Generating Functions for Four Classes of Triple Binomial Sums

Abstract

By means of the generating function approach, four classes of triple sums involving circular products of binomial coefficients are investigated. Recurrence relations and rational generating functions are established.

1. Introduction and Motivation

Evaluating binomial sums has always been a fascinating topic in pure mathematics and applied sciences (cf. [1,2,3]). There exist numerous binomial identities (cf. [4,5,6,7,8]) and their multifold counterparts (cf. [9,10,11]), scattered in the literature. Different techniques (cf. Spivey [12]) have been utilized so far to treat binomial identities, such as integral representations [13], the hypergeometric series approach ([14] §5.5), and the creative telescoping [15].

For many combinatorial structures, the related counting and enumeration problems can efficiently be resolved by generating function techniques (cf. [16,17,18,19]). In particular, the generating function method is also powerful in computing convolution sums (see for example [20,21,22]). Kilic and Prodinger [23] found, by making use of generating functions and the Lagrange expansion theorem (cf. Comtet [1] §3.8), the following elegant binomial formulas (see [23] Theorems 1–3):

$$\begin{array}{cc}\hfill F_{4n-1}& =\sum _{i=0}^n\sum _{j=0}^n\left(\genfrac{}{}{0pt}{}{n+i}{2j}\right)\left(\genfrac{}{}{0pt}{}{n+j}{2i}\right),\hfill \end{array}$$

(1)

$$\begin{array}{cc}\hfill F_{4n-3}& =\sum _{i=0}^n\sum _{j=0}^n\left(\genfrac{}{}{0pt}{}{n+i}{2j+1}\right)\left(\genfrac{}{}{0pt}{}{n+j}{2i+1}\right),\hfill \end{array}$$

(2)

$$\begin{array}{cc}\hfill F_{4n}& =\sum _{i=0}^n\sum _{j=1}^n\left(\genfrac{}{}{0pt}{}{n+i}{2j-1}\right)\left(\genfrac{}{}{0pt}{}{n+j}{2i}\right);\hfill \end{array}$$

(3)

where $F_n$ stands for the Fibonacci number (cf. [24]). By carrying out the same procedure of Kilic and Prodinger, it is not hard to show the following further results:

• There is an extra analogous identity

  $$F_{4n-2}=\sum _{i=0}^n\sum _{j=0}^n\left(\genfrac{}{}{0pt}{}{n+i}{2j}\right)\left(\genfrac{}{}{0pt}{}{n+j}{2i+1}\right).$$
• For the three alternating sums defined by

  $$\begin{array}{cc}\hfill {\mathsf{\Phi }}_a\left(n\right)& =\sum _{i=0}^n\sum _{j=0}^n{(-1)}^{i+j}\left(\genfrac{}{}{0pt}{}{n+i}{2j}\right)\left(\genfrac{}{}{0pt}{}{n+j}{2i}\right),\hfill \\ \hfill {\mathsf{\Phi }}_b\left(n\right)& =\sum _{i=0}^n\sum _{j=0}^n{(-1)}^{i+j}\left(\genfrac{}{}{0pt}{}{n+i}{2j}\right)\left(\genfrac{}{}{0pt}{}{n+j+1}{2i+1}\right),\hfill \\ \hfill {\mathsf{\Phi }}_c\left(n\right)& =\sum _{i=0}^n\sum _{j=0}^n{(-1)}^{i+j}\left(\genfrac{}{}{0pt}{}{n+i+1}{2j+1}\right)\left(\genfrac{}{}{0pt}{}{n+j+1}{2i+1}\right);\hfill \end{array}$$

  they satisfy the common recurrence relation

  $$\mathsf{\Phi }\left(n\right)=3\mathsf{\Phi }(n-1)+3\mathsf{\Phi }(n-2)+4\mathsf{\Phi }(n-3)$$

  with different initial values

  $$\begin{array}{ccc}{\mathsf{\Phi }}_a\left(0\right)=1,\hfill & {\mathsf{\Phi }}_a\left(1\right)=2,\hfill & {\mathsf{\Phi }}_a\left(2\right)=9;\hfill \\ {\mathsf{\Phi }}_b\left(0\right)=1,\hfill & {\mathsf{\Phi }}_b\left(1\right)=3,\hfill & {\mathsf{\Phi }}_b\left(2\right)=11;\hfill \\ {\mathsf{\Phi }}_c\left(0\right)=1,\hfill & {\mathsf{\Phi }}_c\left(1\right)=5,\hfill & {\mathsf{\Phi }}_c\left(2\right)=18.\hfill \end{array}$$
• For the three (partially alternating) sums defined by

  $$\begin{array}{cc}\hfill {\mathsf{\Psi }}_a\left(n\right)& =\sum _{i=0}^n\sum _{j=0}^n{(-1)}^i\left(\genfrac{}{}{0pt}{}{n+i}{2j}\right)\left(\genfrac{}{}{0pt}{}{n+j}{2i}\right),\hfill \\ \hfill {\mathsf{\Psi }}_b\left(n\right)& =\sum _{i=0}^n\sum _{j=0}^n{(-1)}^i\left(\genfrac{}{}{0pt}{}{n+i}{2j}\right)\left(\genfrac{}{}{0pt}{}{n+j+1}{2i+1}\right),\hfill \\ \hfill {\mathsf{\Psi }}_c\left(n\right)& =\sum _{i=0}^n\sum _{j=0}^n{(-1)}^i\left(\genfrac{}{}{0pt}{}{n+i+1}{2j+1}\right)\left(\genfrac{}{}{0pt}{}{n+j+1}{2i+1}\right);\hfill \end{array}$$

  they satisfy the same recurrence relation

  $$\mathsf{\Psi }\left(n\right)=\mathsf{\Psi }(n-1)-13\mathsf{\Psi }(n-2)-4\mathsf{\Psi }(n-3)$$

  with different initial values

  $$\begin{array}{ccc}{\mathsf{\Psi }}_a\left(0\right)=1,\hfill & {\mathsf{\Psi }}_a\left(1\right)=0,\hfill & {\mathsf{\Psi }}_a\left(2\right)=-7;\hfill \\ {\mathsf{\Psi }}_b\left(0\right)=1,\hfill & {\mathsf{\Psi }}_b\left(1\right)=1,\hfill & {\mathsf{\Psi }}_b\left(2\right)=-5;\hfill \\ {\mathsf{\Psi }}_c\left(0\right)=1,\hfill & {\mathsf{\Psi }}_c\left(1\right)=3,\hfill & {\mathsf{\Psi }}_c\left(2\right)=-6.\hfill \end{array}$$

As demanded by an anonymous referee, a sample proof for the recurrence relation satisfied by ${\mathsf{\Phi }}_c\left(n\right)$ is appended at the end of Section 5. For more circular binomial sums, the reader can consult the papers [10,25,26].

Motivated by these remarkable results, we shall examine four classes of triple circular sums by means of the generating function approach (cf. Wilf [27] and [28,29,30]). They will be divided into four separate sections. What remarkable is that the four triple sums in each class satisfy the same recurrence relation. The rational generating functions for these triple sums will be determined explicitly. In order to ensure accuracy, all the displayed mathematical expressions are numerically verified throughout the paper by appropriately devised “Mathematica” commands. In addition, we declare that all the power series appearing in this paper are convergent in the neighborhood of zero.

2. Positive Triple Sums

This section will be devoted to the four positive triple sums defined by

$$\begin{array}{cc}\hfill T_a\left(n\right)& :=\sum _{i=0}^n\sum _{j=0}^n\sum _{k=0}^n\left(\genfrac{}{}{0pt}{}{n+i}{2j}\right)\left(\genfrac{}{}{0pt}{}{n+j}{2k}\right)\left(\genfrac{}{}{0pt}{}{n+k}{2i}\right),\hfill \end{array}$$

(4)

$$\begin{array}{cc}\hfill T_b\left(n\right)& :=\sum _{i=0}^n\sum _{j=0}^n\sum _{k=0}^n\left(\genfrac{}{}{0pt}{}{n+i}{2j}\right)\left(\genfrac{}{}{0pt}{}{n+j}{2k}\right)\left(\genfrac{}{}{0pt}{}{n+k+1}{2i+1}\right),\hfill \end{array}$$

(5)

$$\begin{array}{cc}\hfill T_c\left(n\right)& :=\sum _{i=0}^n\sum _{j=0}^n\sum _{k=0}^n\left(\genfrac{}{}{0pt}{}{n+i}{2j}\right)\left(\genfrac{}{}{0pt}{}{n+j+1}{2k+1}\right)\left(\genfrac{}{}{0pt}{}{n+k+1}{2i+1}\right),\hfill \end{array}$$

(6)

$$\begin{array}{cc}\hfill T_d\left(n\right)& :=\sum _{i=0}^n\sum _{j=0}^n\sum _{k=0}^n\left(\genfrac{}{}{0pt}{}{n+i+1}{2j+1}\right)\left(\genfrac{}{}{0pt}{}{n+j+1}{2k+1}\right)\left(\genfrac{}{}{0pt}{}{n+k+1}{2i+1}\right).\hfill \end{array}$$

(7)

These sums satisfy a common recurrence relation and admit compactly rational generating functions. The main results are enunciated as follows.

