}=\sum\limits_{j=0}^{n}F^{(p)}_{j}F^{(m+1)}_{n-j}-\sum\limits_{j=0}^{n}F^{(p)}_{j}F^{(m)}_{n-j}.$

###### Proof.

Use identity (3.5). ∎

We apply Theorem 3.4 with $m=2$ and $p=q=1$. This choice results in several
different functional equations, the penultimate one coming directly from the
theorem.

$\displaystyle x^{5}F(x)T(x)Q(x)$ $\displaystyle=x^{2}F(x)x^{3}T(x)Q(x)$
$\displaystyle=x^{2}F(x)(Q(x)-T(x))$ $\displaystyle=x^{2}F(x)Q(x)-T(x)+F(x)$
$\displaystyle=x^{2}F(x)Q(x)+x^{3}T(x)Q(x)-Q(x)+F(x).$

######  Theorem 3.6.

We have for each $n\geq 0$,

$\sum\limits_{K(3,n)}F_{k_{1}}T_{k_{2}}Q_{k_{3}}=Q_{n+4}+Q_{n+2}-T_{n+5}+F_{n+3}.$
(3.7)

###### Proof.

Work with

$x^{5}F(x)T(x)Q(x)=x^{2}F(x)Q(x)-T(x)+F(x).$

When passing to the power series and comparing the coefficients of $x^{n}$ use
the identity (2.11) and simplify. ∎

Working with $m=3$ and $p=q=1$ in Theorem 3.4 results in the functional
equations, where again, the penultimate one coming directly from the theorem.

$\displaystyle x^{7}T(x)Q(x)P(x)$ $\displaystyle=x^{3}T(x)x^{4}Q(x)P(x)$
$\displaystyle=x^{3}T(x)(P(x)-Q(x))$ $\displaystyle=x^{3}T(x)P(x)-Q(x)+T(x)$
$\displaystyle=x^{3}T(x)P(x)+x^{4}P(x)Q(x)-P(x)+T(x),$

and this gives the next convolution.

######  Theorem 3.7.

We have

$\sum_{K(3,n-7)}T_{k_{1}}Q_{k_{2}}P_{k_{3}}=\frac{1}{2}(P_{n}+P_{n-2}+P_{n-4}+T_{n}-T_{n-2})-Q_{n}$

valid for each $n\geq 7$.

###### Proof.

Using the identity

$x^{7}T(x)Q(x)P(x)=x^{3}T(x)P(x)-Q(x)+T(x)$

in conjunction with (2.13) we obtain, after minor simplification, the desired
result. ∎

In the following two results we find convolution sums of the remaining triples
with $2\leq m\leq 5$ using Theorem 3.5.

######  Theorem 3.8.

We have the following identity:

$\sum\limits_{K(3,n-5)}F_{k_{1}}Q_{k_{2}}P_{k_{3}}=-Q_{n}-Q_{n-2}+\frac{1}{2}\big{(}P_{n+1}+P_{n-2}-F_{n+1}\big{)}+F_{n}.$
(3.8)

###### Proof.

Use Theorem 3.5 with $m=4$ and $p=2$, pass to power series, apply (2.11) and
(2.12), and simplify. ∎

######  Theorem 3.9.

We have the following identity:

$\sum\limits_{K(3,n-5)}F_{k_{1}}T_{k_{2}}P_{k_{3}}=\frac{1}{2}\big{(}P_{n}-P_{n-1}+P_{n-2}-T_{n}-T_{n-2}+F_{n-1}\big{)}.$

###### Proof.

Use Theorem 3.5 with $m=2$ and $p=5$, pass to power series, apply (2.12) and
(2.13), and simplify. ∎

In the proofs of Theorem 3.8 and Theorem 3.9 we used different approach.
Instead of utilizing Theorem 3.4, we applied Theorem 3.5. This is necessary
here, since, otherwise, the obtained formula would produce an identity with
multiple triple convolution terms instead of one.

