Triple Symmetric Sums of Circular Binomial Products

Abstract

By employing the generating function approach, 16 triple sums for circular binomial products of binomial coefficients are examined. Recurrence relations and generating functions are explicitly determined. These symmetric sums may find potential applications in the analysis of algorithms, symbolic calculus, and computations in theoretical physics.

1. Introduction and Outline

There has been prolonged interest in binomial sums for their wide applications in mathematics, physics, and applied sciences (see [1,2,3,4]). Up to now, thousands of binomial identities are scattered in the literature (see [5,6,7,8,9]). Some of them admit particular beauty and symmetry. Denote by $\left\{F_n\right\}$ the Fibonacci numbers as usual. We record the following sample identities:

• Kilic and Belbachir [10]

  $$\sum _{i=0}^n\sum _{j=0}^n\left({\displaystyle \genfrac{}{}{0.0pt}{}{n-i}{j}}\right)\left({\displaystyle \genfrac{}{}{0.0pt}{}{n-j}{i}}\right)=F_{2n+2}.$$
• Kilic and Prodinger [11]

  $$\sum _{i=0}^n\sum _{j=0}^n\left({\displaystyle \genfrac{}{}{0.0pt}{}{n+i}{2j}}\right)\left({\displaystyle \genfrac{}{}{0.0pt}{}{n+j}{2i}}\right)=F_{4n-1}.$$
• Kilic and Arikan [12]

  $$\sum _{i=0}^n\sum _{j=0}^n\left({\displaystyle \genfrac{}{}{0.0pt}{}{n-i}{j}}\right)\left({\displaystyle \genfrac{}{}{0.0pt}{}{i+j}{j}}\right){(-1)}^i=F_{n+1}.$$

Motivated by these binomial sums and the circular sums treated in [13,14,15], we are going to examine the following “quasi" symmetric triple sums of circular binomial products (i.e., triple sums of symmetric summands except for alternating factors $\Delta (i,j,k)$):

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

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

where the sign function factor $\Delta (i,j,k)$ is given by

$$\Delta (i,j,k)\in \left\{1,{(-1)}^i,{(-1)}^{i+j},{(-1)}^{i+j+k}\right\}.$$

A generating function is just an alternative way of writing a sequence of numbers. The interest of this notion is that certain natural operations on generating functions lead to powerful methods for dealing with recurrences. Therefore, generating functions (cf. [16,17,18,19]) are very useful in combinatorial enumeration problems (for example, the recent solution of Gessel’s lattice path problem (see [20,21]).

By employing the generating function approach, we explicitly determine recurrence relations and generating functions for these 16 triples sums. The contents of the remaining part of the paper are divided into four sections according to the presence of the sign function factor $\Delta (i,j,k)$.

2. Positive Triple Sums $\Delta (\mathit{i},\mathit{j},\mathit{k})=\mathbf{1}$

For the four positive triple sums defined by

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

(1)

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

(2)

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

(3)

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

(4)

we are going to show that they satisfy common recurrence relations and admit compactly rational generating functions. Let $\left[x^m\right]f\left(x\right)$ stand for the coefficient of $x^m$ in the formal power series $f\left(x\right)$. The main results are summarized as follows.

Lemma 1

(Coefficient expressions: $\sigma :=\sqrt{1+x}$).

$$\begin{array}{cc}\hfill \left(\mathrm{a}\right){\mathsf{\Omega }}_a\left(n\right)& =\left[x^{2n}\right]\left\{{\omega }_a(x,\sigma )+{\omega }_a(x,-\sigma )\right\}:{\omega }_a(x,\sigma ):={\displaystyle \frac{(1-x^2){(1+x+\sigma )}^n}{2(1-2x^2-x^3(1-\sigma ))}},\hfill \end{array}$$

(5)

$$\begin{array}{cc}\hfill \left(\mathrm{b}\right){\mathsf{\Omega }}_b\left(n\right)& =\left[x^{2n+1}\right]\left\{{\omega }_b(x,\sigma )+{\omega }_b(x,-\sigma )\right\}:{\omega }_b(x,\sigma ):={\displaystyle \frac{(1-x^2)(1+x){(1+x+\sigma )}^n}{2(1-2x^2-x^3(1-\sigma ))}},\hfill \end{array}$$

(6)

$$\begin{array}{cc}\hfill \left(\mathrm{c}\right){\mathsf{\Omega }}_c\left(n\right)& =\left[x^{2n+1}\right]\left\{{\omega }_c(x,\sigma )+{\omega }_c(x,-\sigma )\right\}:{\omega }_c(x,\sigma ):={\displaystyle \frac{\sigma (1-x^2){(1+x+\sigma )}^{n+1}}{2(1-2x^2-x^3(1-\sigma ))}},\hfill \end{array}$$

(7)

$$\begin{array}{cc}\hfill \left(\mathrm{d}\right){\mathsf{\Omega }}_d\left(n\right)& =\left[x^{2n+1}\right]\left\{{\omega }_d(x,\sigma )+{\omega }_d(x,-\sigma )\right\}:{\omega }_d(x,\sigma ):={\displaystyle \frac{\sigma {(1+x+\sigma )}^{n+1}}{2(1-2x^2-x^3(1-\sigma ))}}.\hfill \end{array}$$

(8)

Proposition 1

(Recurrence relations). The four triple sums $\{{\mathsf{\Omega }}_a\left(n\right),{\mathsf{\Omega }}_b\left(n\right),{\mathsf{\Omega }}_c\left(n\right),{\mathsf{\Omega }}_d\left(n\right)\}$ satisfy the same recurrence relation

$$\mathsf{\Omega }\left(n\right)=6\mathsf{\Omega }(n-1)-3\mathsf{\Omega }(n-2)-\mathsf{\Omega }(n-3)-\mathsf{\Omega }(n-4)$$

with different initial values

$$\begin{array}{cc}& {\left\{{\mathsf{\Omega }}_a\left(n\right)\right\}}_{n=0}^3=\left\{1,1,4,20\right\},\hfill \\ & {\left\{{\mathsf{\Omega }}_b\left(n\right)\right\}}_{n=0}^3=\left\{1,2,9,46\right\},\hfill \\ & {\left\{{\mathsf{\Omega }}_c\left(n\right)\right\}}_{n=0}^3=\left\{1,4,20,106\right\},\hfill \\ & {\left\{{\mathsf{\Omega }}_d\left(n\right)\right\}}_{n=0}^3=\left\{1,8,45,245\right\}.\hfill \end{array}$$

Theorem 1

(Generating functions).

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

(9)

$$\begin{array}{cc}\hfill \left(\mathrm{b}\right)\sum _{n=0}^{\infty }{\mathsf{\Omega }}_b\left(n\right)y^n& ={\displaystyle \frac{1-4y-y^3}{(1-y)(1-5y-2y^2-y^3)}},\hfill \end{array}$$

(10)

$$\begin{array}{cc}\hfill \left(\mathrm{c}\right)\sum _{n=0}^{\infty }{\mathsf{\Omega }}_c\left(n\right)y^n& ={\displaystyle \frac{1-2y-y^2-y^3}{(1-y)(1-5y-2y^2-y^3)}},\hfill \end{array}$$

(11)

$$\begin{array}{cc}\hfill \left(\mathrm{d}\right)\sum _{n=0}^{\infty }{\mathsf{\Omega }}_d\left(n\right)y^n& ={\displaystyle \frac{1+2y}{(1-y)(1-5y-2y^2-y^3)}}.\hfill \end{array}$$

(12)

We take ${\mathsf{\Omega }}_a\left(n\right)$ as a show case to demonstrate how to derive the expression (5) in Lemma 1, the recurrence relation in Proposition 1, and the generating function (9) in Theorem 1. The remaining three triple sums can be treated in similar manners.