Lemma 1 (Coefficient expressions).

$$\begin{array}{cc}\hfill \left(\mathrm{a}\right)T_a\left(n\right)& =\left[x^{7n}\right]\frac{(1-2x^2){(1+x)}^n{(1-x^2)}^{2n+1}}{(1-x-x^2)(1+x-2x^2-x^3+x^4+x^6)},\hfill \end{array}$$

(8)

$$\begin{array}{cc}\hfill \left(\mathrm{b}\right)T_b\left(n\right)& =\left[x^{7n}\right]\frac{(1-2x^2){(1+x)}^{n+1}{(1-x^2)}^{2n+1}}{(1-x-x^2)(1+x-2x^2-x^3+x^4+x^6)},\hfill \end{array}$$

(9)

$$\begin{array}{cc}\hfill \left(\mathrm{c}\right)T_c\left(n\right)& =\left[x^{7n}\right]\frac{{(1+x)}^{n+1}{(1-x^2)}^{2n+3}}{(1-x-x^2)(1+x-2x^2-x^3+x^4+x^6)},\hfill \end{array}$$

(10)

$$\begin{array}{cc}\hfill \left(\mathrm{d}\right)T_d\left(n\right)& =\left[x^{7n}\right]\frac{{(1+x)}^{n+1}{(1-x^2)}^{2n+2}}{(1-x-x^2)(1+x-2x^2-x^3+x^4+x^6)}.\hfill \end{array}$$

(11)

Proposition 1. 

The four triple sums $\{T_a\left(n\right),T_b\left(n\right),T_c\left(n\right),T_d\left(n\right)\}$ satisfy the same recurrence relation

$$T\left(n\right)=17T(n-1)+16T(n-2)+17T(n-3)-T(n-4)$$

with different initial values

$$\begin{array}{cc}& {\left\{T_a\left(n\right)\right\}}_{n=0}^3=\left\{1,2,41,745\right\},\hfill \\ & {\left\{T_b\left(n\right)\right\}}_{n=0}^3=\left\{1,3,67,1205\right\},\hfill \\ & {\left\{T_c\left(n\right)\right\}}_{n=0}^3=\left\{1,5,109,1949\right\},\hfill \\ & {\left\{T_d\left(n\right)\right\}}_{n=0}^3=\left\{1,9,176,3153\right\}.\hfill \end{array}$$

Theorem 1 (Generating functions).

$$\begin{array}{cc}\hfill \left(\mathrm{a}\right)\sum _{n=0}^{\infty }{T}_a\left(n\right)y^n& =\frac{1-15y-9y^2-y^3}{(1+y+y^2)(1-18y+y^2)},\hfill \end{array}$$

(12)

$$\begin{array}{cc}\hfill \left(\mathrm{b}\right)\sum _{n=0}^{\infty }{T}_b\left(n\right)y^n& =\frac{1-14y+y^3}{(1+y+y^2)(1-18y+y^2)},\hfill \end{array}$$

(13)

$$\begin{array}{cc}\hfill \left(\mathrm{c}\right)\sum _{n=0}^{\infty }{T}_c\left(n\right)y^n& =\frac{1-12y+8y^2-y^3}{(1+y+y^2)(1-18y+y^2)},\hfill \end{array}$$

(14)

$$\begin{array}{cc}\hfill \left(\mathrm{d}\right)\sum _{n=0}^{\infty }{T}_d\left(n\right)y^n& =\frac{(1-y)(1-7y)}{(1+y+y^2)(1-18y+y^2)}.\hfill \end{array}$$

(15)

By examining two differences of the above generating functions

$$\begin{array}{cc}\hfill \left(12\right)+\left(13\right)-\left(14\right)& =\frac{1-y^2}{1-y^3},\hfill \\ \hfill \left(13\right)+\left(14\right)-\left(15\right)& =\frac{1-y}{1-y^3};\hfill \end{array}$$

we find that the triples sums $\{T_a\left(n\right),T_b\left(n\right),T_c\left(n\right),T_d\left(n\right)\}$ satisfy the following remarkable relations

$$\begin{array}{c}\hfill T_a\left(n\right)+T_b\left(n\right)-T_c\left(n\right)=\left\{\begin{array}{cc}1,\hfill & n=0(mod3);\hfill \\ 0,\hfill & n=1(mod3);\hfill \\ -1,\hfill & n=2(mod3);\hfill \end{array}\right.\end{array}$$

(16)

$$\begin{array}{c}\hfill T_b\left(n\right)+T_c\left(n\right)-T_d\left(n\right)=\left\{\begin{array}{cc}1,\hfill & n=0(mod3);\hfill \\ -1,\hfill & n=1(mod3);\hfill \\ 0,\hfill & n=2(mod3).\hfill \end{array}\right.\end{array}$$

(17)

However, there exist no such simple relations for other triple sums with alternating signs treated in the remaining next sections of this paper.

We take $T_a\left(n\right)$ as an example to show how to derive expression (8) in Lemma 1, recurrence relation in Proposition 1 and generating function (12) in Theorem 1. Other results can be found analogously.

2.1. Proof of Lemma 1

For a non-negative integer k, let $\left[x^k\right]\varphi \left(x\right)$ stand for the coefficient of $x^k$ in the formal power series $\varphi \left(x\right)$. By means of the following two relations

$$\begin{array}{cc}\hfill \left(\genfrac{}{}{0pt}{}{n+i}{2j}\right)& =\left[x^{2n+2i}\right]\frac{x^{4j}}{{(1-x^2)}^{2j+1}},\hfill \\ \hfill \left(\genfrac{}{}{0pt}{}{n+k}{2i}\right)& =\left[x^{n+k-2i}\right]{(1+x)}^{n+k},\hfill \end{array}$$

the binomial sum with respect to i in $T_a\left(n\right)$ can be expressed as

$$T_a^i\left(n\right):=\sum _{i=0}^n\left(\genfrac{}{}{0pt}{}{n+i}{2j}\right)\left(\genfrac{}{}{0pt}{}{n+k}{2i}\right)=\left[x^{3n+k}\right]\frac{x^{4j}{(1+x)}^{n+k}}{{(1-x^2)}^{2j+1}}.$$

Then, we can proceed with the binomial sum with respect to j in $T_a\left(n\right)$

$$\begin{array}{cc}\hfill T_a^j\left(n\right)& :=\sum _{j=0}^n\left(\genfrac{}{}{0pt}{}{n+j}{2k}\right)T_a^i\left(n\right)\hfill \\ & =\left[x^{3n+k}\right]\frac{{(1+x)}^{n+k}}{1-x^2}\sum _{j=0}^{\infty }\left(\genfrac{}{}{0pt}{}{n+j}{2k}\right)\frac{x^{4j}}{{(1-x^2)}^{2j}},\hfill \end{array}$$

where the upper limit of the sum is released to ∞ because the corresponding coefficient of $x^{3n+k}$ becomes 0 when $j>n$.

For the last series, performing the replacement $j\to m-n$ on the summation index and then writing $y=\frac{x^4}{{(1-x^2)}^2}$, we can evaluate it as follows:

$$\begin{array}{cc}\hfill & \sum _{j=0}^{\infty }\left(\genfrac{}{}{0pt}{}{n+j}{2k}\right)\frac{x^{4j}}{{(1-x^2)}^{2j}}=\sum _{m=n}^{\infty }\left(\genfrac{}{}{0pt}{}{m}{2k}\right)y^{m-n}\hfill \\ \hfill =& \sum _{m=2k}^{\infty }\left(\genfrac{}{}{0pt}{}{m}{2k}\right)y^{m-n}-\sum _{m=2k}^{n-1}\left(\genfrac{}{}{0pt}{}{m}{2k}\right)y^{m-n}\hfill \\ \hfill =& \frac{y^{2k-n}}{{(1-y)}^{2k+1}}-\sum _{m=2k}^{n-1}\left(\genfrac{}{}{0pt}{}{m}{2k}\right)y^{m-n}\hfill \\ \hfill =& \frac{x^{8k-4n}{(1-x^2)}^{2n+2}}{{(1-2x^2)}^{2k+1}}-\sum _{m=2k}^{n-1}\left(\genfrac{}{}{0pt}{}{m}{2k}\right)\frac{x^{4m-4n}}{{(1-x^2)}^{2m-2n}}.\hfill \end{array}$$