We finish this part of the article by showing that one can use a completely
different approach to prove the results in this section. Namely, we can deduce
the fundamental functional equations from other relations of the respective
generating functions.

######  Remark 3.10.

Notice that one can prove Theorem 3.8 and Theorem 3.9 using other
transformations of generating functions. For the first theorem, we can use the
following identities:

$P(x)-Q(x)=x^{4}P(x)Q(x)\qquad\text{and}\qquad
Q(x)=\frac{F(x)}{1+x^{2}-xF(x)},$

where the formula for $Q$ is substituted to the right-hand-side presence of
$Q$ in the first identity. For the second theorem, we start from the identity

$\frac{x}{P(x)}=\frac{x}{F(x)}+\frac{x^{4}}{F(x)}-2x^{3}$

to obtain

$P(x)-F(x)=2x^{2}F(x)P(x)-x^{3}P(x).$ (3.9)

Then we substitute $F(x)=\frac{T(x)}{1+x^{2}T(x)}$ and obtain a functional
equation.

###  3.3 Convolutions of four Fibonacci $m$-step sequences

The functional equations from Theorem 3.4 (as well as any other functional
equations derived in this paper) can be multiplied by arbitrary $F^{(r)}(x)$
or even by any product of the form

$\prod_{u=1}^{N}\left(F^{(r_{u})}(x)\right)^{s_{u}},\qquad
s_{1},\ldots,s_{N},r_{1},\ldots,r_{N}\,\,\mbox{positive integers}.$

In particular cases these functional equations can be used for a different
proof of Theorems 3.6–3.9. For example, using (3.4) we immediately get the
following.

######  Theorem 3.11.

For any $m\geq 1$ we have

$\displaystyle x^{2m+1}F^{(m)}(x)F^{(m+1)}(x)F^{(m+2)}(x)F^{(m+3)}(x)$
$\displaystyle\qquad\qquad=x^{m}F^{(m)}(x)F^{(m+2)}(x)F^{(m+3)}(x)-F^{(m+1)}(x)F^{(m+3)}(x)+F^{(m)}(x)F^{(m+3)}(x).$
(3.10)

This theorem implies a particular closed formula for the convolution of
Fibonacci, Tribonacci, Tetranacci and Pentanacci numbers.

######  Theorem 3.12.

We have

$\sum\limits_{K(4,n-5)}F_{k_{1}}T_{k_{2}}Q_{k_{3}}P_{k_{4}}=\frac{1}{2}\big{(}P_{n+4}-P_{n+3}+P_{n+2}+T_{n+3}+T_{n+1}-F_{n}\big{)}-Q_{n+3}-Q_{n+1}.$

###### Proof.

Set $m=2$ in Equation (3.10) to get

$x^{5}F(x)T(x)Q(x)P(x)=x^{2}F(x)Q(x)P(x)-T(x)P(x)+F(x)P(x).$

Pass to power series, use (3.8), (2.13) and (2.12) and simplify. ∎

######  Remark 3.13.

We note that identity (3.10) can be (after multiplication by $x^{2}$) further
simplified using Theorem 3.5 so that the right-hand-side contains convolution
sums of two sequences.

Other interesting identities involving Fibonacci, Tribonacci, Tetranacci and
Pentanacci numbers can be derived. To be particular, the following hold true.

$\displaystyle x^{4}F(x)T(x)Q(x)P(x)$ $\displaystyle=F(x)T(x)x^{4}Q(x)P(x)$
$\displaystyle=F(x)T(x)(P(x)-Q(x))$
$\displaystyle=F(x)T(x)P(x)-xF(x)T(x)Q(x),$

or

$\displaystyle
x^{6}F(x)T(x)Q(x)P(x)=x^{2}F(x)T(x)x^{4}Q(x)P(x)=(T(x)-F(x))(P(x)-Q(x)).$
(3.11)

The last identity leads to the following, surprising identity.