2.1. Proof of Lemma 1(a)

According to the binomial expressions

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

the following sum with respect to i in ${\mathsf{\Omega }}_a\left(n\right)$ can be written as

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

Then we can proceed with the binomial sum with respect to k in ${\mathsf{\Omega }}_a\left(n\right)$:

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

Consequently, the remaining binomial sum with respect to j in ${\mathsf{\Omega }}_a\left(n\right)$ reads as

$$\begin{array}{cc}\hfill {\mathsf{\Omega }}_a\left(n\right)=& \sum _{j=0}^n{\mathsf{\Omega }}_a^k\left(n\right)=\sum _{j=0}^n\left[x^{2n}\right]{\displaystyle \frac{x^{4j}{(1+x)}^n}{2{(1-x^2)}^{2j+1}}}\left\{{\left({\displaystyle \frac{\sqrt{1+x}+1}{\sqrt{1+x}}}\right)}^{n-j}+{\left({\displaystyle \frac{\sqrt{1+x}-1}{\sqrt{1+x}}}\right)}^{n-j}\right\}\hfill \\ \hfill =& \left[x^{2n}\right]{\displaystyle \frac{{(1+x)}^n}{2(1-x^2)}}{\left({\displaystyle \frac{\sqrt{1+x}+1}{\sqrt{1+x}}}\right)}^n\sum _{j=0}^{\infty }{\displaystyle \frac{x^{4j}}{{(1-x^2)}^{2j}}}{\left({\displaystyle \frac{\sqrt{1+x}}{\sqrt{1+x}+1}}\right)}^j\hfill \\ \hfill +& \left[x^{2n}\right]{\displaystyle \frac{{(1+x)}^n}{2(1-x^2)}}{\left({\displaystyle \frac{\sqrt{1+x}-1}{\sqrt{1+x}}}\right)}^n\sum _{j=0}^{\infty }{\displaystyle \frac{x^{4j}}{{(1-x^2)}^{2j}}}{\left({\displaystyle \frac{\sqrt{1+x}}{\sqrt{1+x}-1}}\right)}^j.\hfill \end{array}$$

For the sum in the last two lines, the upper limit of the sums has been released to ∞, since the contribution of the terms corresponding to $j>n$ to the coefficient $\left[x^{2n}\right]$ results in zero. Evaluating the last series gives rise to

$${\mathsf{\Omega }}_a\left(n\right)=\left[x^{2n}\right]{\mathrm{P}}_n\left(x\right)=\left\{{\mathrm{P}}_n^{+}\left(x\right)+{\mathrm{P}}_n^{-}\left(x\right)\right\},$$

where ${\mathrm{P}}_n^{+}\left(x\right)$ and ${\mathrm{P}}_n^{-}\left(x\right)$ are rational functions defined respectively by

$$\begin{array}{cc}\hfill {\mathrm{P}}_n^{+}\left(x\right)& :={\displaystyle \frac{(1-x^2){(1+x+\sqrt{1+x})}^n}{2(1-2x^2+x^3(\sqrt{1+x}-1))}}={\omega }_a(x,\sigma ),\hfill \\ \\ \hfill {\mathrm{P}}_n^{-}\left(x\right)& :={\displaystyle \frac{(1-x^2){(1+x-\sqrt{1+x})}^n}{2(1-2x^2-x^3(\sqrt{1+x}+1))}}={\omega }_a(x,-\sigma ).\hfill \end{array}$$

This validates the first expression (5) in Lemma 1.

2.2. Proof of Proposition 1(a)

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

$$\begin{array}{cc}\hfill {\Delta }_a\left(n\right):& ={\mathsf{\Omega }}_a\left(n\right)-6{\mathsf{\Omega }}_a(n-1)+3{\mathsf{\Omega }}_a(n-2)+{\mathsf{\Omega }}_a(n-3)+{\mathsf{\Omega }}_a(n-4)\hfill \\ & =\left[x^{2n}\right]\left\{{\mathrm{P}}_n\left(x\right)-6x^2{\mathrm{P}}_{n-1}\left(x\right)+3x^4{\mathrm{P}}_{n-2}\left(x\right)+x^6{\mathrm{P}}_{n-3}\left(x\right)+x^8{\mathrm{P}}_{n-4}\left(x\right)\right\},\hfill \end{array}$$

which can be expressed shortly as

$${\Delta }_a\left(n\right)=\left[x^{2n}\right]\left\{{\mathcal{P}}_{+}\left(x\right)+{\mathcal{P}}_{-}\left(x\right)\right\},$$

where

$$\begin{array}{cc}& {\mathcal{P}}_{+}\left(x\right)={\mathrm{P}}_n^{+}\left(x\right)-6x^2{\mathrm{P}}_{n-1}^{+}\left(x\right)+3x^4{\mathrm{P}}_{n-2}^{+}\left(x\right)+x^6{\mathrm{P}}_{n-3}^{+}\left(x\right)+x^8{\mathrm{P}}_{n-4}^{+}\left(x\right),\hfill \\ & {\mathcal{P}}_{-}\left(x\right)={\mathrm{P}}_n^{-}\left(x\right)-6x^2{\mathrm{P}}_{n-1}^{-}\left(x\right)+3x^4{\mathrm{P}}_{n-2}^{-}\left(x\right)+x^6{\mathrm{P}}_{n-3}^{-}\left(x\right)+x^8{\mathrm{P}}_{n-4}^{-}\left(x\right).\hfill \end{array}$$

Under the change of variable $\boxed{x\to y^2-1}$, it can be verified that both ${\mathcal{P}}_{+}(y^2-1)$ and ${\mathcal{P}}_{-}(y^2-1)$ become polynomials:

$$\begin{array}{cc}& {\mathcal{P}}_{+}(y^2-1)={\displaystyle \frac{1}{2}}{(1+y)}^3(y^2-2){(y^2+y)}^{n-3}(1-2y+3y^2-2y^3-3y^4+3y^5-y^6),\hfill \\ & {\mathcal{P}}_{-}(y^2-1)={\displaystyle \frac{1}{2}}{(1-y)}^3(y^2-2){(y^2-y)}^{n-3}(1+2y+3y^2+2y^3-3y^4-3y^5-y^6).\hfill \end{array}$$

Observe further that ${\mathcal{P}}_{+}(y^2-1)+{\mathcal{P}}_{-}(y^2-1)$ is, in fact, a polynomial in $y^2$ since it is not only a polynomial but also an even function in y. By sending back $\boxed{y\to \sqrt{1+x}}$, we conclude that ${\mathcal{P}}_{+}\left(x\right)+{\mathcal{P}}_{-}\left(x\right)$ is a polynomial in x of degree “$n+2$”. This confirms the recurrence relation for ${\mathsf{\Omega }}_a\left(n\right)$ stated in Proposition 1:

$${\Delta }_a\left(n\right)=\left[x^{2n}\right]\left\{{\mathcal{P}}_{+}\left(x\right)+{\mathcal{P}}_{-}\left(x\right)\right\}=0\mathrm{when}n>3.$$

2.3. Proof of Theorem 1(a)

According to the relation between the recurrent sequence and the generating function (see Stanley [22] Theorem 4.1.1), there exists a polynomial $\rho \left(y\right)$ of degree less than 4 such that

$$\sum _{n=0}^{\infty }{\mathsf{\Omega }}_a\left(n\right)y^n={\displaystyle \frac{\rho \left(y\right)}{1-6y+3y^2+y^3+y^4}}={\displaystyle \frac{\rho \left(y\right)}{(1-y)(1-5y-2y^2-y^3)}}.$$

Finally, by computing the initial values

$${\left\{{\mathsf{\Omega }}_a\left(n\right)\right\}}_{n=0}^3=\left\{1,1,4,20\right\}$$

and then resolving the linear equations concerning the coefficient of $\rho \left(y\right)$, we can determine explicitly

$$\rho \left(y\right)=1-5y+y^2.$$

This completes the proof of Theorem 1(a).