(18)

From this, we obtain the following closed expression:

$$\begin{array}{cc}\hfill T_a^j\left(n\right)& =\left[x^{3n+k}\right]\frac{x^{8k-4n}{(1+x)}^{n+k}{(1-x^2)}^{2n+1}}{{(1-2x^2)}^{2k+1}}\hfill \\ & =\left[x^{7n}\right]\frac{x^{7k}{(1+x)}^{n+k}{(1-x^2)}^{2n+1}}{{(1-2x^2)}^{2k+1}}.\hfill \end{array}$$

This is justified by the fact that the contribution from the sum in (18) is annihilated by $\left[x^{7n}\right]$, since for $n>2k$, the following corresponding sum results in a Laurent polynomial with its highest exponent equal to $n+k-2<3n+k$:

$$\frac{{(1+x)}^{n+k}}{1-x^2}\sum _{m=2k}^{n-1}\left(\genfrac{}{}{0pt}{}{m}{2k}\right)\frac{x^{4m-4n}}{{(1-x^2)}^{2m-2n}}=\sum _{m=2k}^{n-1}\left(\genfrac{}{}{0pt}{}{m}{2k}\right)\frac{x^{4m-4n}{(1+x)}^{n+k}}{{(1-x^2)}^{2m-2n+1}}.$$

Consequently, the remaining binomial sum with respect to k in $T_a\left(n\right)$ reads as

$$\begin{array}{cc}\hfill T_a\left(n\right)& =\sum _{k=0}^n{T}_a^j\left(n\right)=\left[x^{7n}\right]\frac{{(1+x)}^n{(1-x^2)}^{2n+1}}{1-2x^2}\sum _{k=0}^{\infty }\frac{x^{7k}{(1+x)}^k}{{(1-2x^2)}^{2k}}.\hfill \end{array}$$

Evaluating the last series gives rise to

$$T_a\left(n\right)=\left[x^{7n}\right]{\mathrm{P}}_n\left(x\right),$$

where ${\mathrm{P}}_n\left(x\right)$ is a rational function defined by

$${\mathrm{P}}_n\left(x\right):=\frac{(1-2x^2){(1+x)}^n{(1-x^2)}^{2n+1}}{(1-x-x^2)(1+x-2x^2-x^3+x^4+x^6)}.$$

This validates the first expression (8) in Lemma 1.

2.2. Proof of Proposition 1

To prove the recurrence relation in Proposition 1, we rewrite it alternatively by

$$\begin{array}{cc}\hfill {\Delta }_a\left(n\right)& :=T_a\left(n\right)-17T_a(n-1)-16T_a(n-2)-17T_a(n-3)+T_a(n-4)\hfill \\ & =\left[x^{7n}\right]\left\{{\mathrm{P}}_n\left(x\right)-17x^7{\mathrm{P}}_{n-1}\left(x\right)-16x^{14}{\mathrm{P}}_{n-2}\left(x\right)-17x^{21}{\mathrm{P}}_{n-3}\left(x\right)+x^{28}{\mathrm{P}}_{n-4}\left(x\right)\right\}.\hfill \end{array}$$

By factorizing the expression inside the braces “$\{\cdots \}$”, we obtain

$${\Delta }_a\left(n\right)=\left[x^{7n}\right]\mathbf{P}\left(x\right),$$

where $\mathbf{P}\left(x\right)$ is a Laurent polynomial given explicitly by

$$\begin{array}{cc}\hfill \mathbf{P}\left(x\right)& :={(1-x)}^{2n-7}{(1+x)}^{3n-11}(1-2x^2)(1+x-2x^2-3x^3+x^4+2x^5+x^8)\hfill \\ & \times \left(1+3x+x^2-4x^3-x^4+7x^5+8x^6-11x^7-24x^8+3x^9+16x^{10}+x^{11}-x^{12}\right).\hfill \end{array}$$

When $n\ge 4$, it is obvious that $\mathbf{P}\left(x\right)$ results in a polynomial of degree $4+5n<7n$. Therefore the coefficient $\left[x^{7n}\right]\mathbf{P}\left(x\right)$ vanishes, which confirms the recurrence relation ${\Delta }_a\left(n\right)=0$ as in Proposition 1.

2.3. Proof of Theorem 1

According to Proposition 1, we can manipulate the following generating function:

$$\begin{array}{cc}\hfill \mathcal{P}\left(y\right)& :=\sum _{n=0}^{\infty }{T}_a\left(n\right)y^n=T_a\left(0\right)+T_a\left(1\right)y+T_a\left(2\right)y^2+T_a\left(3\right)y^3+\sum _{n=4}^{\infty }{T}_a\left(n\right)y^n\hfill \\ & =T_a\left(0\right)+T_a\left(1\right)y+T_a\left(2\right)y^2+T_a\left(3\right)y^3\hfill \\ & +\sum _{n=4}^{\infty }{y}^n\left\{17T_a(n-1)+16T_a(n-2)+17T_a(n-3)-T_a(n-4)\right\}\hfill \\ & =T_a\left(0\right)+T_a\left(1\right)y+T_a\left(2\right)y^2+T_a\left(3\right)y^3+17y\sum _{n=3}^{\infty }{T}_a\left(n\right)y^n\hfill \\ & +16y^2\sum _{n=2}^{\infty }{T}_a\left(n\right)y^n+17y^3\sum _{n=1}^{\infty }{T}_a\left(n\right)y^n-y^4\sum _{n=0}^{\infty }{T}_a\left(n\right)y^n\hfill \\ & =T_a\left(0\right)+T_a\left(1\right)y+T_a\left(2\right)y^2+T_a\left(3\right)y^3+\mathcal{P}\left(y\right)\left\{17y+16y^2+17y^3-y^4\right\}\hfill \\ & -17y\left\{T_a\left(0\right)+T_a\left(1\right)y+T_a\left(2\right)y^2\right\}-16y^2\left\{T_a\left(0\right)+T_a\left(1\right)y\right\}-17y^3{T}_a\left(0\right).\hfill \end{array}$$

Finally, by computing the initial values

$${\left\{T_a\left(n\right)\right\}}_{n=0}^3=\left\{1,2,41,745\right\}$$

and then simplifying the last expression, we have a functional equation

$$\mathcal{P}\left(y\right)=1-15y-9y^2-y^3+\mathcal{P}\left(y\right)\left\{17y+16y^2+17y^3-y^4\right\},$$

which is equivalent to the following generating function as in Theorem 1:

$$\mathcal{P}\left(y\right)=\frac{1-15y-9y^2-y^3}{1-17y-16y^2-17y^3+y^4}=\frac{1-15y-9y^2-y^3}{(1+y+y^2)(1-18y+y^2)}.$$

3. Alternating Triple Sums

For the four alternating triple sums defined by

$$\begin{array}{cc}\hfill U_a\left(n\right)& :=\sum _{i=0}^n\sum _{j=0}^n\sum _{k=0}^n{(-1)}^{i+j+k}\left(\genfrac{}{}{0pt}{}{n+i}{2j}\right)\left(\genfrac{}{}{0pt}{}{n+j}{2k}\right)\left(\genfrac{}{}{0pt}{}{n+k}{2i}\right),\hfill \end{array}$$

(19)

$$\begin{array}{cc}\hfill U_b\left(n\right)& :=\sum _{i=0}^n\sum _{j=0}^n\sum _{k=0}^n{(-1)}^{i+j+k}\left(\genfrac{}{}{0pt}{}{n+i}{2j}\right)\left(\genfrac{}{}{0pt}{}{n+j}{2k}\right)\left(\genfrac{}{}{0pt}{}{n+k+1}{2i+1}\right),\hfill \end{array}$$

(20)

$$\begin{array}{cc}\hfill U_c\left(n\right)& :=\sum _{i=0}^n\sum _{j=0}^n\sum _{k=0}^n{(-1)}^{i+j+k}\left(\genfrac{}{}{0pt}{}{n+i}{2j}\right)\left(\genfrac{}{}{0pt}{}{n+j+1}{2k+1}\right)\left(\genfrac{}{}{0pt}{}{n+k+1}{2i+1}\right),\hfill \end{array}$$

(21)

$$\begin{array}{cc}\hfill U_d\left(n\right)& :=\sum _{i=0}^n\sum _{j=0}^n\sum _{k=0}^n{(-1)}^{i+j+k}\left(\genfrac{}{}{0pt}{}{n+i+1}{2j+1}\right)\left(\genfrac{}{}{0pt}{}{n+j+1}{2k+1}\right)\left(\genfrac{}{}{0pt}{}{n+k+1}{2i+1}\right);\hfill \end{array}$$

(22)

the related results about recurrence relations and generating functions are simpler compared with those in the last section. They are highlighted as follows.

