Multiple Sums of Circular Binomial Products

Abstract

Five classes of multiple sums about circular products of binomial coefficients are investigated. Their generating functions are explicitly expressed as rational functions, with coefficients being Fibonacci and Lucas numbers. This is fulfilled by a novel approach called “recursive construction”, which also permits us to establish, for the circular sums, both generating functions and recurrence relations through the Lagrange expansion formula.

1. Introduction and Outline

The Fibonacci and Lucas numbers are well known mathematical sequences. They are defined as follows:

• Recurrence relations:

  $$F_n=F_{n-1}+F_{n-2}\mathrm{and}{L}_n=L_{n-1}+L_{n-2}.$$
• Boundary conditions:

  $$F_0=0,F_1=1\mathrm{and}{L}_0=2,L_1=1.$$
• Generating functions:

  $$\sum _{n=0}^{\infty }{F}_n{x}^n=\frac{x}{1-x-x^2}\mathrm{and}\sum _{n=0}^{\infty }{L}_n{x}^n=\frac{2-x}{1-x-x^2}.$$

There exist numerous identities about Fibonacci and Lucas numbers in the mathematical literature (see [1,2,3,4,5]). About a half century ago, Carlitz [6] initiated and discovered, by examining the characteristic polynomial of a certain binomial matrix, the following beautiful identity for the circular sum of binomial coefficients:

$$\sum _{0\le k_1,k_2,\cdots ,k_m\le n}\left(\genfrac{}{}{0pt}{}{n-k_1}{k_m}\right)\prod _{i=1}^{m-1}\left(\genfrac{}{}{0pt}{}{n-k_{i+1}}{k_i}\right)=\frac{F_{mn+m}}{F_m}.$$

(1)

Despite abundant studies on binomial sums, research articles on multiple circular sums are quite rare. Apart from Carlitz’s pioneering work, the only related studies are about different proofs of (1), published recently in [5,7,8]. In their monograph [3], Benjamin and Quinn recorded a few variants (identities 138–141). For example, we have

$$\sum _{0\le k_1,k_2,\cdots ,k_m\le n}\left(\genfrac{}{}{0pt}{}{n-\lambda }{k_m}\right)\prod _{i=1}^{m-1}\left(\genfrac{}{}{0pt}{}{n-k_{i+1}}{k_i}\right)=\left\{\begin{array}{cc}{F}_{m+1}^{\lambda }{F}_{m+2}^{n-\lambda },\hfill & 0\le \lambda \le n;\hfill \\ \frac{F_{m+1}^{n+1}+{(-1)}^n{F}_m^{n+1}}{F_{m+2}}\hfill & \lambda =n+1.\hfill \end{array}\right.$$

(2)

For this identity, the case for $0\le \lambda \le n$ corresponds to identity 139 in [3]. As a warm up, we present a detailed proof for (2) by “the recursive construction” approach. Denoting the multiple sum in question by ${\mathrm{W}}_m(n,\lambda )$, we can compute it recursively as follows:

$$\begin{array}{cc}\hfill {\mathrm{W}}_m(n,\lambda )& =\sum _{0\le k_2,\cdots ,k_m\le n}\left(\genfrac{}{}{0pt}{}{n-\lambda }{k_m}\right)\prod _{i=2}^{m-1}\left(\genfrac{}{}{0pt}{}{n-k_{i+1}}{k_i}\right)\sum _{k_1=0}^n\left(\genfrac{}{}{0pt}{}{n-k_2}{k_1}\right)\hfill \\ & =\sum _{0\le k_3,\cdots ,k_m\le n}{F}_3^n\left(\genfrac{}{}{0pt}{}{n-\lambda }{k_m}\right)\prod _{i=3}^{m-1}\left(\genfrac{}{}{0pt}{}{n-k_{i+1}}{k_i}\right)\sum _{k_2=0}^n\left(\genfrac{}{}{0pt}{}{n-k_3}{k_2}\right)\frac{F_2^{k_2}}{F_3^{k_2}}\hfill \\ & =\sum _{0\le k_4,\cdots ,k_m\le n}{F}_4^n\left(\genfrac{}{}{0pt}{}{n-\lambda }{k_m}\right)\prod _{i=4}^{m-1}\left(\genfrac{}{}{0pt}{}{n-k_{i+1}}{k_i}\right)\sum _{k_3=0}^n\left(\genfrac{}{}{0pt}{}{n-k_4}{k_3}\right)\frac{F_3^{k_3}}{F_4^{k_3}}\hfill \\ & ⋮\hfill \\ & =\sum _{0\le k_m\le n}{F}_{m+1}^n\left(\genfrac{}{}{0pt}{}{n-\lambda }{k_m}\right)\frac{F_m^{k_m}}{F_{m+1}^{k_m}}=\left\{\begin{array}{cc}{F}_{m+1}^{\lambda }{F}_{m+2}^{n-\lambda },\hfill & 0\le \lambda \le n;\hfill \\ \frac{F_{m+1}^{n+1}+{(-1)}^n{F}_m^{n+1}}{F_{m+2}}\hfill & \lambda =n+1.\hfill \end{array}\right.\hfill \end{array}$$

The aim of the present paper is to examine, by computing generating functions (see [9] (chapter 1) and [10]), the following circular sums:

$$\sum _{0\le k_1,k_2,\cdots ,k_m\le n}{\Delta }_n(k_1,k_m)\prod _{ı=1}^{m-1}\left(\genfrac{}{}{0pt}{}{n-k_{ı+1}}{k_{ı}}\right),$$

(3)

where the “weight function” ${\Delta }_n(k_1,k_m)$ is assigned to one of the five coefficients below:

$$\left\{\left(\genfrac{}{}{0pt}{}{1+n-k_1}{k_m}\right)\frac{1+n-k_1+k_m}{1+n-k_1},\left(\genfrac{}{}{0pt}{}{n-k_1}{2k_m}\right),\left(\genfrac{}{}{0pt}{}{n-k_1+1}{2k_m+1}\right),\left(\genfrac{}{}{0pt}{}{n+k_1}{2k_m}\right),\left(\genfrac{}{}{0pt}{}{n+k_1+1}{2k_m+1}\right)\right\}.$$

(4)

For the multiple sum corresponding to the first case, we shall establish, in the next section, not only the generating function, but also the explicit formula in closed form. Then, in Section 3 and Section 4, the generating functions will be derived for the multiple sums of the second and the third cases. For the fourth and fifth cases, the generating functions will be shown for the corresponding multiple sums in Section 5 and Section 6. Finally, the paper will end with further observation and comments.

Throughout the paper, the following two simple binomial series will play a crucial role in our derivation:

$$\begin{array}{cccc}& {(1+x)}^m=\sum _{k=0}^m\left(\genfrac{}{}{0pt}{}{m}{k}\right)x^k:\hfill & \hfill & \left[x^k\right]{(1+x)}^m=\left(\genfrac{}{}{0pt}{}{m}{k}\right);\hfill \\ & \frac{x^m}{{(1-x)}^{m+1}}=\sum _{k=m}^{\infty }\left(\genfrac{}{}{0pt}{}{k}{m}\right)x^k:\hfill & & \left[x^k\right]\frac{x^m}{{(1-x)}^{m+1}}=\left(\genfrac{}{}{0pt}{}{k}{m}\right);\hfill \end{array}$$

where $\left[x^k\right]f\left(x\right)$ stands for the coefficient of $x^k$ in the formal power series $f\left(x\right)$. They will be utilized recursively to construct generating functions for the multiple sums displayed in (3). Compared with the three existing proofs of (1) given by Carlitz [6] through matrix trace evaluations, by Mikic [5] via the induction principle, and by combinatorial bijections in [3,7,11], our recursive construction approach is more transparent and flexible in dealing with a variety of multiple binomial sums.