3. Alternating Triple Sums with $\Delta (\mathit{i},\mathit{j},\mathit{k})={(-\mathbf{1})}^{\mathit{i}}$

For the four triple sums with an alternating sign defined by

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

(13)

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

(14)

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

(15)

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

(16)

their recurrence relations and generating functions are summarized in the following statements.

Lemma 2

(Coefficient expressions: $\sigma :=\sqrt{1+x}$).

$$\begin{array}{cc}\hfill \left(\mathrm{a}\right){\mathsf{\Phi }}_a\left(n\right)& =\left[x^{2n}\right]\left\{{\varphi }_a(x,\sigma )+{\varphi }_a(x,-\sigma )\right\}:{\varphi }_a(x,\sigma ):={\displaystyle \frac{{(-1)}^n(1+x^2){(1+x+\sigma )}^n}{2(1+2x^2-x^3(1-\sigma ))}},\hfill \end{array}$$

(17)

$$\begin{array}{cc}\hfill \left(\mathrm{b}\right){\mathsf{\Phi }}_b\left(n\right)& =\left[x^{2n+1}\right]\left\{{\varphi }_b(x,\sigma )+{\varphi }_b(x,-\sigma )\right\}:{\varphi }_b(x,\sigma ):={\displaystyle \frac{{(-1)}^n(1+x)(1+x^2){(1+x+\sigma )}^n}{2(1+2x^2-x^3(1-\sigma ))}},\hfill \end{array}$$

(18)

$$\begin{array}{cc}\hfill \left(\mathrm{c}\right){\mathsf{\Phi }}_c\left(n\right)& =\left[x^{2n+1}\right]\left\{{\varphi }_c(x,\sigma )+{\varphi }_c(x,-\sigma )\right\}:{\varphi }_c(x,\sigma ):={\displaystyle \frac{{(-1)}^n\sigma (1+x^2){(1+x+\sigma )}^{n+1}}{2(1+2x^2-x^3(1-\sigma ))}},\hfill \end{array}$$

(19)

$$\begin{array}{cc}\hfill \left(\mathrm{d}\right){\mathsf{\Phi }}_d\left(n\right)& =\left[x^{2n+1}\right]\left\{{\varphi }_d(x,\sigma )+{\varphi }_d(x,-\sigma )\right\}:{\varphi }_d(x,\sigma ):={\displaystyle \frac{{(-1)}^n\sigma {(1+x+\sigma )}^{n+1}}{2(1+2x^2-x^3(1-\sigma ))}}.\hfill \end{array}$$

(20)

Proposition 2.

The four triple sums $\{{\mathsf{\Phi }}_a\left(n\right),{\mathsf{\Phi }}_b\left(n\right),{\mathsf{\Phi }}_c\left(n\right),{\mathsf{\Phi }}_d\left(n\right)\}$ satisfy the same recurrence relation

$$\begin{array}{cc}\hfill \mathsf{\Phi }\left(n\right)& =8\mathsf{\Phi }(n-1)-24\mathsf{\Phi }(n-2)+21\mathsf{\Phi }(n-3)-14\mathsf{\Phi }(n-4)\hfill \\ & +3\mathsf{\Phi }(n-5)+3\mathsf{\Phi }(n-6)-4\mathsf{\Phi }(n-7)\hfill \end{array}$$

with different initial values

$$\begin{array}{cc}\hfill {\left\{{\mathsf{\Phi }}_a\left(n\right)\right\}}_{n=0}^6& =\left\{1,1,2,0,-34,-243,-1150\right\},\hfill \\ \hfill {\left\{{\mathsf{\Phi }}_b\left(n\right)\right\}}_{n=0}^6& =\left\{1,2,7,22,44,-61,-1167\right\},\hfill \\ \hfill {\left\{{\mathsf{\Phi }}_c\left(n\right)\right\}}_{n=0}^6& =\left\{1,4,14,40,53,-292,-2953\right\},\hfill \\ \hfill {\left\{{\mathsf{\Phi }}_d\left(n\right)\right\}}_{n=0}^6& =\left\{1,8,33,99,150,-592,-6692\right\}.\hfill \end{array}$$

Theorem 2

(Generating functions).

$$\begin{array}{cc}\hfill \left(\mathrm{a}\right)\sum _{n=0}^{\infty }{\mathsf{\Phi }}_a\left(n\right)y^n& ={\displaystyle \frac{1-7y+18y^2-13y^3+7y^4-2y^5}{1-8y+24y^2-21y^3+14y^4-3y^5-3y^6+4y^7}},\hfill \end{array}$$

(21)

$$\begin{array}{cc}\hfill \left(\mathrm{b}\right)\sum _{n=0}^{\infty }{\mathsf{\Phi }}_b\left(n\right)y^n& ={\displaystyle \frac{1-6y+15y^2-7y^3+8y^4-7y^5+4y^6}{1-8y+24y^2-21y^3+14y^4-3y^5-3y^6+4y^7}},\hfill \end{array}$$

(22)

$$\begin{array}{cc}\hfill \left(\mathrm{c}\right)\sum _{n=0}^{\infty }{\mathsf{\Phi }}_c\left(n\right)y^n& ={\displaystyle \frac{1-4y+6y^2+3y^3-y^4+3y^5-4y^6}{1-8y+24y^2-21y^3+14y^4-3y^5-3y^6+4y^7}},\hfill \end{array}$$

(23)

$$\begin{array}{cc}\hfill \left(\mathrm{d}\right)\sum _{n=0}^{\infty }{\mathsf{\Phi }}_d\left(n\right)y^n& ={\displaystyle \frac{1-7y^2+6y^3-4y^4}{1-8y+24y^2-21y^3+14y^4-3y^5-3y^6+4y^7}}.\hfill \end{array}$$

(24)

We take ${\mathsf{\Phi }}_b\left(n\right)$ as an example to show how to derive the expression (18) in Lemma 2, the recurrence relation in Proposition 2, and the generating function (22) in Theorem 2. Other results can be done analogously.