Lemma 2 (Coefficient expressions).

$$\begin{array}{cc}\hfill \left(\mathrm{a}\right)U_a\left(n\right)& =\left[x^{7n}\right]\frac{(1+2x^2+2x^4){(1+x)}^n{(1+x^2)}^{2n+1}}{(1+x+x^2)(1-x+4x^2-3x^3+7x^4-4x^5+5x^6)},\hfill \end{array}$$

(23)

$$\begin{array}{cc}\hfill \left(\mathrm{b}\right)U_b\left(n\right)& =\left[x^{7n}\right]\frac{(1+2x^2+2x^4){(1+x)}^{n+1}{(1+x^2)}^{2n+1}}{(1+x+x^2)(1-x+4x^2-3x^3+7x^4-4x^5+5x^6)},\hfill \end{array}$$

(24)

$$\begin{array}{cc}\hfill \left(\mathrm{c}\right)U_c\left(n\right)& =\left[x^{7n}\right]\frac{{(1+x)}^{n+1}{(1+x^2)}^{2n+3}}{(1+x+x^2)(1-x+4x^2-3x^3+7x^4-4x^5+5x^6)},\hfill \end{array}$$

(25)

$$\begin{array}{cc}\hfill \left(\mathrm{d}\right)U_d\left(n\right)& =\left[x^{7n}\right]\frac{{(1+x)}^{n+1}{(1+x^2)}^{2n+2}}{(1+x+x^2)(1-x+4x^2-3x^3+7x^4-4x^5+5x^6)}.\hfill \end{array}$$

(26)

Proposition 2. 

The four triple sums $\{U_a\left(n\right),U_b\left(n\right),U_c\left(n\right),U_d\left(n\right)\}$ satisfy the same recurrence relation

$$U\left(n\right)=-24U(n-2)-25U(n-3)$$

with different initial values

$$\begin{array}{cc}\hfill {\left\{U_a\left(n\right)\right\}}_{n=0}^2& =\left\{1,0,-19\right\},\hfill \\ \hfill {\left\{U_b\left(n\right)\right\}}_{n=0}^2& =\left\{1,1,-21\right\},\hfill \\ \hfill {\left\{U_c\left(n\right)\right\}}_{n=0}^2& =\left\{1,3,-25\right\},\hfill \\ \hfill {\left\{U_d\left(n\right)\right\}}_{n=0}^2& =\left\{1,7,-24\right\}.\hfill \end{array}$$

Theorem 2 (Generating functions).

$$\begin{array}{cc}\hfill \left(\mathrm{a}\right)\sum _{n=0}^{\infty }{U}_a\left(n\right)y^n& =\frac{1+5y^2}{1+24y^2+25y^3},\hfill \end{array}$$

(27)

$$\begin{array}{cc}\hfill \left(\mathrm{b}\right)\sum _{n=0}^{\infty }{U}_b\left(n\right)y^n& =\frac{1+y+3y^2}{1+24y^2+25y^3},\hfill \end{array}$$

(28)

$$\begin{array}{cc}\hfill \left(\mathrm{c}\right)\sum _{n=0}^{\infty }{U}_c\left(n\right)y^n& =\frac{1+3y-y^2}{1+24y^2+25y^3},\hfill \end{array}$$

(29)

$$\begin{array}{cc}\hfill \left(\mathrm{d}\right)\sum _{n=0}^{\infty }{U}_d\left(n\right)y^n& =\frac{1+7y}{1+24y^2+25y^3}.\hfill \end{array}$$

(30)

Since proofs of the above results are quite similar, we present only demonstrations for $U_b\left(n\right)$ about expression (24) in Lemma 2, recurrence relation in Proposition 2 and generating function (28) in Theorem 2.

3.1. Proof of Lemma 2

According to the binomial relations

$$\begin{array}{cc}\hfill {(-1)}^i\left(\genfrac{}{}{0pt}{}{n+i}{2j}\right)& =\left[x^{2n+2i}\right]\frac{{(-1)}^n{x}^{4j}}{{(1+x^2)}^{2j+1}},\hfill \\ \hfill \left(\genfrac{}{}{0pt}{}{n+k+1}{2i+1}\right)& =\left[x^{n+k-2i}\right]{(1+x)}^{n+k+1},\hfill \end{array}$$

(31)

we can express the binomial sum with respect to i in $U_b\left(n\right)$ as

$$\begin{array}{cc}\hfill U_b^i\left(n\right):=\sum _{i=0}^n{(-1)}^i\left(\genfrac{}{}{0pt}{}{n+i}{2j}\right)\left(\genfrac{}{}{0pt}{}{n+k+1}{2i+1}\right)& =\left[x^{3n+k}\right]{(-1)}^n\frac{x^{4j}{(1+x)}^{n+k+1}}{{(1+x^2)}^{2j+1}}.\hfill \end{array}$$

Then we can treat the binomial sum with respect to j in $U_b\left(n\right)$

$$\begin{array}{cc}\hfill U_b^j\left(n\right)& :=\sum _{j=0}^n{(-1)}^j\left(\genfrac{}{}{0pt}{}{n+j}{2k}\right)U_b^i\left(n\right)\hfill \\ & =\left[x^{3n+k}\right]{(-1)}^n\frac{{(1+x)}^{n+k+1}}{1+x^2}\sum _{j=0}^{\infty }{(-1)}^j\left(\genfrac{}{}{0pt}{}{n+j}{2k}\right)\frac{x^{4j}}{{(1+x^2)}^{2j}}.\hfill \end{array}$$

Evaluating the last series under $y=\frac{x^4}{{(1+x^2)}^2}$, and then following the same discussion as for (18), we derive the closed expression below

$$\begin{array}{cc}\hfill U_b^j\left(n\right)& =\left[x^{3n+k}\right]\frac{{(1+x)}^{n+k+1}}{1+x^2}\frac{y^{2k-n}}{{(1+y)}^{2k+1}}\hfill \\ & -\left[x^{3n+k}\right]\frac{{(1+x)}^{n+k+1}}{1+x^2}\sum _{m=2k}^{n-1}{(-1)}^m\left(\genfrac{}{}{0pt}{}{m}{2k}\right)y^{m-n}\hfill \\ & =\left[x^{3n+k}\right]\frac{x^{8k-4n}{(1+x)}^{n+k+1}{(1+x^2)}^{2n+1}}{{(1+2x^2+2x^4)}^{2k+1}}\hfill \\ & =\left[x^{7n}\right]\frac{x^{7k}{(1+x)}^{n+k+1}{(1+x^2)}^{2n+1}}{{(1+2x^2+2x^4)}^{2k+1}}.\hfill \end{array}$$

Now, we come to the binomial sum with respect to k in $U_b\left(n\right)$

$$\begin{array}{cc}\hfill U_b\left(n\right)& =\sum _{k=0}^n{(-1)}^k{U}_b^j\left(n\right)\hfill \\ & =\left[x^{7n}\right]\frac{{(1+x)}^{n+1}{(1+x^2)}^{2n+1}}{1+2x^2+2x^4}\sum _{k=0}^{\infty }{(-1)}^k\frac{x^{7k}{(1+x)}^k}{{(1+2x^2+2x^4)}^{2k}}.\hfill \end{array}$$

Evaluating the last series yields

$$U_b\left(n\right)=\left[x^{7n}\right]{\mathrm{F}}_n\left(x\right),$$

where ${\mathrm{F}}_n\left(x\right)$ is a rational function defined by

$${\mathrm{F}}_n\left(x\right):=\frac{(1+2x^2+2x^4){(1+x)}^{n+1}{(1+x^2)}^{2n+1}}{{(1+2x^2+2x^4)}^2+x^7+x^8}.$$

This confirms the first expression (24) in Lemma 2.

3.2. Proof of Proposition 2

The recurrence relation in Proposition 2 follows by considering

$$\begin{array}{cc}\hfill {\Delta }_b\left(n\right)& :=U_b\left(n\right)+24U_b(n-2)+25U_b(n-3)\hfill \\ & =\left[x^{7n}\right]\left\{{\mathrm{F}}_n\left(x\right)+24x^{14}{\mathrm{F}}_{n-2}\left(x\right)+25x^{21}{\mathrm{F}}_{n-3}\left(x\right)\right\}.\hfill \end{array}$$

This can further be simplified by factorization

$${\Delta }_b\left(n\right)=\left[x^{7n}\right]\mathbf{F}\left(x\right),$$

where $\mathbf{F}\left(x\right)$ is given explicitly by

$$\begin{array}{cc}\hfill \mathbf{F}\left(x\right)& :=(1+2x^2+2x^4){(1+x)}^{n-2}{(1+x^2)}^{2n-5}(1+x^2+x^3-x^4+x^5)\hfill \\ & \times (1+3x+4x^2+3x^3-x^4-6x^5-4x^6+4x^7+5x^8).\hfill \end{array}$$

When $n\ge 3$, we find that ${\Delta }_b\left(n\right)=0$ because in this case, $\mathbf{F}\left(x\right)$ is a polynomial of degree $5+5n<7n$.