2. The First Circular Sum ${\mathcal{W}}_{\mathit{m}}\left(\mathit{n}\right)$

In this section, we are going to examine the circular sum defined by

$${\mathcal{W}}_m\left(n\right)=\sum _{0\le k_1,k_2,\cdots ,k_m\le n}\left(\genfrac{}{}{0pt}{}{1+n-k_1}{k_m}\right)\frac{1+n-k_1+k_m}{1+n-k_1}\prod _{ı=1}^{m-1}\left(\genfrac{}{}{0pt}{}{n-k_{ı+1}}{k_{ı}}\right).$$

Then, we have generating functions and the explicit formula as in the following theorem.

Theorem 1.

The circular sum ${\mathcal{W}}_m\left(n\right)$ has generating functions

$$\begin{array}{cc}\hfill {\mathcal{W}}_m\left(n\right)& =\left[x^{n+1}\right]\left\{\frac{(2-x){(F_m+x{F}_{m-1})}^{n+1}}{F_m(1-x-x^2)}-\frac{2}{F_m(1-x{F}_{m-1})}\right\}\hfill \end{array}$$

(5)

$$\begin{array}{cc}& \hspace{1em}\hspace{1em}\hspace{1em}=\left[y^{n+1}\right]\left\{\frac{2-y{L}_m}{F_m(1-y{L}_m+{(-1)}^m{y}^2)}-\frac{2}{F_m(1-y{F}_{m-1})}\right\}\hfill \end{array}$$

(6)

and admits the following explicit formula:

$${\mathcal{W}}_m\left(n\right)=\frac{L_{mn+m}-2F_{m-1}^{n+1}}{F_m}.$$

(7)

The above closed formula provides a counterpart of Carlitz’s (1) in terms of Lucas numbers. By means of domino tilings, Benjamin and Rouse [7] construct a bijective proof for (1). The interested reader is encouraged to find a similar combinatorial explanation for (7) by incorporating the random approaches devised by Benjamin et al. [11].

Proof. 

First, it is not hard to check the following relations:

$$\begin{array}{cc}& \left(\genfrac{}{}{0pt}{}{n-k_2}{k_1}\right)=\left[x^{k_1}\right]{(1+x)}^{n-k_2},\hfill \\ & \left(\genfrac{}{}{0pt}{}{1+n-k_1}{k_m}\right)\frac{1+n-k_1+k_m}{1+n-k_1}=\left[x^{1+n-k_1}\right]\frac{(2-x)x^{k_m}}{{(1-x)}^{1+k_m}}.\hfill \end{array}$$

(8)

We can express the binomial sum with respect to $k_1$ in ${\mathcal{W}}_m\left(n\right)$ as follows:

$$\begin{array}{cc}\hfill {\mathcal{W}}_m^1\left(n\right)& =\sum _{k_1=0}^n\left(\genfrac{}{}{0pt}{}{n-k_2}{k_1}\right)\left(\genfrac{}{}{0pt}{}{1+n-k_1}{k_m}\right)\frac{1+n-k_1+k_m}{1+n-k_1}\hfill \\ & =\left[x^{n+1}\right]\frac{(2-x)x^{k_m}}{{(1-x)}^{1+k_m}}{(1+x)}^{n-k_2}.\hfill \end{array}$$

Next, we can proceed with the binomial sum with respect to $k_2$ in ${\mathcal{W}}_m\left(n\right)$:

$$\begin{array}{cc}\hfill {\mathcal{W}}_m^2\left(n\right)& =\sum _{k_2=0}^n{\mathcal{W}}_m^1\left(n\right)\left(\genfrac{}{}{0pt}{}{n-k_3}{k_2}\right)\hfill \\ & =\left[x^{n+1}\right]\frac{(2-x){(1+x)}^n{x}^{k_m}}{{(1-x)}^{1+k_m}}\sum _{k_2=0}^n\left(\genfrac{}{}{0pt}{}{n-k_3}{k_2}\right){(1+x)}^{-k_2}\hfill \\ & =\left[x^{n+1}\right]\frac{(2-x)x^{k_m}{(1+x)}^n}{{(1-x)}^{1+k_m}}\times {\left\{\frac{2+x}{1+x}\right\}}^{n-k_3}\hfill \\ & =\left[x^{n+1}\right]\frac{(2-x)x^{k_m}{(2+x)}^n}{{(1-x)}^{1+k_m}}\times {\left\{\frac{1+x}{2+x}\right\}}^{k_3}\hfill \\ & =\left[x^{n+1}\right]\frac{(2-x)x^{k_m}{(F_3+x{F}_2)}^n}{{(1-x)}^{1+k_m}}\times {\left\{\frac{F_2+x{F}_1}{F_3+x{F}_2}\right\}}^{k_3}.\hfill \end{array}$$

Iterating this process, we have the binomial sum with respect to $k_{\ell }$ in ${\mathcal{W}}_m\left(n\right)$:

$$\begin{array}{cc}\hfill {\mathcal{W}}_m^{\ell }\left(n\right)& =\sum _{k_{\ell }=0}^n{\mathcal{W}}_m^{\ell -1}\left(n\right)\left(\genfrac{}{}{0pt}{}{n-k_{\ell +1}}{k_{\ell }}\right)\hfill \\ & =\left[x^{n+1}\right]\frac{(2-x)x^{k_m}{(F_{\ell +1}+x{F}_{\ell })}^n}{{(1-x)}^{1+k_m}}\times {\left\{\frac{F_{\ell }+x{F}_{\ell -1}}{F_{\ell +1}+x{F}_{\ell }}\right\}}^{k_{\ell +1}}.\hfill \end{array}$$

Then, the penultimate binomial sum with respect to $k_{m-1}$ in ${\mathcal{W}}_m\left(n\right)$ results in

$$\begin{array}{cc}\hfill {\mathcal{W}}_m^{m-1}\left(n\right)& =\sum _{k_{m-1}=0}^n{\mathcal{W}}_m^{m-2}\left(n\right)\left(\genfrac{}{}{0pt}{}{n-k_m}{k_{m-1}}\right)\hfill \\ & =\left[x^{n+1}\right]\frac{(2-x)x^{k_m}{(F_m+x{F}_{m-1})}^n}{{(1-x)}^{1+k_m}}\times {\left\{\frac{F_{m-1}+x{F}_{m-2}}{F_m+x{F}_{m-1}}\right\}}^{k_m}.\hfill \end{array}$$

Finally, we can express the circular binomial sum ${\mathcal{W}}_m\left(n\right)$ as

$$\begin{array}{cc}\hfill {\mathcal{W}}_m\left(n\right)=& \sum _{k_m=0}^n{\mathcal{W}}_m^{m-1}\left(n\right)=\left[x^{n+1}\right]\frac{(2-x){(F_m+x{F}_{m-1})}^n}{1-x}\hfill \\ \hfill \times & \sum _{k_m=0}^{\infty }\left\{1-\chi (k_m=n+1)\right\}{\left\{\frac{x(F_{m-1}+x{F}_{m-2})}{(1-x)(F_m+x{F}_{m-1})}\right\}}^{k_m}\hfill \\ \hfill =& \left[x^{n+1}\right]\frac{(2-x){(F_m+x{F}_{m-1})}^n}{1-x}\hfill \\ \hfill \times & \left\{{\left(1-\frac{x(F_{m-1}+x{F}_{m-2})}{(1-x)(F_m+x{F}_{m-1})}\right)}^{-1}-{\left(\frac{x(F_{m-1}+x{F}_{m-2})}{(1-x)(F_m+x{F}_{m-1})}\right)}^{n+1}\right\},\hfill \end{array}$$

where $\chi$ denotes the logical function with $\chi \left(\mathrm{true}\right)=1$ and $\chi \left(\mathrm{false}\right)=0$. By simplifying the last expression, we obtain that displayed in (5).