######  Theorem 3.14.

For any $m\geq 1$ we have

$\sum\limits_{K(4,n-2m-2)}F^{(m)}_{k_{1}}F^{(m+1)}_{k_{2}}F^{(m+2)}_{k_{3}}F^{(m+3)}_{k_{3}}=\sum\limits_{j=0}^{n}\big{(}F^{(m+3)}_{j}-F^{(m+2)}_{j}\big{)}\big{(}F^{(m+1)}_{n-j}-F^{(m)}_{n-j}\big{)}.$

###### Proof.

Following identity (3.11) and using (2.2) we have

$\displaystyle x^{2m+2}F^{(m)}(x)F^{(m+1)}(x)F^{(m+2)}(x)F^{(m+3)}(x)$
$\displaystyle\qquad\qquad=\big{(}x^{m}F^{(m)}(x)F^{(m+1)}(x)\big{)}\big{(}x^{m+2}F^{(m+2)}(x)F^{(m+3)}(x)\big{)}$
$\displaystyle\qquad\qquad=\big{(}F^{(m+1)}(x)-F^{(m)}(x)\big{)}\big{(}F^{(m+1)}(x)-F^{(m)}(x)\big{)}$

and the identity appears after passing to power series. ∎

##  4 Concluding results and comments

###  4.1 Higher order convolutions

Identities for even higher order convolutions can also be derived by mimicking
the argument concluding previous section. For instance, Theorem 3.14 has the
following generalization.

######  Theorem 4.1.

Fix $\ell\geq 1$ and an integer $m\geq 1$. Then

$\sum\limits_{K(2\ell,n-\ell(m+\ell-1))}\prod\limits_{j=0}^{2\ell-1}F_{k_{j+1}}^{(m+j)}=\sum\limits_{K(\ell,n)}\prod\limits_{j=0}^{\ell-1}\big{(}F^{(m+2j+1)}_{k_{j}}-F^{(m+2j)}_{k_{j}}\big{)}$
(4.1)

and

$\sum\limits_{K(2\ell+1,n-\ell(m+\ell-1))}\prod\limits_{j=0}^{2\ell}F_{k_{j}}^{(m+j)}=\sum\limits_{K(\ell+1,n)}F_{k_{\ell+1}}^{(m+2\ell)}\left[\prod\limits_{j=0}^{\ell-1}\big{(}F^{(m+2j+1)}_{k_{j}}-F^{(m+2j)}_{k_{j}}\big{)}\right].$
(4.2)

###### Proof.

We sketch the proof of (4.2). Using (3.5), we can write the following
functional equation.

$\displaystyle x^{\ell(m+\ell-1)}F^{(m)}(x)\cdots F^{(m+2\ell-1)}(x)$
$\displaystyle\quad=\big{(}x^{m}F^{(m)}(x)F^{(m+1)}(x)\big{)}\cdots
x^{m+2\ell-2}\big{(}x^{m}F^{(m+2\ell-2)}(x)F^{(m+2\ell-1)}(x)\big{)}$
$\displaystyle\quad=\big{(}F^{(m+1)}(x)-F^{(m)}(x)\big{)}\cdots\big{(}F^{(m+2\ell-1)}(x)-F^{(m+2\ell-2)}(x)\big{)}$

Passing to power series we obtain the desired formula. ∎

Theorem 4.1, in particular, allows us to effectively cut off half of the
sequences from the initial convolution. For instance, it is sufficient to know
the exact formula for the convolution of three among the Fibonacci to
Hexanacci and Heptanacci numbers (so $2\leq m\leq 7$) to know the expression
for the convolution of all of the sequences.

In Theorem 4.1 in identity (4.2) the sequence $F^{(m+2\ell)}_{n}$ does not
take a part in reducing the order of convolution. It is clear that another
sequence can be distinguished in a similar way. The change results in
different summation or product ranges, which are easily adjustable for a
specific example. This implies, for instance, the following.