3.3. Proof of Theorem 2

By making use of the recurrence relation in Proposition 2, we turn to treat the generating function

$$\begin{array}{cc}\hfill \mathcal{F}\left(y\right)& :=\sum _{n=0}^{\infty }{U}_b\left(n\right)y^n=U_b\left(0\right)+U_b\left(1\right)y+U_b\left(2\right)y^2+\sum _{n=3}^{\infty }{U}_b\left(n\right)y^n\hfill \\ & =U_b\left(0\right)+U_b\left(1\right)y+U_b\left(2\right)y^2+\sum _{n=3}^{\infty }{y}^n\left\{-24T_a(n-2)-25T_a(n-3)\right\}\hfill \\ & =U_b\left(0\right)+U_b\left(1\right)y+U_b\left(2\right)y^2-24y^2\sum _{n=1}^{\infty }{U}_b\left(n\right)y^n-25y^3\sum _{n=0}^{\infty }{U}_b\left(n\right)y^n\hfill \\ & =U_b\left(0\right)+U_b\left(1\right)y+U_b\left(2\right)y^2+\mathcal{F}\left(y\right)\left\{-24y^2-25y^3\right\}+24y^2{U}_b\left(0\right).\hfill \end{array}$$

Finally, taking into account of the initial values

$${\left\{U_b\left(n\right)\right\}}_{n=0}^2=\left\{1,1,-21\right\},$$

and then simplifying the last expression, we find that

$$\mathcal{F}\left(y\right)=1+y+3y^2-\mathcal{F}\left(y\right)\left\{24y^2+25y^3\right\},$$

which is equivalent to the following generating function as in Theorem 2:

$$\mathcal{F}\left(y\right)=\frac{1+y+3y^2}{1+24y^2+25y^3}.$$

4. Triple Sums with One Alternating Factor

Further for the four (partially alternating) triple sums defined by

$$\begin{array}{cc}\hfill V_a\left(n\right)& :=\sum _{i=0}^n\sum _{j=0}^n\sum _{k=0}^n{(-1)}^i\left(\genfrac{}{}{0pt}{}{n+i}{2j}\right)\left(\genfrac{}{}{0pt}{}{n+j}{2k}\right)\left(\genfrac{}{}{0pt}{}{n+k}{2i}\right),\hfill \end{array}$$

(32)

$$\begin{array}{cc}\hfill V_b\left(n\right)& :=\sum _{i=0}^n\sum _{j=0}^n\sum _{k=0}^n{(-1)}^i\left(\genfrac{}{}{0pt}{}{n+i}{2j}\right)\left(\genfrac{}{}{0pt}{}{n+j}{2k}\right)\left(\genfrac{}{}{0pt}{}{n+k+1}{2i+1}\right),\hfill \end{array}$$

(33)

$$\begin{array}{cc}\hfill V_c\left(n\right)& :=\sum _{i=0}^n\sum _{j=0}^n\sum _{k=0}^n{(-1)}^i\left(\genfrac{}{}{0pt}{}{n+i}{2j}\right)\left(\genfrac{}{}{0pt}{}{n+j+1}{2k+1}\right)\left(\genfrac{}{}{0pt}{}{n+k+1}{2i+1}\right),\hfill \end{array}$$

(34)

$$\begin{array}{cc}\hfill V_d\left(n\right)& :=\sum _{i=0}^n\sum _{j=0}^n\sum _{k=0}^n{(-1)}^i\left(\genfrac{}{}{0pt}{}{n+i+1}{2j+1}\right)\left(\genfrac{}{}{0pt}{}{n+j+1}{2k+1}\right)\left(\genfrac{}{}{0pt}{}{n+k+1}{2i+1}\right);\hfill \end{array}$$

(35)

the related results about recurrence relations and generating functions are more involved. They are summarized in the following statements.

Lemma 3 (Coefficient expressions).

$$\begin{array}{cc}\hfill \left(\mathrm{a}\right)V_a\left(n\right)& =\left[x^{7n}\right]{(-1)}^n\frac{(1+2x^2){(1+x)}^n{(1+x^2)}^{2n+1}}{1+4x^2+4x^4-x^7-x^8},\hfill \end{array}$$

(36)

$$\begin{array}{cc}\hfill \left(\mathrm{b}\right)V_b\left(n\right)& =\left[x^{7n}\right]{(-1)}^n\frac{(1+2x^2){(1+x)}^{n+1}{(1+x^2)}^{2n+1}}{1+4x^2+4x^4-x^7-x^8},\hfill \end{array}$$

(37)

$$\begin{array}{cc}\hfill \left(\mathrm{c}\right)V_c\left(n\right)& =\left[x^{7n}\right]{(-1)}^n\frac{{(1+x)}^{n+1}{(1+x^2)}^{2n+3}}{1+4x^2+4x^4-x^7-x^8},\hfill \end{array}$$

(38)

$$\begin{array}{cc}\hfill \left(\mathrm{d}\right)V_d\left(n\right)& =\left[x^{7n}\right]{(-1)}^n\frac{{(1+x)}^{n+1}{(1+x^2)}^{2n+2}}{1+4x^2+4x^4-x^7-x^8}.\hfill \end{array}$$

(39)

Proposition 3. 

The four triple sums $\{V_a\left(n\right),V_b\left(n\right),V_c\left(n\right),V_d\left(n\right)\}$ satisfy the same recurrence relation

$$\begin{array}{cc}\hfill V\left(n\right)& =V(n-1)-105V(n-2)-210V(n-3)-174V(n-4)\hfill \\ & -370V(n-5)+175V(n-6)+17V(n-7)-9V(n-8)\hfill \end{array}$$

with different initial values

$$\begin{array}{cc}\hfill {\left\{V_a\left(n\right)\right\}}_{n=0}^7& =\left\{1,0,-27,-113,2624,19915,-227039,-2839488\right\},\hfill \\ \hfill {\left\{V_b\left(n\right)\right\}}_{n=0}^7& =\left\{1,1,-21,-281,1571,35001,-67471,-4015639\right\},\hfill \\ \hfill {\left\{V_c\left(n\right)\right\}}_{n=0}^7& =\left\{1,3,-25,-533,1417,61665,28265,-6641605\right\},\hfill \\ \hfill {\left\{V_d\left(n\right)\right\}}_{n=0}^7& =\left\{1,7,-40,-897,1761,102704,110721,-10870889\right\}.\hfill \end{array}$$

Theorem 3 (Generating functions).

$$\begin{array}{cc}\hfill \left(\mathrm{a}\right)\sum _{n=0}^{\infty }{V}_a\left(n\right)y^n& =\frac{1-y+78y^2+124y^3+76y^4+126y^5-37y^6-3y^7}{1-y+105y^2+210y^3+174y^4+370y^5-175y^6-17y^7+9y^8},\hfill \end{array}$$

(40)

$$\begin{array}{cc}\hfill \left(\mathrm{b}\right)\sum _{n=0}^{\infty }{V}_b\left(n\right)y^n& =\frac{1+83y^2+55y^3+31y^4+59y^5+14y^6-9y^7}{1-y+105y^2+210y^3+174y^4+370y^5-175y^6-17y^7+9y^8},\hfill \end{array}$$

(41)

$$\begin{array}{cc}\hfill \left(\mathrm{c}\right)\sum _{n=0}^{\infty }{V}_c\left(n\right)y^n& =\frac{1+2y+77y^2+17y^3+129y^4-75y^5+40y^6-9y^7}{1-y+105y^2+210y^3+174y^4+370y^5-175y^6-17y^7+9y^8},\hfill \end{array}$$

(42)

$$\begin{array}{cc}\hfill \left(\mathrm{d}\right)\sum _{n=0}^{\infty }{V}_d\left(n\right)y^n& =\frac{1+6y+58y^2+88y^3+102y^4-54y^5+7y^6}{1-y+105y^2+210y^3+174y^4+370y^5-175y^6-17y^7+9y^8}.\hfill \end{array}$$

(43)

In what follows, we are going to illustrate for $V_c\left(n\right)$ about how to establish expression (38) in Lemma 3, recurrence relation in Proposition 3 and generating function (42) in Theorem 3. The proofs for the remaining results are omitted for similarities.