Recalling the Binet formulae

$$\left.\begin{array}{c}{F}_n=\frac{{\alpha }^n - {\beta }^n}{\alpha - \beta }\hfill \\ L_n={\alpha }^n+{\beta }^n\hfill \end{array}\right\}\mathrm{where}\alpha ,\beta =\frac{1\pm \sqrt{5}}{2},$$

we can confirm the closed formula (7) as follows:

$$\begin{array}{cc}\hfill \left[x^{n+1}\right]& \frac{(2-x){(F_m+x{F}_{m-1})}^{n+1}}{1-x-x^2}\hfill \\ \hfill =& \sum _{k=0}^{n+1}{L}_k\left(\genfrac{}{}{0pt}{}{n+1}{k}\right)F_m^k{F}_{m-1}^{1+n-k}\hfill \\ \hfill =& \sum _{k=0}^{n+1}({\alpha }^k+{\beta }^k)\left(\genfrac{}{}{0pt}{}{n+1}{k}\right)F_m^k{F}_{m-1}^{1+n-k}\hfill \\ \hfill =& {(\alpha F_m+F_{m-1})}^{n+1}+{(\beta F_m+F_{m-1})}^{n+1}\hfill \\ \hfill =& {\alpha }^{m(n+1)}+{\beta }^{m(n+1)}=L_{mn+m}.\hfill \end{array}$$

Furthermore, taking into account that

$$\begin{array}{cc}\hfill L_{mn+m}={\alpha }^{mn+m}+{\beta }^{mn+m}& =\left[y^{n+1}\right]\left\{\frac{1}{1-y{\alpha }^m}+\frac{1}{1-y{\beta }^m}\right\}\hfill \\ & =\left[y^{n+1}\right]\frac{2-y{L}_m}{1-y{L}_m+{(-1)}^m{y}^2},\hfill \end{array}$$

we deduce another generating function (6) and the proof is complete.      □

3. The Second Circular Sum ${\mathcal{U}}_{\mathit{m}}\left(\mathit{n}\right)$

This section will be devoted to the circular sum defined by

$${\mathcal{U}}_m\left(n\right)=\sum _{0\le k_1,k_2,\cdots ,k_m\le n}\left(\genfrac{}{}{0pt}{}{n-k_1}{2k_m}\right)\prod _{ı=1}^{m-1}\left(\genfrac{}{}{0pt}{}{n-k_{ı+1}}{k_{ı}}\right).$$

Its generating functions are determined by the following theorem.

Theorem 2

(Generating functions).

$$\begin{array}{cc}\hfill {\mathcal{U}}_m\left(n\right)& =\left[x^n\right]\frac{(1-x){(F_m+x{F}_{m-1})}^{n+1}}{{(1-x)}^2(F_m+x{F}_{m-1})-x^2(F_{m-1}+x{F}_{m-2})}\hfill \end{array}$$

(9)

$$\begin{array}{cc}& \hspace{1em}\hspace{1em}\hspace{1em}=\left[y^n\right]\frac{1-y{F}_{m+1}}{1-2y{F}_{m+1}+2y^2{F}_m^2+{(-1)}^m{y}^2+{(-1)}^m{y}^3{F}_m}.\hfill \end{array}$$

(10)

The first five examples of this theorem are highlighted as in the corollary below, where for $m=1$ and $m=2$ the corresponding sequences are recorded, respectively, as “A005251” and “A085810” in [12].

Corollary 1

(Generating functions).

$$\begin{array}{cc}\hfill {\mathcal{U}}_1\left(n\right)& =\left[y^n\right]\frac{1-y}{1-2y+y^2-y^3};\hfill \\ \hfill {\mathcal{U}}_2\left(n\right)& =\left[y^n\right]\frac{1-2y}{1-4y+3y^2+y^3};\hfill \\ \hfill {\mathcal{U}}_3\left(n\right)& =\left[y^n\right]\frac{1-3y}{1-6y+7y^2-2y^3};\hfill \\ \hfill {\mathcal{U}}_4\left(n\right)& =\left[y^n\right]\frac{1-5y}{1-10y+19y^2+3y^3};\hfill \\ \hfill {\mathcal{U}}_5\left(n\right)& =\left[y^n\right]\frac{1-8y}{1-16y+49y^2-5y^3}.\hfill \end{array}$$

Proof of Theorem 2. 

By making use of (8) and the binomial relation

$$\begin{array}{cc}& \left(\genfrac{}{}{0pt}{}{n-k_1}{2k_m}\right)=\left[x^{n-k_1}\right]\frac{x^{2k_m}}{{(1-x)}^{1+2k_m}},\hfill \end{array}$$

we can express the binomial sum with respect to $k_1$ in ${\mathcal{U}}_m\left(n\right)$ as follows:

$${\mathcal{U}}_m^1\left(n\right)=\sum _{k_1=0}^n\left(\genfrac{}{}{0pt}{}{n-k_2}{k_1}\right)\left(\genfrac{}{}{0pt}{}{n-k_1}{2k_m}\right)=\left[x^n\right]\frac{x^{2k_m}{(1+x)}^n}{{(1-x)}^{1+2k_m}{(1+x)}^{k_2}}.$$

Then, we can proceed with the binomial sum with respect to $k_2$ in ${\mathcal{U}}_m\left(n\right)$:

$$\begin{array}{cc}\hfill {\mathcal{U}}_m^2\left(n\right)& =\sum _{k_2=0}^n{\mathcal{U}}_m^1\left(n\right)\left(\genfrac{}{}{0pt}{}{n-k_3}{k_2}\right)\hfill \\ & =\left[x^n\right]\frac{x^{2k_m}{(1+x)}^n}{{(1-x)}^{1+2k_m}}\sum _{k_2=0}^n\left(\genfrac{}{}{0pt}{}{n-k_3}{k_2}\right){\left(\frac{1}{1+x}\right)}^{k_2}\hfill \\ & =\left[x^n\right]\frac{x^{2k_m}{(2+x)}^n}{{(1-x)}^{1+2k_m}}\times {\left\{\frac{1+x}{2+x}\right\}}^{k_3}\hfill \\ & =\left[x^n\right]\frac{x^{2k_m}{(F_3+x{F}_2)}^n}{{(1-x)}^{1+2k_m}}\times {\left\{\frac{F_2+x{F}_1}{F_3+x{F}_2}\right\}}^{k_3}.\hfill \end{array}$$

Iterating this process, we have the binomial sum with respect to $k_{\ell }$ in ${\mathcal{U}}_m\left(n\right)$:

$$\begin{array}{cc}\hfill {\mathcal{U}}_m^{\ell }\left(n\right)& =\sum _{k_{\ell }=0}^n{\mathcal{U}}_m^{\ell -1}\left(n\right)\left(\genfrac{}{}{0pt}{}{n-k_{\ell +1}}{k_{\ell }}\right)\hfill \\ & =\left[x^n\right]\frac{x^{2k_m}{(F_{\ell +1}+x{F}_{\ell })}^n}{{(1-x)}^{1+2k_m}}\times {\left\{\frac{F_{\ell }+x{F}_{\ell -1}}{F_{\ell +1}+x{F}_{\ell }}\right\}}^{k_{\ell +1}}.\hfill \end{array}$$

The penultimate binomial sum with respect to $k_{m-1}$ in ${\mathcal{U}}_m\left(n\right)$ results in

$$\begin{array}{cc}\hfill {\mathcal{U}}_m^{m-1}\left(n\right)& =\sum _{k_{m-1}=0}^n{\mathcal{U}}_m^{m-2}\left(n\right)\left(\genfrac{}{}{0pt}{}{n-k_m}{k_{m-1}}\right)\hfill \\ & =\left[x^n\right]\frac{x^{2k_m}{(F_m+x{F}_{m-1})}^n}{{(1-x)}^{1+2k_m}}\times {\left\{\frac{F_{m-1}+x{F}_{m-2}}{F_m+x{F}_{m-1}}\right\}}^{k_m}.\hfill \end{array}$$