######  Corollary 4.2.

We have

$\displaystyle\sum\limits_{K(3,n-2)}F_{k_{1}}T_{k_{2}}Q_{k_{3}}$
$\displaystyle=\sum\limits_{j=0}^{n}(T_{j}-F_{j})Q_{n-j}=\sum\limits_{j=0}^{n+1}F_{n+1-j}(Q_{j}-T_{j}).$

The above observation also leads to yet another look at the ”switch” effect
described in Section 2. Namely, this effect turns out to be the two different
versions of the simplified convolution sum of three sequences.

Finally, we also note that identities and can be stated in a more general
setting. Namely, the left-hand-side of either identity can be any finite
product of the terms

$F^{(m)}_{a}F^{(m+1)}_{b},$

even with repetitions, and the general formula can be adjusted for that case
as well. We leave the derivation of such a convoluted formula to the reader.

###  4.2 Some general cases of convolution of two sequences

In Section 2, we derived convolution identities that involve all pairs of
sequences up to $m=5$. However, following the proof of Corollary 2.9 we can do
more. In fact, we can deliver the general algorithm for finding a simple and
closed form of the convolution of $F^{(m)}$ with $F^{(m+p)}$ in the following
cases.
Case 1: $p|m$.
Case 2: $p|m+1$.
Case 3: $p=2m+2$.

We start with Case 1. So let $m=\ell\cdot p$ for some integer $\ell\geq 1$.
Applying Theorem 2.3 we get

$F^{((\ell+1)p)}(x)-F^{(\ell\cdot p)}(x)=\left(x^{\ell\cdot
p}+\cdots+x^{(\ell+1)p-1}\right)F^{(\ell\cdot p)}(x)F^{((\ell+1)p)}(x).$ (4.3)

Multiplying (4.3) repeatedly by $x^{p}$ we stack up a total of $\ell$
equalities:

$\displaystyle x^{p}F^{((\ell+1)p)}(x)-x^{p}F^{(\ell\cdot p)}(x)$
$\displaystyle=\left(x^{(\ell+1)p}+\cdots+x^{(\ell+2)p-1}\right)F^{(\ell\cdot
p)}(x)F^{((\ell+1)p)}(x),$ $\displaystyle\vdots$ $\displaystyle\qquad\vdots$
$\displaystyle x^{(\ell-1)p}F^{((\ell+1)p)}(x)-x^{(\ell-1)p}F^{(\ell\cdot
p)}(x)$ $\displaystyle=\left(x^{(2\ell-1)p}+\cdots+x^{2\ell\cdot
p-1}\right)F^{(\ell\cdot p)}(x)F^{((\ell+1)p)}(x).$

Adding everything up we get

$\sum\limits_{j=0}^{\ell-1}x^{j\cdot p}\big{(}F^{((\ell+1)p)}(x)-F^{(\ell\cdot
p)}(x)\big{)}=\left(x^{\ell\cdot p}+\cdots+x^{2\ell\cdot
p-1}\right)F^{(\ell\cdot p)}(x)F^{((\ell+1)p)}(x).$ (4.4)

In the next part of our computation we use the following convention. Whenever
we combine several sums into one with fixed summation range, the remaining
terms that are not included in the combined sum are called other terms and are
referred as $o.t.$ Depending in the exact case, these terms can be explicitly
derived. We do not do that in the below computation as it makes the formula
presented in the algorithm difficult to follow. Thus, only the important terms
are explicit.