4.1. Proof of Lemma 3

Since the binomial sum with respect to i in $V_c\left(n\right)$ is the same as $U_b^i\left(n\right)$, we can immediately write down

$$\begin{array}{c}\hfill V_c^i\left(n\right):=\sum _{i=0}^n{(-1)}^i\left(\genfrac{}{}{0pt}{}{n+i}{2j}\right)\left(\genfrac{}{}{0pt}{}{n+k+1}{2i+1}\right)=\left[x^{3n+k}\right]{(-1)}^n\frac{x^{4j}{(1+x)}^{n+k+1}}{{(1+x^2)}^{2j+1}}.\end{array}$$

Then we can deal with the binomial sum with respect to j in $V_c\left(n\right)$

$$\begin{array}{cc}\hfill V_c^j\left(n\right)& :=\sum _{j=0}^n\left(\genfrac{}{}{0pt}{}{n+j+1}{2k+1}\right)V_c^i\left(n\right)\hfill \\ & =\left[x^{3n+k}\right]{(-1)}^n\frac{{(1+x)}^{n+k+1}}{1+x^2}\sum _{j=0}^{\infty }\left(\genfrac{}{}{0pt}{}{n+j+1}{2k+1}\right)\frac{x^{4j}}{{(1+x^2)}^{2j}}.\hfill \end{array}$$

Evaluating the last series under $y=\frac{x^4}{{(1+x^2)}^2}$, and then making a similar reasoning as for (18), we have the following closed expression

$$\begin{array}{cc}\hfill V_c^j\left(n\right)& =\left[x^{3n+k}\right]{(-1)}^n\frac{{(1+x)}^{n+k+1}}{1+x^2}\frac{y^{2k-n}}{{(1-y)}^{2k+2}}\hfill \\ & -\left[x^{3n+k}\right]{(-1)}^n\frac{{(1+x)}^{n+k+1}}{1+x^2}\sum _{m=2k}^{n-1}\left(\genfrac{}{}{0pt}{}{m+1}{2k+1}\right)y^{m-n}\hfill \\ & =\left[x^{3n+k}\right]{(-1)}^n\frac{x^{8k-4n}{(1+x)}^{n+k+1}{(1+x^2)}^{2n+3}}{{(1+2x^2)}^{2k+2}}\hfill \\ & =\left[x^{7n}\right]{(-1)}^n\frac{x^{7k}{(1+x)}^{n+k+1}{(1+x^2)}^{2n+3}}{{(1+2x^2)}^{2k+2}}.\hfill \end{array}$$

Consequently, the binomial sum with respect to k in $V_c\left(n\right)$ becomes

$$\begin{array}{cc}\hfill V_c\left(n\right)& =\sum _{k=0}^n{V}_c^j\left(n\right)=\left[x^{7n}\right]{(-1)}^n\frac{{(1+x)}^{n+1}{(1+x^2)}^{2n+3}}{{(1+2x^2)}^2}\sum _{k=0}^{\infty }\frac{x^{7k}{(1+x)}^k}{{(1+2x^2)}^{2k}}.\hfill \end{array}$$

By evaluating the last series, we have further

$$V_c\left(n\right)=\left[x^{7n}\right]{\mathrm{G}}_n\left(x\right),$$

where ${\mathrm{G}}_n\left(x\right)$ is a rational function defined by

$${\mathrm{G}}_n\left(x\right):={(-1)}^n\frac{{(1+x)}^{n+1}{(1+x^2)}^{2n+3}}{1+4x^2+4x^4-x^7-x^8}.$$

This proves the first expression (37) as in Lemma 3.

4.2. Proof of Proposition 3

The recurrence relation in Proposition 3 follows by examining

$$\begin{array}{c}{\Delta }_c\left(n\right):=V_c\left(n\right)-V_c(n-1)+105V_c(n-2)+210V_c(n-3)+174V_c(n-4)\hfill \\ \hspace{1em}\hspace{1em}+370V_c(n-5)-175V_c(n-6)-17V_c(n-7)+9V_c(n-8)\hfill \\ =\left[x^{7n}\right]\{{\mathrm{G}}_n\left(x\right)-x^7{\mathrm{G}}_{n-1}\left(x\right)+105x^{14}{\mathrm{G}}_{n-2}\left(x\right)+210x^{21}{\mathrm{G}}_{n-3}\left(x\right)+174x^{28}{\mathrm{G}}_{n-4}\left(x\right)\hfill \\ \hspace{1em}\hspace{1em}\hspace{1em}+370x^{35}{\mathrm{G}}_{n-5}\left(x\right)-175x^{42}{\mathrm{G}}_{n-6}\left(x\right)-17x^{49}{\mathrm{G}}_{n-7}\left(x\right)+9x^{56}{\mathrm{G}}_{n-8}\left(x\right)\}.\hfill \end{array}$$

The above expression can be simplified into

$${\Delta }_c\left(n\right)={(-1)}^n\left[x^{7n}\right]\mathbf{G}\left(x\right),$$

where $\mathbf{G}\left(x\right)$ is a Laurent polynomial given by

$$\begin{array}{cc}\hfill \mathbf{G}\left(x\right)& :={(1+x)}^{n-7}{(1+x^2)}^{2n-13}\{1+8x+40x^2+152x^3+474x^4+1272x^5\hfill \\ & +3012x^6+6410x^7+12421x^8+22123x^9+36497x^{10}+56098x^{11}+80734x^{12}\hfill \\ & +109244x^{13}+139575x^{14}+169157x^{15}+195689x^{16}+218068x^{17}+236920x^{18}\hfill \\ & +254914x^{19}+275629x^{20}+301957x^{21}+334535x^{22}+369778x^{23}+400796x^{24}\hfill \\ & +418950x^{25}+416333x^{26}+389845x^{27}+341563x^{28}+278638x^{29}+211582x^{30}\hfill \\ & +149708x^{31}+99658x^{32}+63991x^{33}+40997x^{34}+27276x^{35}+19162x^{36}\hfill \\ & +13660x^{37}+9361x^{38}+6059x^{39}+3444x^{40}+1733x^{41}+832x^{42}+256x^{43}\hfill \\ & +96x^{44}+9x^{45}-26x^{46}+9x^{47}-9x^{48}\}.\hfill \end{array}$$

When $n\ge 8$, this is a polynomial of degree $15+5n<7n$, which implies the recurrence relation ${\Delta }_c\left(n\right)=0$ as in Proposition 3.

4.3. Proof of Theorem 3

Finally, for the generating function

$$\begin{array}{cc}\hfill \mathcal{G}\left(y\right):=\sum _{n=0}^{\infty }{V}_c\left(n\right)y^n& =V_c\left(0\right)+V_c\left(1\right)y+V_c\left(2\right)y^2+V_c\left(3\right)y^3+V_c\left(4\right)y^4\hfill \\ & +V_c\left(5\right)y^5+V_c\left(6\right)y^6+V_c\left(7\right)y^7+\sum _{n=8}^{\infty }{V}_c\left(n\right)y^n,\hfill \end{array}$$

replacing $V_c\left(n\right)$ by the recurrence relation in Proposition 3 and then applying the initial values

$${\left\{V_c\left(n\right)\right\}}_{n=0}^7=\left\{1,3,-25,-533,1417,61665,28265,-6641605\right\},$$

we obtain the following simplified equation

$$\begin{array}{cc}\hfill \mathcal{G}\left(y\right)& =1+2y+77y^2+17y^3+129y^4-75y^5+40y^6-9y^7\hfill \\ & +\mathcal{G}\left(y\right)\left\{y-105y^2-210y^3-174y^4-370y^5+175y^6+17y^7-9y^8\right\},\hfill \end{array}$$

which confirms the generating function as in Theorem 3

$$\mathcal{G}\left(y\right)=\frac{1+2y+77y^2+17y^3+129y^4-75y^5+40y^6-9y^7}{1-y+105y^2+210y^3+174y^4+370y^5-175y^6-17y^7+9y^8}.$$

5. Triple Sums with Two Alternating Factors

Finally, define four further variants of alternating triple sums

$$\begin{array}{cc}\hfill W_a\left(n\right)& :=\sum _{i=0}^n\sum _{j=0}^n\sum _{k=0}^n{(-1)}^{i+j}\left(\genfrac{}{}{0pt}{}{n+i}{2j}\right)\left(\genfrac{}{}{0pt}{}{n+j}{2k}\right)\left(\genfrac{}{}{0pt}{}{n+k}{2i}\right),\hfill \end{array}$$

(44)

$$\begin{array}{cc}\hfill W_b\left(n\right)& :=\sum _{i=0}^n\sum _{j=0}^n\sum _{k=0}^n{(-1)}^{i+j}\left(\genfrac{}{}{0pt}{}{n+i}{2j}\right)\left(\genfrac{}{}{0pt}{}{n+j}{2k}\right)\left(\genfrac{}{}{0pt}{}{n+k+1}{2i+1}\right),\hfill \end{array}$$

(45)

$$\begin{array}{cc}\hfill W_c\left(n\right)& :=\sum _{i=0}^n\sum _{j=0}^n\sum _{k=0}^n{(-1)}^{i+j}\left(\genfrac{}{}{0pt}{}{n+i}{2j}\right)\left(\genfrac{}{}{0pt}{}{n+j+1}{2k+1}\right)\left(\genfrac{}{}{0pt}{}{n+k+1}{2i+1}\right),\hfill \end{array}$$

(46)

$$\begin{array}{cc}\hfill W_d\left(n\right)& :=\sum _{i=0}^n\sum _{j=0}^n\sum _{k=0}^n{(-1)}^{i+j}\left(\genfrac{}{}{0pt}{}{n+i+1}{2j+1}\right)\left(\genfrac{}{}{0pt}{}{n+j+1}{2k+1}\right)\left(\genfrac{}{}{0pt}{}{n+k+1}{2i+1}\right).\hfill \end{array}$$

(47)

We have the following results about recurrence relations and generating functions.