Finally, we can express the circular binomial sum ${\mathcal{U}}_m\left(n\right)$ as

$$\begin{array}{cc}\hfill {\mathcal{U}}_m\left(n\right)=& \sum _{k_m=0}^n{\mathcal{U}}_m^{m-1}\left(n\right)\hfill \\ \hfill =& \left[x^n\right]\frac{{(F_m+x{F}_{m-1})}^n}{1-x}\sum _{k_m=0}^{\infty }{\left\{\frac{x^2(F_{m-1}+x{F}_{m-2})}{{(1-x)}^2(F_m+x{F}_{m-1})}\right\}}^{k_m}\hfill \\ \hfill =& \left[x^n\right]\frac{(1-x){(F_m+x{F}_{m-1})}^{n+1}}{{(1-x)}^2(F_m+x{F}_{m-1})-x^2(F_{m-1}+x{F}_{m-2})},\hfill \end{array}$$

which validates the first expression (9) in the theorem.

In order to derive the alternative generating function (10), we recall one of Lagrange’s expansion formulae [13] (cf. Chu [14,15] and Comtet [9] (§3.8)). For a formal power series $\phi \left(x\right)$ subject to the condition $\phi \left(0\right)\ne 0$, the functional equation $y=x/\phi \left(x\right)$ determines x as an implicit function of y. Then, for another formal power series $G\left(x\right)$ in the variable x, we have the following expansion formula for the composite series:

$$\frac{G\left(x\right(y\left)\right)}{1-x{\phi }^{\prime }\left(x\right)/\phi \left(x\right)}=\sum _{n=0}^{\infty }{y}^n\left[x^n\right]\left\{G\left(x\right){\phi }^n\left(x\right)\right\}.$$

(11)

Now, specifying in the above equality by

$$\phi \left(x\right)=F_m+x{F}_{m-1}:y=\frac{x}{F_m+x{F}_{m-1}}⟺x=\frac{y{F}_m}{1-y{F}_{m-1}},$$

(12)

we can deduce from (9) the corresponding function

$$G\left(x\right)=\frac{(1-x)(F_m+x{F}_{m-1})}{{(1-x)}^2(F_m+x{F}_{m-1})-x^2(F_{m-1}+x{F}_{m-2})}.$$

Under the change in variable $x\to {\displaystyle \frac{y{F}_m}{1-y{F}_{m-1}}}$, the generating function in (10) becomes

$$\begin{array}{cc}\hfill \frac{G\left(x\right)}{1-x{\phi }^{\prime }\left(x\right)/\phi \left(x\right)}& ⟹\frac{1-y(F_m+y{F}_{m-1})}{1-2y(F_m+F_{m-1})+y^2(F_m^2+F_{m-1}^2+F_m{F}_{m-1})+y^3{F}_m(F_{m-1}^2-F_m{F}_{m-2})}\hfill \\ & ==\frac{1-y{F}_{m+1}}{1-2y{F}_{m+1}+2y^2{F}_m^2+{(-1)}^m{y}^2+{(-1)}^m{y}^3{F}_m},\hfill \end{array}$$

where we have invoked the Fibonacci recurrence and Cassini’s formula ([4] (theorem 5.3))

$$F_m=F_{m-1}+F_{m-2}\mathrm{and}{F}_{m-1}^2={(-1)}^m+F_m{F}_{m-2}.$$

(13)

This completes the proof of Theorem 2.          □

4. The Third Circular Sum ${\mathcal{V}}_{\mathit{m}}\left(\mathit{n}\right)$

Analogously, for the circular sum defined by

$${\mathcal{V}}_m\left(n\right)=\sum _{0\le k_1,k_2,\cdots ,k_m\le n}\left(\genfrac{}{}{0pt}{}{n-k_1+1}{2k_m+1}\right)\prod _{ı=1}^{m-1}\left(\genfrac{}{}{0pt}{}{n-k_{ı+1}}{k_{ı}}\right),$$

we are going to show in this section the following theorem.

Theorem 3

(Generating functions).

$$\begin{array}{cc}\hfill {\mathcal{V}}_m\left(n\right)& =\left[x^n\right]\frac{{(F_{m-1}+x{F}_m)}^{n+1}}{{(1-x)}^2(F_{m-1}+x{F}_m)-x^2(F_{m-2}+x{F}_{m-1})}\hfill \end{array}$$

(14)

$$\begin{array}{cc}& \hspace{1em}\hspace{1em}\hspace{1em}=\left[y^n\right]\frac{1-y{F}_{m-1}}{1-2y{F}_{m+1}+2y^2{F}_m^2+{(-1)}^m{y}^2+{(-1)}^m{y}^3{F}_m}.\hfill \end{array}$$

(15)

We record below the first five particular cases, where the first three sequences correspond to “A005314”, “A116423”, and “107839” in [12].

Corollary 2

(Generating functions).

$$\begin{array}{cc}\hfill {\mathcal{V}}_1\left(n\right)& =\left[y^n\right]\frac{1}{1-2y+y^2-y^3};\hfill \\ \hfill {\mathcal{V}}_2\left(n\right)& =\left[y^n\right]\frac{1-y}{1-4y+3y^2+y^3};\hfill \\ \hfill {\mathcal{V}}_3\left(n\right)& =\left[y^n\right]\frac{1-y}{1-6y+7y^2-2y^3};\hfill \\ \hfill {\mathcal{V}}_4\left(n\right)& =\left[y^n\right]\frac{1-2y}{1-10y+19y^2+3y^3};\hfill \\ \hfill {\mathcal{V}}_5\left(n\right)& =\left[y^n\right]\frac{1-3y}{1-16y+49y^2-5y^3}.\hfill \end{array}$$

Proof of Theorem 3. 

According to (8) and the binomial relation

$$\begin{array}{cc}& \left(\genfrac{}{}{0pt}{}{n-k_1+1}{2k_m+1}\right)=\left[x^{n-k_1}\right]\frac{x^{2k_m}}{{(1-x)}^{2+2k_m}},\hfill \end{array}$$

we can express the binomial sum with respect to $k_1$ in ${\mathcal{V}}_m\left(n\right)$ as

$${\mathcal{V}}_m^1\left(n\right)=\sum _{k_1=0}^n\left(\genfrac{}{}{0pt}{}{n-k_2}{k_1}\right)\left(\genfrac{}{}{0pt}{}{n-k_1+1}{2k_m+1}\right)=\left[x^n\right]\frac{x^{2k_m}{(1+x)}^n}{{(1-x)}^{2+2k_m}{(1+x)}^{k_2}}.$$

Then, we can proceed with the binomial sum with respect to $k_2$ in ${\mathcal{V}}_m\left(n\right)$:

$$\begin{array}{cc}\hfill {\mathcal{V}}_m^2\left(n\right)& =\sum _{k_2=0}^n{\mathcal{V}}_m^1\left(n\right)\left(\genfrac{}{}{0pt}{}{n-k_3}{k_2}\right)\hfill \\ & =\left[x^n\right]\frac{x^{2k_m}{(1+x)}^n}{{(1-x)}^{2+2k_m}}\sum _{k_2=0}^n\left(\genfrac{}{}{0pt}{}{n-k_3}{k_2}\right){\left(\frac{1}{1+x}\right)}^{k_2}\hfill \\ & =\left[x^n\right]\frac{x^{2k_m}{(2+x)}^n}{{(1-x)}^{2+2k_m}}\times {\left\{\frac{1+x}{2+x}\right\}}^{k_3}\hfill \\ & =\left[x^n\right]\frac{x^{2k_m}{(F_3+x{F}_2)}^n}{{(1-x)}^{2+2k_m}}\times {\left\{\frac{F_2+x{F}_1}{F_3+x{F}_2}\right\}}^{k_3}.\hfill \end{array}$$