We go back to (4.4) and rewrite this into power series to obtain

$\displaystyle\sum\limits_{j=0}^{\ell-1}\big{(}F^{((\ell+1)p)}_{n-j\cdot
p}-F^{(\ell\cdot p)}_{n-j\cdot p}\big{)}$
$\displaystyle=\sum\limits_{k=0}^{n-(2\ell\cdot
p-1)}F^{((\ell+1)p)}_{k}\left(\sum\limits_{j=0}^{\ell\cdot p-1}F^{(\ell\cdot
p)}_{n-(2\ell\cdot p-1)-k-j}\right)+o.t.$
$\displaystyle=\sum\limits_{k=0}^{n-(2\ell\cdot
p-1)}F^{((\ell+1)p)}_{k}F^{(\ell\cdot p)}_{n-(2\ell\cdot p)-k}+o.t.$

From this we can derive the closed formula for the desired convolution.

######  Example 4.3.

Set $m=4$ and $p=2$ and denote $s_{n}=F^{(6)}_{n}$. It follows that

$s_{n-2}-Q_{n-2}+s_{n}-Q_{n}=\sum\limits_{j=0}^{n-7}s_{j}Q_{n-3-j}+o.t.$ (4.5)

It is now easy to find that

$o.t.=2s_{n-6}+s_{n-5},$

and this finishes the formula.

Setting $m=p=2$ restores (2.8).

We now move to Case 2: $p|m+1$. So, let $m=\ell\cdot p-1$ for some positive
integer $\ell$. The further reasoning is similar to case $p|m$. Applying
Theorem 2.3 we get

$F^{((\ell+1)p-1)}(x)-F^{(\ell\cdot p-1)}(x)=\left(x^{\ell\cdot
p-1}+\cdots+x^{(\ell+1)p-2}\right)F^{(\ell\cdot p-1)}(x)F^{((\ell+1)p-1)}(x).$
(4.6)

We multiply (4.6) again by $x^{p}$ to get

$\displaystyle x^{p}F^{((\ell+1)p-1)}(x)-x^{p}F^{(\ell\cdot p-1)}(x)$
$\displaystyle=\left(x^{(\ell+1)p-1}+\cdots+x^{(\ell+2)p-2}\right)F^{(\ell\cdot
p)}(x)F^{((\ell+1)p)}(x),$ $\displaystyle\vdots$ $\displaystyle\qquad\vdots$
$\displaystyle x^{(\ell-1)p}F^{((\ell+1)p-1)}(x)-x^{(\ell-1)p}F^{(\ell\cdot
p-1)}(x)$ $\displaystyle=\left(x^{(2\ell-1)p-1}+\cdots+x^{2\ell\cdot
p-2}\right)F^{(\ell\cdot p-1)}(x)F^{((\ell+1)p-1)}(x).$

Adding everything up and passing to power series, we have

$\displaystyle\sum\limits_{j=0}^{\ell-1}\big{(}F^{((\ell+1)p-1)}_{n-j\cdot
p}-F^{(\ell\cdot p)-1}_{n-j\cdot p}\big{)}$
$\displaystyle=\sum\limits_{k=0}^{n-(2\ell\cdot
p-2)}F^{((\ell+1)p-1)}_{k}\left(\sum\limits_{j=0}^{\ell\cdot p-1}F^{(\ell\cdot
p-1)}_{n-(2\ell\cdot p-2)-k-j}\right)+o.t.$
$\displaystyle=2\sum\limits_{k=0}^{n-(2\ell\cdot
p-2)}F^{((\ell+1)p)-1}_{k}F^{(\ell\cdot p-1)}_{n-(2\ell\cdot p-1)-k}+o.t.$

######  Example 4.4.

Setting $p=3$ and $m=2$ we reproduce the proof of (2.12). Setting $p=2$ and
$m=3$ we reproduce the identity (2.13).

Finally, consider Case 3 and $p=2m+2$. The key feature of this case is the
following simple lemma.