Lemma 4 (Coefficient expressions).

$$\begin{array}{cc}\hfill \left(\mathrm{a}\right)W_a\left(n\right)& =\left[x^{7n}\right]\frac{(1+2x^2+2x^4){(1+x)}^n{(1+x^2)}^{2n+1}}{1+4x^2+8x^4+8x^6-x^7+3x^8},\hfill \end{array}$$

(48)

$$\begin{array}{cc}\hfill \left(\mathrm{b}\right)W_b\left(n\right)& =\left[x^{7n}\right]\frac{(1+2x^2+2x^4){(1+x)}^{n+1}{(1+x^2)}^{2n+1}}{1+4x^2+8x^4+8x^6-x^7+3x^8},\hfill \end{array}$$

(49)

$$\begin{array}{cc}\hfill \left(\mathrm{c}\right)W_c\left(n\right)& =\left[x^{7n}\right]\frac{{(1+x)}^{n+1}{(1+x^2)}^{2n+3}}{1+4x^2+8x^4+8x^6-x^7+3x^8},\hfill \end{array}$$

(50)

$$\begin{array}{cc}\hfill \left(\mathrm{d}\right)W_d\left(n\right)& =\left[x^{7n}\right]\frac{{(1+x)}^{n+1}{(1+x^2)}^{2n+2}}{1+4x^2+8x^4+8x^6-x^7+3x^8}.\hfill \end{array}$$

(51)

Proposition 4. 

The four triple sums $\{W_a\left(n\right),W_b\left(n\right),W_c\left(n\right),W_d\left(n\right)\}$ satisfy the same recurrence relation

$$\begin{array}{cc}\hfill W\left(n\right)& =15W(n-1)-17W(n-2)-334W(n-3)-318W(n-4)\hfill \\ & +2370W(n-5)-1889W(n-6)-449W(n-7)-225W(n-8)\hfill \end{array}$$

with different initial values

$$\begin{array}{cc}\hfill {\left\{W_a\left(n\right)\right\}}_{n=0}^7& =\left\{1,2,25,209,1906,17569,156641,1402914\right\},\hfill \\ \hfill {\left\{W_b\left(n\right)\right\}}_{n=0}^7& =\left\{1,3,27,249,2067,18103,150513,1238083\right\},\hfill \\ \hfill {\left\{W_c\left(n\right)\right\}}_{n=0}^7& =\left\{1,5,39,373,2949,25223,201417,1571445\right\},\hfill \\ \hfill {\left\{W_d\left(n\right)\right\}}_{n=0}^7& =\left\{1,9,64,657,5737,52544,469953,4165193\right\}.\hfill \end{array}$$

Theorem 4 (Generating functions).

$$\begin{array}{cc}\hfill \left(\mathrm{a}\right)\sum _{n=0}^{\infty }{W}_a\left(n\right)y^n& =\frac{1-13y+12y^2+202y^3+182y^4-852y^5+413y^6+15y^7}{1-15y+17y^2+334y^3+318y^4-2370y^5+1889y^6+449y^7+225y^8},\hfill \end{array}$$

(52)

$$\begin{array}{cc}\hfill \left(\mathrm{b}\right)\sum _{n=0}^{\infty }{W}_b\left(n\right)y^n& =\frac{1-12y-y^2+229y^3+111y^4-1067y^5+638y^6-175y^7}{1-15y+17y^2+334y^3+318y^4-2370y^5+1889y^6+449y^7+225y^8},\hfill \end{array}$$

(53)

$$\begin{array}{cc}\hfill \left(\mathrm{c}\right)\sum _{n=0}^{\infty }{W}_c\left(n\right)y^n& =\frac{1-10y-19y^2+207y^3+5y^4-425y^5+228y^6+25y^7}{1-15y+17y^2+334y^3+318y^4-2370y^5+1889y^6+449y^7+225y^8},\hfill \end{array}$$

(54)

$$\begin{array}{cc}\hfill \left(\mathrm{d}\right)\sum _{n=0}^{\infty }{W}_d\left(n\right)y^n& =\frac{1-6y-54y^2+184y^3+294y^4-474y^5-329y^6}{1-15y+17y^2+334y^3+318y^4-2370y^5+1889y^6+449y^7+225y^8}.\hfill \end{array}$$

(55)

For the triple sum $W_d\left(n\right)$, we give a detailed proof for expression (51) in Lemma 4, recurrence relation in Proposition 4 and generating function (55) in Theorem 4. The other results for triple sums $W_a\left(n\right),W_b\left(n\right)$ and $W_c\left(n\right)$ can be confirmed likewise.

5.1. Proof of Lemma 4

By combining (31) with

$${(-1)}^i\left(\genfrac{}{}{0pt}{}{n+i+1}{2j+1}\right)=\left[x^{2n+2i}\right]\frac{{(-1)}^n{x}^{4j}}{{(1+x^2)}^{2j+2}},$$

we can rewrite the binomial sum with respect to i in $W_d\left(n\right)$ as

$$\begin{array}{cc}\hfill W_d^i\left(n\right):=\sum _{i=0}^n{(-1)}^i\left(\genfrac{}{}{0pt}{}{n+i+1}{2j+1}\right)\left(\genfrac{}{}{0pt}{}{n+k+1}{2i+1}\right)& =\left[x^{3n+k}\right]{(-1)}^n\frac{x^{4j}{(1+x)}^{n+k+1}}{{(1+x^2)}^{2j+2}}.\hfill \end{array}$$

Then, the binomial sum with respect to j in $W_d\left(n\right)$ can be expressed as

$$\begin{array}{cc}\hfill W_d^j\left(n\right)& :=\sum _{j=0}^n{(-1)}^j\left(\genfrac{}{}{0pt}{}{n+j+1}{2k+1}\right)W_d^i\left(n\right)\hfill \\ & =\left[x^{3n+k}\right]{(-1)}^n\frac{{(1+x)}^{n+k+1}}{{(1+x^2)}^2}\sum _{j=0}^{\infty }{(-1)}^j\left(\genfrac{}{}{0pt}{}{n+j+1}{2k+1}\right)\frac{x^{4j}}{{(1+x^2)}^{2j}}.\hfill \end{array}$$

Evaluating the last series under $y=\frac{x^4}{{(1+x^2)}^2}$, and then following a similar argument as for (18), we have the following closed expression:

$$\begin{array}{cc}\hfill W_d^j\left(n\right)& =\left[x^{3n+k}\right]\frac{{(1+x)}^{n+k+1}}{{(1+x^2)}^2}\frac{y^{2k-n}}{{(1+y)}^{2k+2}}\hfill \\ & -\left[x^{3n+k}\right]\frac{{(1+x)}^{n+k+1}}{{(1+x^2)}^2}\sum _{m=2k}^{n-1}{(-1)}^m\left(\genfrac{}{}{0pt}{}{m+1}{2k+1}\right)y^{m-n}\hfill \\ & =\left[x^{3n+k}\right]\frac{x^{8k-4n}{(1+x)}^{n+k+1}{(1+x^2)}^{2n+2}}{{(1+2x^2+2x^4)}^{2k+2}}\hfill \\ & =\left[x^{7n}\right]\frac{x^{7k}{(1+x)}^{n+k+1}{(1+x^2)}^{2n+2}}{{(1+2x^2+2x^4)}^{2k+2}}.\hfill \end{array}$$

Next, we turn to handle the binomial sum with respect to k in $W_d\left(n\right)$

$$\begin{array}{cc}\hfill W_d\left(n\right)& =\sum _{k=0}^n{W}_d^j\left(n\right)=\left[x^{7n}\right]\frac{{(1+x)}^{n+1}{(1+x^2)}^{2n+2}}{{(1+2x^2+2x^4)}^2}\sum _{k=0}^{\infty }\frac{x^{7k}{(1+x)}^k}{{(1+2x^2+2x^4)}^{2k}}.\hfill \end{array}$$

Evaluating the last series, we have

$$W_d\left(n\right)=\left[x^{7n}\right]{\mathrm{H}}_n\left(x\right),$$

where ${\mathrm{H}}_n\left(x\right)$ is a rational function defined by

$${\mathrm{H}}_n\left(x\right):=\frac{{(1+x)}^{n+1}{(1+x^2)}^{2n+2}}{{(1+2x^2+2x^4)}^2-x^7-x^8}.$$

This validates the first expression (49) in Lemma 4.