Iterating this process, we have the binomial sum with respect to $k_{\ell }$ in ${\mathcal{V}}_m\left(n\right)$:

$$\begin{array}{cc}\hfill {\mathcal{V}}_m^{\ell }\left(n\right)& =\sum _{k_{\ell }=0}^n{\mathcal{V}}_m^{\ell -1}\left(n\right)\left(\genfrac{}{}{0pt}{}{n-k_{\ell +1}}{k_{\ell }}\right)\hfill \\ & =\left[x^n\right]\frac{x^{2k_m}{(F_{\ell +1}+x{F}_{\ell })}^n}{{(1-x)}^{2+2k_m}}\times {\left\{\frac{F_{\ell }+x{F}_{\ell -1}}{F_{\ell +1}+x{F}_{\ell }}\right\}}^{k_{\ell +1}}.\hfill \end{array}$$

Then, the penultimate binomial sum with respect to $k_{m-1}$ in ${\mathcal{V}}_m\left(n\right)$ results in

$$\begin{array}{cc}\hfill {\mathcal{V}}_m^{m-1}\left(n\right)& =\sum _{k_{m-1}=0}^n{\mathcal{V}}_m^{m-2}\left(n\right)\left(\genfrac{}{}{0pt}{}{n-k_m}{k_{m-1}}\right)\hfill \\ & =\left[x^n\right]\frac{x^{2k_m}{(F_m+x{F}_{m-1})}^n}{{(1-x)}^{2+2k_m}}\times {\left\{\frac{F_{m-1}+x{F}_{m-2}}{F_m+x{F}_{m-1}}\right\}}^{k_m}.\hfill \end{array}$$

Finally, we can express the circular binomial sum ${\mathcal{V}}_m\left(n\right)$ as

$$\begin{array}{cc}\hfill {\mathcal{V}}_m\left(n\right)=& \sum _{k_m=0}^n{\mathcal{V}}_m^{m-1}\left(n\right)\hfill \\ \hfill =& \left[x^n\right]\frac{{(F_m+x{F}_{m-1})}^n}{{(1-x)}^2}\sum _{k_m=0}^{\infty }{\left\{\frac{x^2(F_{m-1}+x{F}_{m-2})}{{(1-x)}^2(F_m+x{F}_{m-1})}\right\}}^{k_m}\hfill \\ \hfill =& \left[x^n\right]\frac{{(F_m+x{F}_{m-1})}^{n+1}}{{(1-x)}^2(F_m+x{F}_{m-1})-x^2(F_{m-1}+x{F}_{m-2})},\hfill \end{array}$$

which is exactly the first expression (14) in Theorem 3.

To derive another generating function (15), by combining (11) with (12), we can deduce from (14) the corresponding function

$$G\left(x\right)=\frac{F_m+x{F}_{m-1}}{{(1-x)}^2(F_m+x{F}_{m-1})-x^2(F_{m-1}+x{F}_{m-2})}.$$

Under the change of variable $x\to {\displaystyle \frac{y{F}_m}{1-y{F}_{m-1}}}$, we can convert the generating function in (15) into the following one

$$\begin{array}{cc}\hfill \frac{G\left(x\right)}{1-x{\phi }^{\prime }\left(x\right)/\phi \left(x\right)}& ⟹\frac{1-y{F}_{m-1}}{1-2y(F_m+F_{m-1})+y^2(F_m^2+F_{m-1}^2+F_m{F}_{m-1})+y^3{F}_m(F_{m-1}^2-F_m{F}_{m-2})}\hfill \\ & ==\frac{1-y{F}_{m-1}}{1-2y{F}_{m+1}+2y^2{F}_m^2+{(-1)}^m{y}^2+{(-1)}^m{y}^3{F}_m},\hfill \end{array}$$

where the last step is justified again by (13).              □

5. The Fourth Circular Sum ${\mathrm{U}}_{\mathit{m}}\left(\mathit{n}\right)$

This section will be devoted to the circular sum defined by

$${\mathrm{U}}_m\left(n\right)=\sum _{0\le k_1,k_2,\cdots ,k_m\le n}\left(\genfrac{}{}{0pt}{}{n+k_1}{2k_m}\right)\prod _{ı=1}^{m-1}\left(\genfrac{}{}{0pt}{}{n-k_{ı+1}}{k_{ı}}\right).$$

Its generating functions are determined by the following theorem.

Theorem 4

(Generating functions).

$$\begin{array}{cc}\hfill {\mathrm{U}}_m\left(n\right)& =\left[x^{2n}\right]\frac{(1-x){(F_{m-1}+x{F}_m)}^{n+1}}{{(1-x)}^2(F_{m-1}+x{F}_m)-x^2(F_{m-2}+x{F}_{m-1})}\hfill \end{array}$$

(16)

$$\begin{array}{cc}& \hspace{1em}\hspace{1em}\hspace{1em}=\left[y^n\right]\frac{1-2y{F}_m-{(-1)}^m{y}^2}{1-4y{F}_m-y^2{F}_m^2+5y^2{F}_m{F}_{m-2}+{(-1)}^m{y}^3{F}_{m-2}}.\hfill \end{array}$$

(17)

The first five examples of this theorem are highlighted as in the corollary below, where for $m=1$ and $m=2$ the corresponding sequences are Fibonacci numbers “A001519” and “A014445” in [12].

Corollary 3

(Generating functions).

$$\begin{array}{cc}\hfill {\mathrm{U}}_1\left(n\right)& =\left[y^n\right]\frac{1-y}{1-3y+y^2}=F_{2n+1};\hfill \\ \hfill {\mathrm{U}}_2\left(n\right)& =\left[y^n\right]\frac{1-2y-y^2}{1-4y-y^2}=F_{3n};\hfill \\ \hfill {\mathrm{U}}_3\left(n\right)& =\left[y^n\right]\frac{1-4y+y^2}{1-8y+6y^2-y^3};\hfill \\ \hfill {\mathrm{U}}_4\left(n\right)& =\left[y^n\right]\frac{1-6y-y^2}{1-12y+6y^2+y^3};\hfill \\ \hfill {\mathrm{U}}_5\left(n\right)& =\left[y^n\right]\frac{1-10y+y^2}{1-20y+25y^2-2y^3}.\hfill \end{array}$$

Proof of Theorem 4. 