3.1. Proof of Lemma 2(b)

According to the binomial expressions

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

the binomial sum with respect to i in ${\mathsf{\Phi }}_b\left(n\right)$ can be reformulated as

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

Then we can proceed with the binomial sum with respect to k in ${\mathsf{\Phi }}_b\left(n\right)$:

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

Consequently, the remaining binomial sum with respect to j in ${\mathsf{\Phi }}_b\left(n\right)$ reads as

$$\begin{array}{cc}\hfill {\mathsf{\Phi }}_b\left(n\right)=& \sum _{j=0}^n{\mathsf{\Phi }}_b^k\left(n\right)\hfill \\ \hfill =& \sum _{j=0}^n\left[x^{2n+1}\right]{(-1)}^n{\displaystyle \frac{x^{4j}{(1+x)}^{n+1}}{2{(1+x^2)}^{2j+1}}}\left\{{\left({\displaystyle \frac{\sqrt{1+x}+1}{\sqrt{1+x}}}\right)}^{n-j}+{\left({\displaystyle \frac{\sqrt{1+x}-1}{\sqrt{1+x}}}\right)}^{n-j}\right\}\hfill \\ \hfill =& \left[x^{2n+1}\right]{(-1)}^n{\displaystyle \frac{{(1+x)}^{n+1}}{2(1+x^2)}}{\left({\displaystyle \frac{\sqrt{1+x}+1}{\sqrt{1+x}}}\right)}^n\sum _{j=0}^{\infty }{\displaystyle \frac{x^{4j}}{{(1+x^2)}^{2j}}}{\left({\displaystyle \frac{\sqrt{1+x}}{\sqrt{1+x}+1}}\right)}^j\hfill \\ \hfill +& \left[x^{2n+1}\right]{(-1)}^n{\displaystyle \frac{{(1+x)}^{n+1}}{2(1+x^2)}}{\left({\displaystyle \frac{\sqrt{1+x}-1}{\sqrt{1+x}}}\right)}^n\sum _{j=0}^{\infty }{\displaystyle \frac{x^{4j}}{{(1+x^2)}^{2j}}}{\left({\displaystyle \frac{\sqrt{1+x}}{\sqrt{1+x}-1}}\right)}^j,\hfill \end{array}$$

where the upper limit of the sums has been released to ∞, since the contribution of the terms corresponding to $j>n$ to the coefficient $\left[x^{2n+1}\right]$ results in zero. Evaluating the last series gives rise to

$${\mathsf{\Phi }}_b\left(n\right)=\left[x^{2n+1}\right]{\mathrm{P}}_n\left(x\right)=\left[x^{2n+1}\right]\left\{{\mathrm{P}}_n^{+}\left(x\right)+{\mathrm{P}}_n^{-}\left(x\right)\right\},$$

where ${\mathrm{P}}_n^{+}\left(x\right)$ and ${\mathrm{P}}_n^{-}\left(x\right)$ are respectively rational functions defined by

$$\begin{array}{cc}\hfill {\mathrm{P}}_n^{+}\left(x\right)& :={(-1)}^n{\displaystyle \frac{(1+x)(1+x^2){(1+x+\sqrt{1+x})}^n}{2(1+2x^2+x^3(\sqrt{1+x}-1))}}={\varphi }_b(x,\sigma ),\hfill \\ \\ \hfill {\mathrm{P}}_n^{-}\left(x\right)& :={(-1)}^n{\displaystyle \frac{(1+x)(1+x^2){(1+x-\sqrt{1+x})}^n}{2(1+2x^2-x^3(\sqrt{1+x}+1))}}={\varphi }_b(x,-\sigma ).\hfill \end{array}$$

This validates the second expression (18) in Lemma 2.

3.2. Proof of Proposition 2(b)

To prove the recurrence relation in Proposition 2, we write it alternatively by

$$\begin{array}{cc}\hfill {\Delta }_b\left(n\right):=& {\mathsf{\Phi }}_b\left(n\right)-8{\mathsf{\Phi }}_b(n-1)+24{\mathsf{\Phi }}_b(n-2)-21{\mathsf{\Phi }}_b(n-3)\hfill \\ \hfill +& 14{\mathsf{\Phi }}_b(n-4)-3{\mathsf{\Phi }}_b(n-5)-3{\mathsf{\Phi }}_b(n-6)+4{\mathsf{\Phi }}_b(n-7)\hfill \\ \hfill =& \left[x^{2n+1}\right]\{{\mathrm{P}}_n\left(x\right)-8x^2{\mathrm{P}}_{n-1}\left(x\right)+24x^4{\mathrm{P}}_{n-2}\left(x\right)-21x^6{\mathrm{P}}_{n-3}\left(x\right)\hfill \\ \hfill +& 14x^8{\mathrm{P}}_{n-4}\left(x\right)-3x^{10}{\mathrm{P}}_{n-5}\left(x\right)-3x^{12}{\mathrm{P}}_{n-6}\left(x\right)+4x^{14}{\mathrm{P}}_{n-7}\left(x\right)\},\hfill \end{array}$$

which can be expressed shortly as

$${\Delta }_b\left(n\right)=\left[x^{2n+1}\right]\left\{{\mathcal{P}}_{+}\left(x\right)+{\mathcal{P}}_{-}\left(x\right)\right\},$$

where

$$\begin{array}{cc}\hfill {\mathcal{P}}_{+}\left(x\right)& ={\mathrm{P}}_n^{+}\left(x\right)-8x^2{\mathrm{P}}_{n-1}^{+}\left(x\right)+24x^4{\mathrm{P}}_{n-2}^{+}\left(x\right)-21x^6{\mathrm{P}}_{n-3}^{+}\left(x\right)\hfill \\ & +14x^8{\mathrm{P}}_{n-4}^{+}\left(x\right)-3x^{10}{\mathrm{P}}_{n-5}^{+}\left(x\right)-3x^{12}{\mathrm{P}}_{n-6}^{+}\left(x\right)+4x^{14}{\mathrm{P}}_{n-7}^{+}\left(x\right),\hfill \\ \hfill {\mathcal{P}}_{-}\left(x\right)& ={\mathrm{P}}_n^{-}\left(x\right)-8x^2{\mathrm{P}}_{n-1}^{-}\left(x\right)+24x^4{\mathrm{P}}_{n-2}^{-}\left(x\right)-21x^6{\mathrm{P}}_{n-3}^{-}\left(x\right)\hfill \\ & +14x^8{\mathrm{P}}_{n-4}^{-}\left(x\right)-3x^{10}{\mathrm{P}}_{n-5}^{-}\left(x\right)-3x^{12}{\mathrm{P}}_{n-6}^{-}\left(x\right)+4x^{14}{\mathrm{P}}_{n-7}^{-}\left(x\right).\hfill \end{array}$$

Under the change of variable $\boxed{x\to y^2-1}$, it can be verified that both ${\mathcal{P}}_{+}(y^2-1)$ and ${\mathcal{P}}_{-}(y^2-1)$ become polynomials:

$$\begin{array}{cc}\hfill {\mathcal{P}}_{+}(y^2-1)=& {\displaystyle \frac{{(-1)}^{n-1}}{2}}{(1+y)}^5(2-2y^2+y^4){(y+y^2)}^{n-5}(1-6y+9y^2+12y^3-52y^4\hfill \\ \hfill +& 33y^5+73y^6-126y^7+16y^8+117y^9-98y^{10}-7y^{11}+47y^{12}-24y^{13}+4y^{14}),\hfill \\ \hfill {\mathcal{P}}_{-}(y^2-1)=& {\displaystyle \frac{{(-1)}^{n-1}}{2}}{(1-y)}^5(2-2y^2+y^4){(y^2-y)}^{n-5}(1+6y+9y^2-12y^3-52y^4\hfill \\ \hfill -& 33y^5+73y^6+126y^7+16y^8-117y^9-98y^{10}+7y^{11}+47y^{12}+24y^{13}+4y^{14}).\hfill \end{array}$$

Observe further that ${\mathcal{P}}_{+}(y^2-1)+{\mathcal{P}}_{-}(y^2-1)$ is, in fact, a polynomial in $y^2$ since it is not only a polynomial but also an even function in y. By sending back $\boxed{y\to \sqrt{1+x}}$, we conclude that ${\mathcal{P}}_{+}\left(x\right)+{\mathcal{P}}_{-}\left(x\right)$ is a polynomial in x of degree “$n+6$”. This confirms the recurrence relation for ${\mathsf{\Phi }}_b\left(n\right)$ stated in Proposition 2:

$${\Delta }_b\left(n\right)=\left[x^{2n+1}\right]\left\{{\mathcal{P}}_{+}\left(x\right)+{\mathcal{P}}_{-}\left(x\right)\right\}=0\mathrm{when}n>6.$$