######  Lemma 4.5.

For any $m\geq 2$ and any $n\geq 0$ we have

$\sum\limits_{k=0}^{2m+1}F_{n+k}^{(m)}=4F_{n+2m}^{(m)}.$ (4.7)

###### Proof.

Write

$\displaystyle 4F_{n+2m}^{(m)}$
$\displaystyle=\big{(}2F_{n+m}^{(m)}+\cdots+2F_{n+2m-1}^{(m)}\big{)}+2F_{n+2m}^{(m)}$
$\displaystyle=2F_{n+m}^{(m)}+\big{(}2F_{n+m+1}^{(m)}+\cdots+2F_{n+2m}^{(m)}\big{)}$
$\displaystyle=F_{m}^{(m)}+\cdots+F_{n+m-1}^{(m)}+F_{n+m}^{(m)}$
$\displaystyle\qquad+F_{n+m+1}^{(m)}+\cdots+F_{n+2m}^{(m)}+F_{n+2m+1}^{(m)}.$

∎

We now apply Theorem 2.3 to obtain

$F^{3m+2}(x)-F^{(m)}(x)=\big{(}x^{m}+\cdots+x^{3m+1}\big{)}F^{(m)}(x)F^{(3m+2)}(x).$

This implies, using (4.7), that

$\displaystyle F_{n}^{(3m+2)}-F_{n}^{(m)}$
$\displaystyle=\sum\limits_{k=0}^{n-(3m+1)}F_{k}^{(3m+2)}\big{(}F_{n-(3m-1)-k}^{(m)}+\cdots+F_{n-m-k}^{(m)}\big{)}+o.t.$
$\displaystyle=4\sum\limits_{k=0}^{n-(3m+1)}F_{k}^{(3m+2)}F_{n-m-1-k}^{(m)}+o.t.$

######  Example 4.6.

Set $m=2$ and let $\mathcal{O}=F^{(8)}$. Then we have

$\displaystyle\mathcal{O}_{n}-F_{n}$
$\displaystyle=4\sum\limits_{j=0}^{n-7}\mathcal{O}_{j}F_{n-3-j}+\mathcal{O}_{n-3}+2\mathcal{O}_{n-4}+4\mathcal{O}_{n-5}+7\mathcal{O}_{n-6}$
$\displaystyle=4\sum\limits_{j=0}^{n-3}\mathcal{O}_{j}F_{n-3-j}+\mathcal{O}_{n-3}-2\mathcal{O}_{n-4}-\mathcal{O}_{n-6}.$

It follows after some calculation that

$\sum\limits_{j=0}^{n}\mathcal{O}_{k}F_{n-j}=\frac{1}{4}\left(\mathcal{O}_{n+3}-\mathcal{O}_{n}+2\mathcal{O}_{n-1}+\mathcal{O}_{n-3}-F_{n+3}\right).$
(4.8)

The methods provided to this point allow us to find a closed form of the
convolution of two different sequences out of the set of Fibonacci,
Tribonacci, Tetranacci and Pentanacci numbers. If we include Hexanacci numbers
(i.e. $s=F^{(6)}$), then we can compute all convolutions but the convolution
of Fibonacci and Hexanacci numbers. This is at first glance surprising, but in
fact the presented algorithm does not allow us to deal with that case, even
though we can compute the explicit form of
$\sum_{j=0}^{n}F^{(20)}_{j}F^{(27)}_{n-j}$.

We now show how to find the sum $\sum_{j=0}^{n}s_{j}F_{n-j}$. In order to deal
with this problem, we have to use another approach.

######  Theorem 4.7.

For $n\geq 3$ we have

$\sum\limits_{j=0}^{n}s_{j}F_{n-j}=\frac{1}{5}\left(s_{n+3}+s_{n+1}-s_{n}+3s_{n-1}+s_{n-3}-F_{n+3}-F_{n+1}\right).$
(4.9)