By making use of the following relations

$$\begin{array}{cc}& \left(\genfrac{}{}{0pt}{}{n-k_2}{k_1}\right)=\left[x^{n-k_1-k_2}\right]{(1+x)}^{n-k_2},\hfill \\ & \left(\genfrac{}{}{0pt}{}{n+k_1}{2k_m}\right)=\left[x^{n+k_1}\right]\frac{x^{2k_m}}{{(1-x)}^{1+2k_m}};\hfill \end{array}$$

(18)

we can express the binomial sum with respect to $k_1$ in ${\mathrm{U}}_m\left(n\right)$ as follows:

$${\mathrm{U}}_m^1\left(n\right)=\sum _{k_1=0}^n\left(\genfrac{}{}{0pt}{}{n-k_2}{k_1}\right)\left(\genfrac{}{}{0pt}{}{n+k_1}{2k_m}\right)=\left[x^{2n}\right]\frac{x^{2k_m}{(1+x)}^n}{{(1-x)}^{1+2k_m}}\frac{x^{k_2}}{{(1+x)}^{k_2}}.$$

Then, we can proceed with the binomial sum with respect to $k_2$ in ${\mathrm{U}}_m\left(n\right)$:

$$\begin{array}{cc}\hfill {\mathrm{U}}_m^2\left(n\right)& =\sum _{k_2=0}^n{\mathrm{U}}_m^1\left(n\right)\left(\genfrac{}{}{0pt}{}{n-k_3}{k_2}\right)\hfill \\ & =\left[x^{2n}\right]\frac{x^{2k_m}{(1+x)}^n}{{(1-x)}^{1+2k_m}}\sum _{k_2=0}^n\left(\genfrac{}{}{0pt}{}{n-k_3}{k_2}\right){\left(\frac{x}{1+x}\right)}^{k_2}\hfill \\ & =\left[x^{2n}\right]\frac{x^{2k_m}{(1+2x)}^n}{{(1-x)}^{1+2k_m}}\times {\left\{\frac{1+x}{1+2x}\right\}}^{k_3}\hfill \\ & =\left[x^{2n}\right]\frac{x^{2k_m}{(F_2+x{F}_3)}^n}{{(1-x)}^{1+2k_m}}\times {\left\{\frac{F_1+x{F}_2}{F_2+x{F}_3}\right\}}^{k_3}.\hfill \end{array}$$

Iterating this process, we have the binomial sum with respect to $k_{\ell }$ in ${\mathrm{U}}_m\left(n\right)$:

$$\begin{array}{cc}\hfill {\mathrm{U}}_m^{\ell }\left(n\right)& =\sum _{k_{\ell }=0}^n{\mathrm{U}}_m^{\ell -1}\left(n\right)\left(\genfrac{}{}{0pt}{}{n-k_{\ell +1}}{k_{\ell }}\right)\hfill \\ & =\left[x^{2n}\right]\frac{x^{2k_m}{(F_{\ell }+x{F}_{\ell +1})}^n}{{(1-x)}^{1+2k_m}}\times {\left\{\frac{F_{\ell -1}+x{F}_{\ell }}{F_{\ell }+x{F}_{\ell +1}}\right\}}^{k_{\ell +1}}.\hfill \end{array}$$

The penultimate binomial sum with respect to $k_{m-1}$ in ${\mathrm{U}}_m\left(n\right)$ results in

$$\begin{array}{cc}\hfill {\mathrm{U}}_m^{m-1}\left(n\right)& =\sum _{k_{m-1}=0}^n{\mathrm{U}}_m^{m-2}\left(n\right)\left(\genfrac{}{}{0pt}{}{n-k_m}{k_{m-1}}\right)\hfill \\ & =\left[x^{2n}\right]\frac{x^{2k_m}{(F_{m-1}+x{F}_m)}^n}{{(1-x)}^{1+2k_m}}\times {\left\{\frac{F_{m-2}+x{F}_{m-1}}{F_{m-1}+x{F}_m}\right\}}^{k_m}.\hfill \end{array}$$

Finally, we can express the circular binomial sum ${\mathrm{U}}_m\left(n\right)$ as

$$\begin{array}{cc}\hfill {\mathrm{U}}_m\left(n\right)=& \sum _{k_m=0}^n{\mathrm{U}}_m^{m-1}\left(n\right)\hfill \\ \hfill =& \left[x^{2n}\right]\frac{{(F_{m-1}+x{F}_m)}^n}{1-x}\sum _{k_m=0}^{\infty }{\left\{\frac{x^2(F_{m-2}+x{F}_{m-1})}{{(1-x)}^2(F_{m-1}+x{F}_m)}\right\}}^{k_m}\hfill \\ \hfill =& \left[x^{2n}\right]\frac{(1-x){(F_{m-1}+x{F}_m)}^{n+1}}{{(1-x)}^2(F_{m-1}+x{F}_m)-x^2(F_{m-2}+x{F}_{m-1})},\hfill \end{array}$$

which validates the first expression (16) in the theorem.

In order to derive the alternative expression (17), we define the polynomial

$${\varphi }_n\left(x\right)=(1-x){(F_{m-1}+x{F}_m)}^{n+1}$$

and consider the linear combination

$${\mathrm{U}}_m\left(n\right)-4F_m{\mathrm{U}}_m(n-1)-F_m(F_m-5F_{m-2}){\mathrm{U}}_m(n-2)+{(-1)}^m{F}_{m-2}{\mathrm{U}}_m(n-3)=\left[x^{2n}\right]\Phi \left(x\right).$$

According to (16), the rational function $\Phi \left(x\right)$ is given by

$$\Phi \left(x\right)=\frac{{\varphi }_n\left(x\right)-4F_m{x}^2{\varphi }_{n-1}\left(x\right)-F_m(F_m-5F_{m-2})x^4{\varphi }_{n-2}\left(x\right)+{(-1)}^m{F}_{m-2}{x}^6{\varphi }_{n-3}\left(x\right)}{{(1-x)}^2(F_{m-1}+x{F}_m)-x^2(F_{m-2}+x{F}_{m-1})}.$$

Keeping in mind (13), we can rewrite the denominator of $\Phi \left(x\right)$

$$\begin{array}{cc}& {(1-x)}^2(F_{m-1}+x{F}_m)-x^2(F_{m-2}+x{F}_{m-1})\hfill \\ & =F_{m-1}(1-x-x^2)+x{F}_{m-2}(1-3x+x^2)\hfill \end{array}$$

and factorize the numerator of $\Phi \left(x\right)$ into

$$\begin{array}{cc}& (1-x){(F_{m-1}+x{F}_m)}^{n-2}\left\{F_{m-1}(1-x-x^2)+x{F}_{m-2}(1-3x+x^2)\right\}\hfill \\ \hfill \times & \left\{(1+x)(1+x+x^2)F_{m-1}^2+x(2+x+x^2)F_m{F}_{m-1}+(1-x)x^2{F}_m^2\right\}.\hfill \end{array}$$

Therefore, $\Phi \left(x\right)$ turns out to be a polynomial of degree $n+2$ in x with the coefficient of $x^{2n}$ equal to zero when $n>2$. Consequently, ${\mathrm{U}}_m\left(n\right)$ satisfies for $n>2$ the following three-term recurrence relation:

$${\mathrm{U}}_m\left(n\right)=4F_m{\mathrm{U}}_m(n-1)+F_m(F_m-5F_{m-2}){\mathrm{U}}_m(n-2)-{(-1)}^m{F}_{m-2}{\mathrm{U}}_m(n-3).$$

By applying (16), we can also determine the initial values:

$$\begin{array}{cc}\hfill {\mathrm{U}}_m\left(0\right)& =1;\hfill \\ \hfill {\mathrm{U}}_m\left(1\right)& =2F_m;\hfill \\ \hfill {\mathrm{U}}_m\left(2\right)& =5F_m{F}_{m+1}-F_{m+1}{F}_{m-1}.\hfill \end{array}$$

Then, it is not hard to establish the generating function

$$\begin{array}{cc}\hfill G\left(y\right)& =\sum _{n\ge 0}{\mathrm{U}}_m\left(n\right)y^n={\mathrm{U}}_m\left(0\right)+y{\mathrm{U}}_m\left(1\right)+y^2{\mathrm{U}}_m\left(2\right)+\sum _{n\ge 3}{\mathrm{U}}_m\left(n\right)y^n\begin{array}{|c|}\hline n\to k+3\\ \hline\end{array}\hfill \\ & =1+2y{F}_m+y^2{F}_{m+1}(5F_m-F_{m-1})\hfill \\ & +\sum _{k\ge 0}{y}^{k+3}\left\{4F_m{\mathrm{U}}_m(k+2)+F_m(F_m-5F_{m-2}){\mathrm{U}}_m(k+1)-{(-1)}^m{F}_{m-2}{\mathrm{U}}_m\left(k\right)\right\}\hfill \\ & =1+2y{F}_m+y^2{F}_{m+1}(5F_m-F_{m-1})-\left\{4y{F}_m+8y^2{F}_m^2+y^2{F}_m(F_m-5F_{m-2})\right\}\hfill \\ & +G\left(y\right)\left\{4y{F}_m+y^2{F}_m(F_m-5F_{m-2})-{(-1)}^m{y}^3{F}_{m-2}\right\}\hfill \\ & =\frac{1-2y{F}_m+y^2(5F_m{F}_{m+1}+5F_m{F}_{m-2}-F_{m+1}{F}_{m-1}-9F_m^2)}{1-4y{F}_m-y^2{F}_m(F_m-5F_{m-2})+{(-1)}^m{y}^3{F}_{m-2}}.\hfill \end{array}$$