3.3. Proof of Theorem 2(b)

According to the relation between the recurrent sequence and the generating function, there exists a polynomial $\rho \left(y\right)$ of a degree less than 7 such that

$$\sum _{n=0}^{\infty }{\mathsf{\Phi }}_b\left(n\right)y^n={\displaystyle \frac{\rho \left(y\right)}{1-8y+24y^2-21y^3+14y^4-3y^5-3y^6+4y^7}}.$$

Finally, by computing the initial values

$${\left\{{\mathsf{\Phi }}_b\left(n\right)\right\}}_{n=0}^6=\left\{1,2,7,22,44,-61,-1167\right\}$$

and then resolving the linear equations concerning the coefficient of $\rho \left(y\right)$, we can determine explicitly

$$\rho \left(y\right)=1-6y+15y^2-7y^3+8y^4-7y^5+4y^6.$$

This completes the proof of Theorem 2(b).□

4. Alternating Triple Sums with $\Delta (\mathit{i},\mathit{j},\mathit{k})={(-\mathbf{1})}^{\mathit{i}+\mathit{j}}$

For the four triple sums with double alternating signs defined by

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

(25)

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

(26)

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

(27)

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

(28)

their recurrence relations and generating functions are highlighted as follows.

Lemma 3

(Coefficient expressions:  $\sigma :=\sqrt{1+x}$).

$$\begin{array}{cc}\hfill \left(\mathrm{a}\right){\mathsf{\Psi }}_a\left(n\right)& =\left[x^{2n}\right]\left\{{\psi }_a(x,\sigma )+{\psi }_a(x,-\sigma )\right\}:{\psi }_a(x,\sigma ):={\displaystyle \frac{{(-1)}^n(1+x^2){(1+x+\sigma )}^n}{2(1+2x^2+x^3(1-\sigma )+2x^4)}},\hfill \end{array}$$

(29)

$$\begin{array}{cc}\hfill \left(\mathrm{b}\right){\mathsf{\Psi }}_b\left(n\right)& =\left[x^{2n+1}\right]\left\{{\psi }_b(x,\sigma )+{\psi }_b(x,-\sigma )\right\}:{\psi }_b(x,\sigma ):={\displaystyle \frac{{(-1)}^n(1+x)(1+x^2){(1+x+\sigma )}^n}{2(1+2x^2+x^3(1-\sigma )+2x^4)}},\hfill \end{array}$$

(30)

$$\begin{array}{cc}\hfill \left(\mathrm{c}\right){\mathsf{\Psi }}_c\left(n\right)& =\left[x^{2n+1}\right]\left\{{\psi }_c(x,\sigma )+{\psi }_c(x,-\sigma )\right\}:{\psi }_c(x,\sigma ):={\displaystyle \frac{{(-1)}^n\sigma (1+x^2){(1+x+\sigma )}^{n+1}}{2(1+2x^2+x^3(1-\sigma )+2x^4)}},\hfill \end{array}$$

(31)

$$\begin{array}{cc}\hfill \left(\mathrm{d}\right){\mathsf{\Psi }}_d\left(n\right)& =\left[x^{2n+1}\right]\left\{{\psi }_d(x,\sigma )+{\psi }_d(x,-\sigma )\right\}:{\psi }_d(x,\sigma ):={\displaystyle \frac{{(-1)}^n\sigma {(1+x+\sigma )}^{n+1}}{2(1+2x^2+x^3(1-\sigma )+2x^4)}}.\hfill \end{array}$$

(32)

Proposition 3.

The four triple sums $\{{\mathsf{\Psi }}_a\left(n\right),{\mathsf{\Psi }}_b\left(n\right),{\mathsf{\Psi }}_c\left(n\right),{\mathsf{\Psi }}_d\left(n\right)\}$ satisfy the same recurrence relation

$$\begin{array}{cc}\hfill \mathsf{\Psi }\left(n\right)& =8\mathsf{\Psi }(n-1)-32\mathsf{\Psi }(n-2)+75\mathsf{\Psi }(n-3)-74\mathsf{\Psi }(n-4)+25\mathsf{\Psi }(n-5)\hfill \\ & -73\mathsf{\Psi }(n-6)+24\mathsf{\Psi }(n-7)-16\mathsf{\Psi }(n-8)\hfill \end{array}$$

with different initial values

$$\begin{array}{cc}\hfill {\left\{{\mathsf{\Psi }}_a\left(n\right)\right\}}_{n=0}^7& =\left\{1,1,0,-4,6,119,428,313,-2803\right\},\hfill \\ \hfill {\left\{{\mathsf{\Psi }}_b\left(n\right)\right\}}_{n=0}^7& =\left\{1,2,1,-10,-30,75,697,1585\right\},\hfill \\ \hfill {\left\{{\mathsf{\Psi }}_c\left(n\right)\right\}}_{n=0}^7& =\left\{1,4,2,-24,-51,236,1595,2941\right\},\hfill \\ \hfill {\left\{{\mathsf{\Psi }}_d\left(n\right)\right\}}_{n=0}^7& =\left\{1,8,21,5,-130,-192,1572,8565\right\}.\hfill \end{array}$$

Theorem 3

(Generating functions).

$$\begin{array}{cc}\hfill \left(\mathrm{a}\right)\sum _{n=0}^{\infty }{\mathsf{\Psi }}_a\left(n\right)y^n& ={\displaystyle \frac{1-7y+24y^2-47y^3+37y^4-8y^5+16y^6}{1-8y+32y^2-75y^3+74y^4-25y^5+73y^6-24y^7+16y^8}},\hfill \end{array}$$

(33)

$$\begin{array}{cc}\hfill \left(\mathrm{b}\right)\sum _{n=0}^{\infty }{\mathsf{\Psi }}_b\left(n\right)y^n& ={\displaystyle \frac{1-6y+17y^2-29y^3+6y^4+43y^5-16y^6+16y^7}{1-8y+32y^2-75y^3+74y^4-25y^5+73y^6-24y^7+16y^8}},\hfill \end{array}$$

(34)

$$\begin{array}{cc}\hfill \left(\mathrm{c}\right)\sum _{n=0}^{\infty }{\mathsf{\Psi }}_c\left(n\right)y^n& ={\displaystyle \frac{1-4y+2y^2+13y^3-21y^4-3y^5-4y^6}{1-8y+32y^2-75y^3+74y^4-25y^5+73y^6-24y^7+16y^8}},\hfill \end{array}$$

(35)

$$\begin{array}{cc}\hfill \left(\mathrm{d}\right)\sum _{n=0}^{\infty }{\mathsf{\Psi }}_d\left(n\right)y^n& ={\displaystyle \frac{1-11y^2+18y^3-24y^4}{1-8y+32y^2-75y^3+74y^4-25y^5+73y^6-24y^7+16y^8}}.\hfill \end{array}$$

(36)

We take ${\mathsf{\Psi }}_c\left(n\right)$ as a sample to show how to derive the expression (31) in Lemma 3, the recurrence relation in Proposition 3, and the generating function (35) in Theorem 3. Other results can be proved similarly.