###### Proof.

Notice that by (3.5) or (2.6) we have

$\displaystyle s_{n}-F_{n}$
$\displaystyle=\sum\limits_{j=0}^{n-2}s_{j}F_{n-j-2}+\sum\limits_{j=0}^{n-3}s_{j}F_{n-j-3}+\sum\limits_{j=0}^{n-4}s_{j}F_{n-j-4}+\sum\limits_{j=0}^{n-5}s_{j}F_{n-j-5},$
$\displaystyle s_{n+2}-F_{n+2}$
$\displaystyle=\sum\limits_{j=0}^{n}s_{j}F_{n-j}+\sum\limits_{j=0}^{n-1}s_{j}F_{n-j-1}+\sum\limits_{j=0}^{n-2}s_{j}F_{n-j-2}+\sum\limits_{j=0}^{n-3}s_{j}F_{n-j-3}.$

Summing up and rearranging we get

$\displaystyle s_{n+2}-F_{n+2}+s_{n}-F_{n}$
$\displaystyle=\sum\limits_{j=0}^{n-5}s_{j}(F_{n-j-5}+F_{n-j-4}+2F_{n-j-3}+2F_{n-j-2}+F_{n-j-1}+F_{n-j})$
$\displaystyle\quad\quad+s_{n-1}+2s_{n-2}+5s_{n-3}+9s_{n-4}.$

To proceed further, we apply the identity

$F_{n}+F_{n+1}+2F_{n+2}+2F_{n+3}+F_{n+4}+F_{n+5}=5F_{n+4}$ (4.10)

valid for any $n\geq 0$ and we substitute $n\to n+1$ to obtain

$\displaystyle s_{n+3}-F_{n+3}+s_{n+1}-F_{n+1}$
$\displaystyle=\sum\limits_{j=0}^{n-5}5s_{j}F_{n-j-1}+s_{n-1}+2s_{n-2}+5s_{n-3}+9s_{n-4}$
$\displaystyle=5\sum\limits_{j=0}^{n}s_{j}F_{n-j}+s_{n}-3s_{n-1}-s_{n-3}.$

Thus, after minor adjustments, we have (4.9). ∎

We note that there are more cases where an identity similar to (4.10) leads to
a closed sum formula. Namely, if we consider $p+m\leq 8$ and let $S=F^{(7)}$
(Heptanacci numbers), then the only missing cases, not following from the
rules described by the three cases, are

$\sum\limits_{j=0}^{n}S_{j}F_{n-j},\qquad\sum\limits_{j=0}^{n}S_{j}Q_{n-j}\qquad\text{and}\qquad\sum\limits_{j=0}^{n}\mathcal{O}_{j}T_{n-j}.$

These sums can be derived using an approach similar to (4.9), but this time
with the aid of the following identities:

$\displaystyle 2F_{n}+2F_{n+1}+3F_{n+2}+3F_{n+3}+3F_{n+4}+F_{n+5}+F_{n+6}$
$\displaystyle=11F_{n+4},$ (4.11) $\displaystyle
Q_{n}+Q_{n+1}+Q_{n+2}+Q_{n+3}+Q_{n+4}+2Q_{n+5}+2Q_{n+6}+2Q_{n+7}+Q_{n+8}$
$\displaystyle=3Q_{n+8},$ (4.12) $\displaystyle
2T_{n}+2T_{n+1}+3T_{n+2}+5T_{n+3}+5T_{n+4}+3T_{n+5}+3T_{n+6}+2T_{n+7}$
$\displaystyle=11T_{n+6}.$ (4.13)

To clarify how to use them, we write, for example,

$\displaystyle
Q_{n}+Q_{n+1}+Q_{n+2}+Q_{n+3}+Q_{n+4}+2Q_{n+5}+2Q_{n+6}+2Q_{n+7}+Q_{n+8}$
$\displaystyle\qquad=(Q_{n}+Q_{n+1}+Q_{n+2})+(Q_{n+3}+Q_{n+4}+Q_{n+5})$
$\displaystyle\qquad\qquad+(Q_{n+5}+Q_{n+6}+Q_{n+7})+(Q_{n+6}+Q_{n+7}+Q_{n+8})$

and it is clear that each bracket can be generated from the identity

$S(x)-Q(x)=(x^{4}+x^{5}+x^{6})Q(x)S(x).$

###  4.3 Open problems

In the previous section we presented the algorithm for computation of the
convolution sum of two sequences under (major) restrictions. We dealt with
missing case $m=2$ and $p=4$ separately so that all convolution sums with
$m+p\leq 8$ for $m\geq 2$ and $p\geq 1$ have their closed forms calculated.
The first case that is not covered by our methods (that is, the case with
smallest possible $m+p$ and smallest possible $p$) is the following
convolution sum (see also Table 1):