5.2. Proof of Proposition 4

The recurrence relation in Proposition 4 follows by considering the difference

$$\begin{gathered} \begin{array}{cc}\hfill {\Delta }_d\left(n\right)& :=W_d\left(n\right)-15W_d(n-1)+17W_d(n-2)+334W_d(n-3)+318W_d(n-4)\hfill \\ & -2370W_d(n-5)+1889W_d(n-6)+449W_d(n-7)+225W_d(n-8)\hfill \\ \hfill =& \left[x^{7n}\right]\left\{{\mathrm{H}}_n\left(x\right)-15x^7{\mathrm{H}}_{n-1}\left(x\right)+17x^{14}{\mathrm{H}}_{n-2}\left(x\right)+334x^{21}{\mathrm{H}}_{n-3}\left(x\right)+318x^{28}{\mathrm{H}}_{n-4}\left(x\right) \\ -2370x^{35}{\mathrm{H}}_{n-5}\left(x\right)+1889x^{42}{\mathrm{H}}_{n-6}\left(x\right)+449x^{49}{\mathrm{H}}_{n-7}\left(x\right)+225x^{56}{\mathrm{H}}_{n-8}\left(x\right)\right\}.\hfill \end{array} \end{gathered}$$

The above equality can be simplified, by factorization, into

$${\Delta }_d\left(n\right):=\left[x^{7n}\right]\mathbf{H}\left(x\right),$$

where $\mathbf{H}\left(x\right)$ is given explicitly by

$$\begin{array}{cc}\hfill \mathbf{H}\left(x\right)& :={(1+x)}^{n-7}{(1+x^2)}^{2n-14}\{1+8x+40x^2+152x^3+470x^4+1240x^5+2860x^6\hfill \\ & +5850x^7+10729x^8+17723x^9+26385x^{10}+35286x^{11}+41870x^{12}+42904x^{13}\hfill \\ & +35419x^{14}+17809x^{15}-8927x^{16}-40984x^{17}-72604x^{18}-97318x^{19}-110163x^{20}\hfill \\ & -108819x^{21}-93717x^{22}-67878x^{23}-35464x^{24}-1374x^{25}+29941x^{26}+54453x^{27}\hfill \\ & +69103x^{28}+74078x^{29}+69970x^{30}+59668x^{31}+47574x^{32}+34003x^{33}+21105x^{34}\hfill \\ & +10076x^{35}-526x^{36}-7396x^{37}-9943x^{38}-9449x^{39}-5444x^{40}-1407x^{41}+748x^{42}\hfill \\ & +1596x^{43}+864x^{44}+69x^{45}-42x^{46}+25x^{47}+75x^{48}\}.\hfill \end{array}$$

When $n\ge 8$, this becomes a polynomial of degree $13+5n<7n$, which proves the recurrence relation ${\Delta }_d\left(n\right)=0$ as in Proposition 4.

5.3. Proof of Theorem 4

Finally, for the generating function

$$\begin{array}{cc}\hfill \mathcal{H}\left(y\right):=\sum _{n=0}^{\infty }{W}_d\left(n\right)y^n& =W_d\left(0\right)+W_d\left(1\right)y+W_d\left(2\right)y^2+W_d\left(3\right)y^3+W_d\left(4\right)y^4\hfill \\ & +W_d\left(5\right)y^5+W_d\left(6\right)y^6+W_d\left(7\right)y^7+\sum _{n=8}^{\infty }{W}_d\left(n\right)y^n,\hfill \end{array}$$

first replacing $W_d\left(n\right)$ by the recurrence relation in Proposition 4 and then invoking the initial values

$${\left\{W_d\left(n\right)\right\}}_{n=0}^7=\left\{1,9,64,657,5737,52544,469953,4165193\right\},$$

we establish the following functional equation

$$\begin{array}{cc}\hfill \mathcal{H}\left(y\right)& =1-6y-54y^2+184y^3+294y^4-474y^5-329y^6\hfill \\ & +\mathcal{H}\left(y\right)\left\{15y-17y^2-334y^3-318y^4+2370y^5-1889y^6-449y^7-225y^8\right\},\hfill \end{array}$$

which confirms the generating function as in Theorem 4

$$\underset{—}{\mathcal{H}\left(y\right)=\frac{1-6y-54y^2+184y^3+294y^4-474y^5-329y^6}{1-15y+17y^2+334y^3+318y^4-2370y^5+1889y^6+449y^7+225y^8}.}$$

As anticipated in the introduction, the double sum ${\mathsf{\Phi }}_c\left(n\right)$ satisfies the following recurrence relation

$$\Delta :={\mathsf{\Phi }}_c\left(n\right)-3{\mathsf{\Phi }}_c(n-1)-3{\mathsf{\Phi }}_c(n-2)-4{\mathsf{\Phi }}_c(n-3)=0\mathrm{for}n\ge 3.$$

A proof of this is offered as follows. By applying the expression for $W_d^i\left(n\right)$ in the proof of Lemma 4, we can write

$$\sum _{i=0}^n{(-1)}^i\left(\genfrac{}{}{0pt}{}{n+i+1}{2j+1}\right)\left(\genfrac{}{}{0pt}{}{n+j+1}{2i+1}\right)=\left[x^{3n}\right]{(-1)}^n\frac{x^{3j}{(1+x)}^{n+j+1}}{{(1+x^2)}^{2j+2}}.$$

Then the double sum ${\mathsf{\Phi }}_c\left(n\right)$ can compactly be expressed as

$$\begin{array}{cc}\hfill {\mathsf{\Phi }}_c\left(n\right)& =\left[x^{3n}\right]\sum _{j=0}^n{(-1)}^{n+j}\frac{x^{3j}{(1+x)}^{n+j+1}}{{(1+x^2)}^{2j+2}}\hfill \\ & =\left[x^{3n}\right]{(-1)}^n\frac{{(1+x)}^{n+1}}{{(1+x^2)}^2}\sum _{j=0}^{\infty }{(-1)}^j\frac{x^{3j}{(1+x)}^j}{{(1+x^2)}^{2j}}\hfill \\ & =\left[x^{3n}\right]R_n\left(x\right),\mathrm{where}{R}_n\left(x\right)=\frac{{(-1)}^n{(1+x)}^{n+1}}{1+2x^2+x^3+2x^4}.\hfill \end{array}$$

Now reformulating the difference

$$\begin{array}{cc}\hfill \Delta & ={\mathsf{\Phi }}_c\left(n\right)-3{\mathsf{\Phi }}_c(n-1)-3{\mathsf{\Phi }}_c(n-2)-4{\mathsf{\Phi }}_c(n-3)\hfill \\ & =\left[x^{3n}\right]\left\{R_n\left(x\right)-3x^3{R}_{n-1}\left(x\right)-3x^6{R}_{n-2}\left(x\right)-4x^9{R}_{n-3}\left(x\right)\right\}\hfill \\ & =\left[x^{3n}\right]\frac{{(-1)}^n{(1+x)}^{n-2}}{1+2x^2+x^3+2x^4}\left\{{(1+x)}^3+3x^3{(1+x)}^2-3x^6(1+x)+4x^9\right\}\hfill \\ & =\left[x^{3n}\right]{(-1)}^n{(1+x)}^{n-2}(1+2x)(1+x-x^2-x^3+x^4),\hfill \end{array}$$

we infer that the difference $\Delta$ vanishes when $n>2$ since the polynomial displayed in the last line is of degree $n+3<3n$. In addition, it is routine to determine the generating function

$$\sum _{n=0}^{\infty }{\mathsf{\Phi }}_c\left(n\right)y^n=\frac{1+2y}{1-3y-3y^2-4y^3}.$$

6. Conclusions

The results for triple binomial sums presented in this paper suggest that one may make further investigations about quadruple and quintuple sums. However, numerical tests show that the corresponding recurrence relations and generating functions are very complicated in spite of their existence. Another problem concerns multiple circular sums analogous to those treated by Carlitz [25] and the second author [26]. It would be tough to determine the general pattern for the related recurrence relations and generating functions. Finally, the results obtained in this paper may potentially find applications in conjunction with Fibonacci and Lucas numbers as conducted in [10,23]. The interested reader is encouraged to further explore these areas.