After some simplifications, this confirms the expression in (17).           □

6. The Fifth Circular Sum ${\mathrm{V}}_{\mathit{m}}\left(\mathit{n}\right)$

Analogously, for the circular sum defined by

$${\mathrm{V}}_m\left(n\right)=\sum _{0\le k_1,k_2,\cdots ,k_m\le n}\left(\genfrac{}{}{0pt}{}{n+k_1+1}{2k_m+1}\right)\prod _{ı=1}^{m-1}\left(\genfrac{}{}{0pt}{}{n-k_{ı+1}}{k_{ı}}\right),$$

we are going to show in this section the following theorem.

Theorem 5

(Generating functions).

$$\begin{array}{cc}\hfill {\mathrm{V}}_m\left(n\right)& =\left[x^{2n}\right]\frac{{(F_{m-1}+x{F}_m)}^{n+1}}{{(1-x)}^2(F_{m-1}+x{F}_m)-x^2(F_{m-2}+x{F}_{m-1})}\hfill \end{array}$$

(19)

$$\begin{array}{cc}& \hspace{1em}\hspace{1em}\hspace{1em}=\left[y^n\right]\frac{1+y{F}_{m-3}}{1-4y{F}_m-y^2{F}_m^2+5y^2{F}_{m-2}{F}_m+{(-1)}^m{y}^3{F}_{m-2}}.\hfill \end{array}$$

(20)

We record below the first five particular cases, where for $m=1$ and $m=2$ the sequences are again Fibonacci numbers, corresponding to “A001906” and “A015448” in [12].

Corollary 4

(Generating functions).

$$\begin{array}{cc}\hfill {\mathrm{V}}_1\left(n\right)& =\left[y^n\right]\frac{1}{1-3y+y^2}=F_{2n+2};\hfill \\ \hfill {\mathrm{V}}_2\left(n\right)& =\left[y^n\right]\frac{1+y}{1-4y-y^2}=F_{3n+2};\hfill \\ \hfill {\mathrm{V}}_3\left(n\right)& =\left[y^n\right]\frac{1}{1-8y+6y^2-y^3};\hfill \\ \hfill {\mathrm{V}}_4\left(n\right)& =\left[y^n\right]\frac{1+y}{1-12y+6y^2+y^3};\hfill \\ \hfill {\mathrm{V}}_5\left(n\right)& =\left[y^n\right]\frac{1+y}{1-20y+25y^2-2y^3}.\hfill \end{array}$$

Proof of Theorem 5. 

In view of (18) and the binomial relation

$$\begin{array}{cc}& \left(\genfrac{}{}{0pt}{}{n+k_1+1}{2k_m+1}\right)=\left[x^{n+k_1}\right]\frac{x^{2k_m}}{{(1-x)}^{2+2k_m}},\hfill \end{array}$$

we can express the binomial sum with respect to $k_1$ in ${\mathrm{V}}_m\left(n\right)$ as

$${\mathrm{V}}_m^1\left(n\right)=\sum _{k_1=0}^n\left(\genfrac{}{}{0pt}{}{n-k_2}{k_1}\right)\left(\genfrac{}{}{0pt}{}{n+k_1+1}{2k_m+1}\right)=\left[x^{2n}\right]\frac{x^{2k_m}{(1+x)}^n}{{(1-x)}^{2+2k_m}}\frac{x^{k_2}}{{(1+x)}^{k_2}}.$$

Then, we can proceed with the binomial sum with respect to $k_2$ in ${\mathrm{V}}_m\left(n\right)$:

$$\begin{array}{cc}\hfill {\mathrm{V}}_m^2\left(n\right)& =\sum _{k_2=0}^n{\mathrm{V}}_m^1\left(n\right)\left(\genfrac{}{}{0pt}{}{n-k_3}{k_2}\right)\hfill \\ & =\left[x^{2n}\right]\frac{x^{2k_m}{(1+x)}^n}{{(1-x)}^{2+2k_m}}\sum _{k_2=0}^n\left(\genfrac{}{}{0pt}{}{n-k_3}{k_2}\right){\left(\frac{x}{1+x}\right)}^{k_2}\hfill \\ & =\left[x^{2n}\right]\frac{x^{2k_m}{(1+2x)}^n}{{(1-x)}^{2+2k_m}}\times {\left\{\frac{1+x}{1+2x}\right\}}^{k_3}\hfill \\ & =\left[x^{2n}\right]\frac{x^{2k_m}{(F_2+x{F}_3)}^n}{{(1-x)}^{2+2k_m}}\times {\left\{\frac{F_1+x{F}_2}{F_2+x{F}_3}\right\}}^{k_3}.\hfill \end{array}$$

Iterating this process, we have the binomial sum with respect to $k_{\ell }$ in ${\mathrm{V}}_m\left(n\right)$:

$$\begin{array}{cc}\hfill {\mathrm{V}}_m^{\ell }\left(n\right)& =\sum _{k_{\ell }=0}^n{\mathrm{V}}_m^{\ell -1}\left(n\right)\left(\genfrac{}{}{0pt}{}{n-k_{\ell +1}}{k_{\ell }}\right)\hfill \\ & =\left[x^{2n}\right]\frac{x^{2k_m}{(F_{\ell }+x{F}_{\ell +1})}^n}{{(1-x)}^{2+2k_m}}\times {\left\{\frac{F_{\ell -1}+x{F}_{\ell }}{F_{\ell }+x{F}_{\ell +1}}\right\}}^{k_{\ell +1}}.\hfill \end{array}$$

Then, the penultimate binomial sum with respect to $k_{m-1}$ in ${\mathrm{V}}_m\left(n\right)$ results in

$$\begin{array}{cc}\hfill {\mathrm{V}}_m^{m-1}\left(n\right)& =\sum _{k_{m-1}=0}^n{\mathrm{V}}_m^{m-2}\left(n\right)\left(\genfrac{}{}{0pt}{}{n-k_m}{k_{m-1}}\right)\hfill \\ & =\left[x^{2n}\right]\frac{x^{2k_m}{(F_{m-1}+x{F}_m)}^n}{{(1-x)}^{2+2k_m}}\times {\left\{\frac{F_{m-2}+x{F}_{m-1}}{F_{m-1}+x{F}_m}\right\}}^{k_m}.\hfill \end{array}$$

Finally, we can express the circular binomial sum ${\mathrm{V}}_m\left(n\right)$ as

$$\begin{array}{cc}\hfill {\mathrm{V}}_m\left(n\right)=& \sum _{k_m=0}^n{\mathrm{V}}_m^{m-1}\left(n\right)\hfill \\ \hfill =& \left[x^{2n}\right]\frac{{(F_{m-1}+x{F}_m)}^n}{{(1-x)}^2}\sum _{k_m=0}^{\infty }{\left\{\frac{x^2(F_{m-2}+x{F}_{m-1})}{{(1-x)}^2(F_{m-1}+x{F}_m)}\right\}}^{k_m}\hfill \\ \hfill =& \left[x^{2n}\right]\frac{{(F_{m-1}+x{F}_m)}^{n+1}}{{(1-x)}^2(F_{m-1}+x{F}_m)-x^2(F_{m-2}+x{F}_{m-1})},\hfill \end{array}$$

which is exactly the first expression (19) in the theorem.