4.1. Proof of Lemma 3(c)

Observe that the binomial sum with respect to i in ${\mathsf{\Psi }}_c\left(n\right)$ is the same as ${\mathsf{\Phi }}_b^i\left(n\right)$:

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

We can proceed with the binomial sum with respect to k in ${\mathsf{\Psi }}_c\left(n\right)$:

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

Consequently, the remaining binomial sum with respect to j in ${\mathsf{\Psi }}_c\left(n\right)$ reads as

$$\begin{array}{cc}\hfill {\mathsf{\Psi }}_c\left(n\right)=& \sum _{j=0}^n{(-1)}^j{\mathsf{\Psi }}_c^k\left(n\right)\hfill \\ \hfill =& \sum _{j=0}^n{(-1)}^j\left[x^{2n+1}\right]{(-1)}^n{\displaystyle \frac{x^{4j}\sqrt{1+x}{(1+x)}^{n+1}}{2{(1+x^2)}^{2j+1}}}\left\{{\left({\displaystyle \frac{\sqrt{1+x}+1}{\sqrt{1+x}}}\right)}^{n-j+1}-{\left({\displaystyle \frac{\sqrt{1+x}-1}{\sqrt{1+x}}}\right)}^{n-j+1}\right\}\hfill \\ \hfill =& \left[x^{2n+1}\right]{(-1)}^n{\displaystyle \frac{\sqrt{1+x}{(1+x)}^{n+1}}{2(1+x^2)}}{\left({\displaystyle \frac{\sqrt{1+x}+1}{\sqrt{1+x}}}\right)}^{n+1}\sum _{j=0}^{\infty }{(-1)}^j{\displaystyle \frac{x^{4j}}{{(1+x^2)}^{2j}}}{\left({\displaystyle \frac{\sqrt{1+x}}{\sqrt{1+x}+1}}\right)}^j\hfill \\ \hfill -& \left[x^{2n+1}\right]{(-1)}^n{\displaystyle \frac{\sqrt{1+x}{(1+x)}^{n+1}}{2(1+x^2)}}{\left({\displaystyle \frac{\sqrt{1+x}-1}{\sqrt{1+x}}}\right)}^{n+1}\sum _{j=0}^{\infty }{(-1)}^j{\displaystyle \frac{x^{4j}}{{(1+x^2)}^{2j}}}{\left({\displaystyle \frac{\sqrt{1+x}}{\sqrt{1+x}-1}}\right)}^j,\hfill \end{array}$$

where the upper limit of the sums has been released to ∞ for the same reason as stated before. Evaluating the last series gives rise to

$${\mathsf{\Psi }}_c\left(n\right)=\left[x^{2n+1}\right]\{{\mathrm{P}}_n\left(x\right)=\left[x^{2n+1}\right]\left\{{\mathrm{P}}_n^{+}\left(x\right)-{\mathrm{P}}_n^{-}\left(x\right)\right\},$$

where ${\mathrm{P}}_n^{+}\left(x\right)$ and ${\mathrm{P}}_n^{-}\left(x\right)$ are rational functions defined respectively by

$$\begin{array}{cc}\hfill {\mathrm{P}}_n^{+}\left(x\right)& :={(-1)}^n{\displaystyle \frac{\sqrt{1+x}(1+x^2){(1+x+\sqrt{1+x})}^{n+1}}{2(1+2x^2-x^3(\sqrt{1+x}-1)+2x^4)}}={\psi }_c(x,\sigma ),\hfill \\ \\ \hfill {\mathrm{P}}_n^{-}\left(x\right)& :={(-1)}^n{\displaystyle \frac{\sqrt{1+x}(1+x^2){(1+x-\sqrt{1+x})}^{n+1}}{2(1+2x^2+x^3(\sqrt{1+x}+1)+2x^4)}}=-{\psi }_c(x,-\sigma ).\hfill \end{array}$$

This validates the third expression (31) in Lemma 3.

4.2. Proof of Proposition 3(c)

To prove the recurrence relation in Proposition 3, we write it alternatively by

$$\begin{array}{cc}\hfill {\Delta }_c\left(n\right):=& {\mathsf{\Psi }}_c\left(n\right)-8{\mathsf{\Psi }}_c(n-1)+32{\mathsf{\Psi }}_c(n-2)-75{\mathsf{\Psi }}_c(n-3)+74{\mathsf{\Psi }}_c(n-4)\hfill \\ \hfill -& 25{\mathsf{\Psi }}_c(n-5)+73{\mathsf{\Psi }}_c(n-6)-24{\mathsf{\Psi }}_c(n-7)+16{\mathsf{\Psi }}_c(n-8)\hfill \\ \hfill =& \left[x^{2n+1}\right]\{{\mathrm{P}}_n\left(x\right)-8x^2{\mathrm{P}}_{n-1}\left(x\right)+32x^4{\mathrm{P}}_{n-2}\left(x\right)-75x^6{\mathrm{P}}_{n-3}\left(x\right)+74x^8{\mathrm{P}}_{n-4}\left(x\right)\hfill \\ \hfill -& 25x^{10}{\mathrm{P}}_{n-5}\left(x\right)+73x^{12}{\mathrm{P}}_{n-6}\left(x\right)-24x^{14}{\mathrm{P}}_{n-7}\left(x\right)+16x^{16}{\mathrm{P}}_{n-8}\left(x\right)\},\hfill \end{array}$$

which can be expressed shortly as

$${\Delta }_c\left(n\right)=\left[x^{2n+1}\right]\left\{{\mathcal{P}}_{+}\left(x\right)-{\mathcal{P}}_{-}\left(x\right)\right\},$$

where

$$\begin{array}{cc}\hfill {\mathcal{P}}_{+}\left(x\right)=& {\mathrm{P}}_n^{+}\left(x\right)-8x^2{\mathrm{P}}_{n-1}^{+}\left(x\right)+32x^4{\mathrm{P}}_{n-2}^{+}\left(x\right)-75x^6{\mathrm{P}}_{n-3}^{+}\left(x\right)+74x^8{\mathrm{P}}_{n-4}^{+}\left(x\right)-25x^{10}{\mathrm{P}}_{n-5}^{+}\left(x\right)\hfill \\ \hfill +& 73x^{12}{\mathrm{P}}_{n-6}^{+}\left(x\right)-24x^{14}{\mathrm{P}}_{n-7}^{+}\left(x\right)+16x^{16}{\mathrm{P}}_{n-8}^{+}\left(x\right),\hfill \\ \hfill {\mathcal{P}}_{-}\left(x\right)=& {\mathrm{P}}_n^{-}\left(x\right)-8x^2{\mathrm{P}}_{n-1}^{-}\left(x\right)+32x^4{\mathrm{P}}_{n-2}^{-}\left(x\right)-75x^6{\mathrm{P}}_{n-3}^{-}\left(x\right)+74x^8{\mathrm{P}}_{n-4}^{-}\left(x\right)-25x^{10}{\mathrm{P}}_{n-5}^{-}\left(x\right)\hfill \\ \hfill +& 73x^{12}{\mathrm{P}}_{n-6}^{-}\left(x\right)-24x^{14}{\mathrm{P}}_{n-7}^{-}\left(x\right)+16x^{16}{\mathrm{P}}_{n-8}^{-}\left(x\right).\hfill \end{array}$$