$\sum\limits_{j=0}^{n}P_{j}F^{(9)}_{n-j}.$

The trick that was used above could also work here but this does not replace a
general approach to these sums (identities (4.10)–(4.13) seem to only work in
the presented form, we do not know if/how they generalize, as they were found
by trial and error). In our opinion, a good starting point is to search for an
identity of the form

$\sum\limits_{k\in K}\sum\limits_{j=0}^{p-1}F_{n+j+k}^{(m)}=N\cdot
F_{n+\ell}^{(m)}$ (4.14)

valid for any $n$, $m\geq 2$, with $N$ and $\ell$ being unknown, $K$ being a
finite set, $\ell$ related to $n$ and $p$. Identity (4.10) is the case $m=2$
and follows that pattern with $K=\\{0,2\\}$, $p=4$, $N=5$ and $\ell=4$. The
identity (4.7) is the simplest example of that form, with $K=\\{0\\}$. Any
identity of the form (4.14) would give us yet another convolution sum. We
believe that finding any other solution (or even an infinite family of
solutions) to that equation is a good motivation for further research in the
topic.

$m\ \backslash\ p$ | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9
---|---|---|---|---|---|---|---|---
2 |  (2.11) |  (2.12) |  (4.9) |  (4.11) |  (4.8) | ? | ? | ?
3 |  (2.13) | $p|m$ | $p|m+1$ |  (4.13) | ? | ? | $p=2m+2$ | ?
4 |  (4.5) |  (4.12) | $p|m$ | $p|m+1$ | ? | ? | ? | ?
5 | $p|m+1$ | $p|m+1$ | ? | $p|m$ | $p|m+1$ | ? | ? | ?
6 | $p|m$ | $p|m$ | ? | ? | $p|m$ | $p|m+1$ | ? | ?
7 | $p|m+1$ | ? | $p|m+1$ | ? | ? | $p|m$ | $p|m+1$ | ?
8 | $p|m$ | $p|m+1$ | $p|m$ | ? | ? | ? | $p|m$ | $p|m+1$
9 | $p|m+1$ | $p|m$ | ? | $p|m+1$ | ? | ? | ? | $p|m$
Table 1: Convolution sums $\sum F^{(m)}_{j}F^{(m+p)}_{n-j}$ with $2\leq
m,p\leq 9$ covered directly or indirectly in this article. The cases $m=1$ and
$p=1$ are covered by (2.10) and (2.1), respectively. Question marks indicate
unsolved cases.

##  5 Conclusion

This article was devoted to study convolutions involving Fibonacci $m$-step
numbers. We have applied the prominent generating function approach to prove
several appealing results that strengthen the understanding of these numbers.
Many known identities for Fibonacci, Tribonacci, Tetranacci and Pentanacci
numbers now follow from our results as special cases. In addition, mixed
convolutions involving Fibonacci $m$-step numbers and Jacobsthal and Pell
numbers were stated. To keep things coherent and streamlined, we focused
exclusively on Fibonacci $m$-step numbers. There is still much work to be
done. Identities for Lucas $m$-step numbers, Pell $m$-step numbers, and
others, and also mixed convolutions of these sequences can be studied in the
future.

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##  Appendix A Appendix

Figure 9: Differences between activities regarding productivity, autonomy,
competence, relatedness, pressure, and conflicts. Red lines represent
99.9%-CIs.