In order to prove the alternative generating function (20), define the polynomial

$${\psi }_n\left(x\right)={(F_{m-1}+x{F}_m)}^{n+1}.$$

Then, we can examine the linear combination

$${\mathrm{V}}_m\left(n\right)-4F_m{\mathrm{V}}_m(n-1)-F_m(F_m-5F_{m-2}){\mathrm{V}}_m(n-2)+{(-1)}^m{F}_{m-2}{\mathrm{V}}_m(n-3)=\left[x^{2n}\right]\mathrm{\Psi }\left(x\right).$$

Taking into account (19), the rational function $\mathrm{\Psi }\left(x\right)$ is given by

$$\mathrm{\Psi }\left(x\right)=\frac{{\psi }_n\left(x\right)-4F_m{x}^2{\psi }_{n-1}\left(x\right)-F_m(F_m-5F_{m-2})x^4{\psi }_{n-2}\left(x\right)+{(-1)}^m{F}_{m-2}{x}^6{\psi }_{n-3}\left(x\right)}{{(1-x)}^2(F_{m-1}+x{F}_m)-x^2(F_{m-2}+x{F}_{m-1})}.$$

Analogous to the proof for (17), we can factorize the numerator of $\mathrm{\Psi }\left(x\right)$ into

$$\begin{array}{cc}& {(F_{m-1}+x{F}_m)}^{n-2}\left\{F_{m-1}(1-x-x^2)+x{F}_{m-2}(1-3x+x^2)\right\}\hfill \\ \hfill \times & \left\{(1+x)(1+x+x^2)F_{m-1}^2+x(2+x+x^2)F_m{F}_{m-1}+(1-x)x^2{F}_m^2\right\},\hfill \end{array}$$

which results in a multiple of the denominator polynomial of $\mathrm{\Psi }\left(x\right)$. Therefore, $\mathrm{\Psi }\left(x\right)$ is, in effect, a polynomial of degree $n+1$ in x with the coefficient of $x^{2n}$ equal to zero when $n>1$. Consequently, ${\mathrm{V}}_m\left(n\right)$ satisfies for $n>2$ the same three-term recurrence relation as ${\mathrm{U}}_m\left(n\right)$:

$${\mathrm{V}}_m\left(n\right)=4F_m{\mathrm{V}}_m(n-1)+F_m(F_m-5F_{m-2}){\mathrm{V}}_m(n-2)-{(-1)}^m{F}_{m-2}{\mathrm{V}}_m(n-3).$$

By means of (19), we can also determine the initial values:

$$\begin{array}{cc}\hfill {\mathrm{V}}_m\left(0\right)& =1;\hfill \\ \hfill {\mathrm{V}}_m\left(1\right)& =3F_m+2F_{m-1};\hfill \\ \hfill {\mathrm{V}}_m\left(2\right)& =8F_m^2+13F_m{F}_{m-1}.\hfill \end{array}$$

Now, it is not hard to establish the generating function

$$\begin{array}{cc}\hfill G\left(y\right)& =\sum _{n\ge 0}{\mathrm{V}}_m\left(n\right)y^n={\mathrm{V}}_m\left(0\right)+y{\mathrm{V}}_m\left(1\right)+y^2{\mathrm{V}}_m\left(2\right)+\sum _{n\ge 3}{\mathrm{V}}_m\left(n\right)y^n\begin{array}{|c|}\hline n\to k+3\\ \hline\end{array}\hfill \\ & =1+3y{F}_m+2y{F}_{m-1}+8y^2{F}_m^2+13y^2{F}_m{F}_{m-1}\hfill \\ & +\sum _{k\ge 0}{y}^{k+3}\left\{4F_m{\mathrm{V}}_m(k+2)+F_m(F_m-5F_{m-2}){\mathrm{V}}_m(k+1)-{(-1)}^m{F}_{m-2}{\mathrm{V}}_m\left(k\right)\right\}\hfill \\ & =1+3y{F}_m+2y{F}_{m-1}+8y^2{F}_m^2+13y^2{F}_m{F}_{m-1}\hfill \\ & -\left\{4y{F}_m+4y^2{F}_m(3F_m+2F_{m-1})+y^2{F}_m(F_m-5F_{m-2})\right\}\hfill \\ & +G\left(y\right)\left\{4y{F}_m+y^2{F}_m(F_m-5F_{m-2})-{(-1)}^m{y}^3{F}_{m-2}\right\}\hfill \\ & =\frac{1+y{F}_{m-3}+5y^2(F_m{F}_{m-1}+F_m{F}_{m-2}-F_m^2)}{1-4y{F}_m-y^2{F}_m(F_m-5F_{m-2})+{(-1)}^m{y}^3{F}_{m-2}}.\hfill \end{array}$$

This is equivalent to (20) since the coefficient of $y^2$ in the numerator vanishes.     □

7. Conclusions and Further Observations

By manipulating binomial series, we have succeeded in deriving generating functions for five classes of multiple sums of circular binomial products. Just like other binomial sums, they may find potential applications in computer science (e.g., the analysis of algorithms), numerical analysis, number theory, and probability theory (e.g., binomial distributions).

The informed reader may notice the differences in showing the generating functions claimed in Theorems 2, 3, 4, and 5. For the former two (10) and (15), we made use of the Lagrange expansion formula, which is very useful in classical analysis, number theory, and combinatorics. In order to illustrate an alternative approach, we just verified (17) and (20) by means of the recurrence relations.

Observing Theorems 2 and 3, we obtain the following functional equality:

$$(1-y{F}_{m-1})\sum _{n=0}^{\infty }{\mathcal{U}}_m\left(n\right)y^n=(1-y{F}_{m+1})\sum _{n=0}^{\infty }{\mathcal{V}}_m\left(n\right)y^n.$$

By comparing the coefficient of $y^n$ across, we find the curious relation below:

$${\mathcal{U}}_m\left(n\right)-F_{m-1}{\mathcal{U}}_m(n-1)={\mathcal{V}}_m\left(n\right)-F_{m+1}{\mathcal{V}}_m(n-1)\mathrm{for}n\ge 1.$$

(21)

Analogously, Theorems 4 and 5 imply the equality

$$(1+y{F}_{m-3})\sum _{n=0}^{\infty }{\mathrm{U}}_m\left(n\right)y^n=\left(1-2y{F}_m-{(-1)}^m{y}^2\right)\sum _{n=0}^{\infty }{\mathrm{V}}_m\left(n\right)y^n.$$

Then, the coefficient of $y^n$ across this equation yields another relation:

$${\mathrm{U}}_m\left(n\right)+F_{m-3}{\mathrm{U}}_m(n-1)={\mathrm{V}}_m\left(n\right)-2F_m{\mathrm{V}}_m(n-1)-{(-1)}^m{\mathrm{V}}_m(n-2)\mathrm{for}n\ge 2.$$

(22)

An intriguing problem is how to construct combinatorial proofs for (21) and (22) as constructed in [7,11]?

Furthermore, replacing the weight functions “${\Delta }_n(k_1,k_m)$ by ${\Delta }_n(k_1,\lambda k_m)$” in (4), it is not difficult to express the corresponding multiple sums (3) as coefficients of rational functions in the variable “x”. However, it would be more involved to determine related generating functions in the variable “y”. In addition, there are also slightly modified multiple binomial sums [8] with their values beyond Fibonacci and Lucas numbers, that suggest different number sequences as cyclic factors in the multiple sums. It is worthwhile making further efforts to delve into these topics.