Under the change of variable $\boxed{x\to y^2-1}$, it can be verified that both ${\mathcal{P}}_{+}(y^2-1)$ and ${\mathcal{P}}_{-}(y^2-1)$ become polynomials:

$$\begin{array}{cc}\hfill {\mathcal{P}}_{+}(y^2-1)=& {\displaystyle \frac{{(-1)}^n}{2}}(y^4-2y^2+2){(1+y)}^7{(y+y^2)}^{n-6}(4-27y+72y^2-63y^3-122y^4+408y^5-361y^6\hfill \\ \hfill -& 295y^7+896y^8-516y^9-425y^{10}+710y^{11}-241y^{12}-157y^{13}+170y^{14}-60x^{15}+8y^{16}),\hfill \\ \hfill {\mathcal{P}}_{-}(y^2-1)=& {\displaystyle \frac{{(-1)}^n}{2}}(y^4-2y^2+2){(y-1)}^7{(y^2-y)}^{n-6}(4+27y+72y^2+63y^3-122y^4-408y^5-361y^6\hfill \\ \hfill +& 295y^7+896y^8+516y^9-425y^{10}-710y^{11}-241y^{12}+157y^{13}+170y^{14}+60x^{15}+8y^{16}).\hfill \end{array}$$

Observe further that ${\mathcal{P}}_{+}(y^2-1)-{\mathcal{P}}_{-}(y^2-1)$ is, in fact, a polynomial in $y^2$ since it is not only a polynomial but also an even function in y. By sending back $\boxed{y\to \sqrt{1+x}}$, we conclude that ${\mathcal{P}}_{+}\left(x\right)-{\mathcal{P}}_{-}\left(x\right)$ is a polynomial in x of degree “$n+7$”. This confirms the recurrence relation for ${\mathsf{\Psi }}_c\left(n\right)$ stated in Proposition 3:

$${\Delta }_c\left(n\right)=\left[x^{2n+1}\right]\left\{{\mathcal{P}}_{+}\left(x\right)-{\mathcal{P}}_{-}\left(x\right)\right\}=0\mathrm{when}n>7.$$

4.3. Proof of Theorem 3(c)

According to the relation between the recurrent sequence and the generating function, there exists a polynomial $\rho \left(y\right)$ of a degree less than 8 such that

$$\sum _{n=0}^{\infty }{\mathsf{\Psi }}_c\left(n\right)y^n={\displaystyle \frac{\rho \left(y\right)}{1-8y+32y^2-75y^3+74y^4-25y^5+73y^6-24y^7+16y^8}}.$$

Finally, by computing the initial values

$${\left\{{\mathsf{\Psi }}_c\left(n\right)\right\}}_{n=0}^7=\left\{1,4,2,-24,-51,236,1595,2941\right\}$$

and then resolving the linear equations concerning the coefficient of $\rho \left(y\right)$, we can determine explicitly

$$\rho \left(y\right)=1-4y+2y^2+13y^3-21y^4-3y^5-4y^6.$$

This completes the proof of Theorem 3(c).□

5. Alternating Triple Sums with $\Delta (\mathit{i},\mathit{j},\mathit{k})={(-\mathbf{1})}^{\mathit{i}+\mathit{j}+\mathit{k}}$

Finally, for the four alternating triple sums defined by

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

(37)

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

(38)

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

(39)

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

(40)

their recurrence relations and generating functions are simpler compared with those produced in the precedent sections.

Lemma 4

(Coefficient expressions:  $\mathbf{i}=\sqrt{-1}$ and  $\sigma :=\sqrt{1+x}$).

$$\begin{array}{cc}\hfill \left(\mathrm{a}\right){\mathsf{\Lambda }}_a\left(n\right)& =\left[x^{2n}\right]\left\{{\lambda }_a(x,\sigma )+{\lambda }_a(x,-\sigma )\right\}:{\lambda }_a(x,\sigma ):={\displaystyle \frac{{(-1)}^n(2+x)(1+x^2){(1+x+\sigma \mathbf{i})}^n}{2\mathbf{i}(2+x+4x^2+2x^3+x^4(3-\sigma \mathbf{i})+2x^5)}},\hfill \end{array}$$

(41)

$$\begin{array}{cc}\hfill \left(\mathrm{b}\right){\mathsf{\Lambda }}_b\left(n\right)& =\left[x^{2n+1}\right]\left\{{\lambda }_b(x,\sigma )+{\lambda }_b(x,-\sigma )\right\}:{\lambda }_b(x,\sigma ):={\displaystyle \frac{{(-1)}^n(1+x)(2+x)(1+x^2){(1+x+\sigma \mathbf{i})}^n}{2\mathbf{i}(2+x+4x^2+2x^3+x^4(3-\sigma \mathbf{i})+2x^5)}},\hfill \end{array}$$

(42)

$$\begin{array}{cc}\hfill \left(\mathrm{c}\right){\mathsf{\Lambda }}_c\left(n\right)& =\left[x^{2n+1}\right]\left\{{\lambda }_c(x,\sigma )+{\lambda }_c(x,-\sigma )\right\}:{\lambda }_c(x,\sigma ):={\displaystyle \frac{{(-1)}^n\sigma (2+x)(1+x^2){(1+x+\sigma \mathbf{i})}^{n+1}}{2\mathbf{i}(2+x+4x^2+2x^3+x^4(3-\sigma \mathbf{i})+2x^5)}},\hfill \end{array}$$

(43)

$$\begin{array}{cc}\hfill \left(\mathrm{d}\right){\mathsf{\Lambda }}_d\left(n\right)& =\left[x^{2n+1}\right]\left\{{\lambda }_d(x,\sigma )+{\lambda }_d(x,-\sigma )\right\}:{\lambda }_d(x,\sigma ):={\displaystyle \frac{{(-1)}^n\sigma (2+x){(1+x+\sigma \mathbf{i})}^{n+1}}{2\mathbf{i}(2+x+4x^2+2x^3+x^4(3-\sigma \mathbf{i})+2x^5)}}.\hfill \end{array}$$

(44)

Proposition 4.

The four triple sums $\{{\mathsf{\Lambda }}_a\left(n\right),{\mathsf{\Lambda }}_b\left(n\right),{\mathsf{\Lambda }}_c\left(n\right),{\mathsf{\Lambda }}_d\left(n\right)\}$ satisfy the same recurrence relation

$$\mathsf{\Lambda }\left(n\right)=2\mathsf{\Lambda }(n-1)-11\mathsf{\Lambda }(n-2)+25\mathsf{\Lambda }(n-3)-37\mathsf{\Lambda }(n-4)-4\mathsf{\Lambda }(n-5)$$

with different initial values

$$\begin{array}{cc}\hfill {\left\{{\mathsf{\Lambda }}_a\left(n\right)\right\}}_{n=0}^4& =\left\{1,1,-2,0,10\right\},\hfill \\ \hfill {\left\{{\mathsf{\Lambda }}_b\left(n\right)\right\}}_{n=0}^4& =\left\{1,2,-3,-6,34\right\},\hfill \\ \hfill {\left\{{\mathsf{\Lambda }}_c\left(n\right)\right\}}_{n=0}^4& =\left\{1,4,-2,-24,37\right\},\hfill \\ \hfill {\left\{{\mathsf{\Lambda }}_d\left(n\right)\right\}}_{n=0}^4& =\left\{1,8,9,-45,-26\right\}.\hfill \end{array}$$

Theorem 4

(Generating functions).

$$\begin{array}{cc}\hfill \left(\mathrm{a}\right)\sum _{n=0}^{\infty }{\mathsf{\Lambda }}_a\left(n\right)y^n& ={\displaystyle \frac{1-y+7y^2-10y^3}{(1-3y+4y^2)(1+y+10y^2+y^3)}},\hfill \end{array}$$

(45)

$$\begin{array}{cc}\hfill \left(\mathrm{b}\right)\sum _{n=0}^{\infty }{\mathsf{\Lambda }}_b\left(n\right)y^n& ={\displaystyle \frac{1+4y^2-3y^3}{(1-3y+4y^2)(1+y+10y^2+y^3)}},\hfill \end{array}$$

(46)

$$\begin{array}{cc}\hfill \left(\mathrm{c}\right)\sum _{n=0}^{\infty }{\mathsf{\Lambda }}_c\left(n\right)y^n& ={\displaystyle \frac{1+2y+y^2-y^3}{(1-3y+4y^2)(1+y+10y^2+y^3)}},\hfill \end{array}$$

(47)

$$\begin{array}{cc}\hfill \left(\mathrm{d}\right)\sum _{n=0}^{\infty }{\mathsf{\Lambda }}_d\left(n\right)y^n& ={\displaystyle \frac{1+6y+4y^2}{(1-3y+4y^2)(1+y+10y^2+y^3)}}.\hfill \end{array}$$

(48)

We take ${\mathsf{\Lambda }}_d\left(n\right)$ to exemplify how to derive the expression (44) in Lemma 4, the recurrence relation in Proposition 4, and the generating function (48) in Theorem 4. The results for the remaining three triple sums can be validated following the same procedure.

5.1. Proof of Lemma 4(d)

According to the binomial expressions

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

the binomial sum with respect to i in ${\mathsf{\Lambda }}_d\left(n\right)$ can be rewritten as

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

Then we can proceed with the binomial sum with respect to k in ${\mathsf{\Lambda }}_d\left(n\right)$:

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

Consequently, the remaining binomial sum with respect to j in ${\mathsf{\Lambda }}_d\left(n\right)$ reads as

$$\begin{array}{cc}\hfill {\mathsf{\Lambda }}_d\left(n\right)=& \sum _{j=0}^n{(-1)}^j{\mathsf{\Lambda }}_d^k\left(n\right)\hfill \\ \hfill =& \sum _{j=0}^n{(-1)}^j\left[x^{2n+1}\right]{(-1)}^n{\displaystyle \frac{x^{4j}\sqrt{1+x}{(1+x)}^{n+1}}{2\mathbf{i}{(1+x^2)}^{2j+2}}}\left\{{\left({\displaystyle \frac{\sqrt{1+x}+\mathbf{i}}{\sqrt{1+x}}}\right)}^{n-j+1}-{\left({\displaystyle \frac{\sqrt{1+x}-\mathbf{i}}{\sqrt{1+x}}}\right)}^{n-j+1}\right\}\hfill \\ \hfill =& \left[x^{2n+1}\right]{(-1)}^n{\displaystyle \frac{\sqrt{1+x}{(1+x)}^{n+1}}{2\mathbf{i}{(1+x^2)}^2}}{\left({\displaystyle \frac{\sqrt{1+x}+\mathbf{i}}{\sqrt{1+x}}}\right)}^{n+1}\sum _{j=0}^{\infty }{(-1)}^j{\displaystyle \frac{x^{4j}}{{(1+x^2)}^{2j}}}{\left({\displaystyle \frac{\sqrt{1+x}}{\sqrt{1+x}+\mathbf{i}}}\right)}^j\hfill \\ \hfill -& \left[x^{2n+1}\right]{(-1)}^n{\displaystyle \frac{\sqrt{1+x}{(1+x)}^{n+1}}{2\mathbf{i}{(1+x^2)}^2}}{\left({\displaystyle \frac{\sqrt{1+x}-\mathbf{i}}{\sqrt{1+x}}}\right)}^{n+1}\sum _{j=0}^{\infty }{(-1)}^j{\displaystyle \frac{x^{4j}}{{(1+x^2)}^{2j}}}{\left({\displaystyle \frac{\sqrt{1+x}}{\sqrt{1+x}-\mathbf{i}}}\right)}^j,\hfill \end{array}$$

where we have replaced the upper limit of the sums by ∞. Evaluating the last series gives rise to

$${\mathsf{\Lambda }}_d\left(n\right)=\left[x^{2n+1}\right]{\mathrm{P}}_n\left(x\right)=\left[x^{2n+1}\right]\left\{{\mathrm{P}}_n^{+}\left(x\right)-{\mathrm{P}}_n^{-}\left(x\right)\right\},$$

where ${\mathrm{P}}_n^{+}\left(x\right)$ and ${\mathrm{P}}_n^{-}\left(x\right)$ are rational functions defined respectively by

$$\begin{array}{cc}\hfill {\mathrm{P}}_n^{+}\left(x\right)& :={\displaystyle \frac{{(-1)}^n(2+x)\sqrt{1+x}{(1+x+\mathbf{i}\sqrt{1+x})}^{n+1}}{2\mathbf{i}(2+x+4x^2+2x^3+x^4(3-\mathbf{i}\sqrt{1+x})+2x^5)}}={\lambda }_d(x,\sigma ),\hfill \\ \hfill {\mathrm{P}}_n^{-}\left(x\right)& :={\displaystyle \frac{{(-1)}^n(2+x)\sqrt{1+x}{(1+x-\mathbf{i}\sqrt{1+x})}^{n+1}}{2\mathbf{i}(2+x+4x^2+2x^3+x^4(3+\mathbf{i}\sqrt{1+x})+2x^5)}}=-{\lambda }_d(x,-\sigma ).\hfill \end{array}$$

This validates the fourth expression (44) in Lemma 4.

5.2. Proof of Proposition 4(d)

To prove the recurrence relation in Proposition 4, we write it alternatively by

$$\begin{array}{cc}\hfill {\Delta }_d\left(n\right)& :={\mathsf{\Lambda }}_d\left(n\right)-2{\mathsf{\Lambda }}_d(n-1)+11{\mathsf{\Lambda }}_d(n-2)-25{\mathsf{\Lambda }}_d(n-3)+37{\mathsf{\Lambda }}_d(n-4)+4{\mathsf{\Lambda }}_d(n-5)\hfill \\ & =\left[x^{2n+1}\right]\left\{{\mathrm{P}}_n\left(x\right)-2x^2{\mathrm{P}}_{n-1}\left(x\right)+11x^4{\mathrm{P}}_{n-2}\left(x\right)-25x^6{\mathrm{P}}_{n-3}\left(x\right)+37x^8{\mathrm{P}}_{n-4}\left(x\right)+4x^{10}{\mathrm{P}}_{n-5}\left(x\right)\right\},\hfill \end{array}$$

which can be expressed shortly as

$${\Delta }_d\left(n\right)=\left[x^{2n+1}\right]\left\{{\mathcal{P}}_{+}\left(x\right)-{\mathcal{P}}_{-}\left(x\right)\right\},$$

where

$$\begin{array}{cc}& {\mathcal{P}}_{+}\left(x\right)={\mathrm{P}}_n^{+}\left(x\right)-2x^2{\mathrm{P}}_{n-1}^{+}\left(x\right)+11x^4{\mathrm{P}}_{n-2}^{+}\left(x\right)-25x^6{\mathrm{P}}_{n-3}^{+}\left(x\right)+37x^8{\mathrm{P}}_{n-4}^{+}\left(x\right)+4x^{10}{\mathrm{P}}_{n-5}^{+}\left(x\right),\hfill \\ & {\mathcal{P}}_{-}\left(x\right)={\mathrm{P}}_n^{-}\left(x\right)-2x^2{\mathrm{P}}_{n-1}^{-}\left(x\right)+11x^4{\mathrm{P}}_{n-2}^{-}\left(x\right)-25x^6{\mathrm{P}}_{n-3}^{-}\left(x\right)+37x^8{\mathrm{P}}_{n-4}^{-}\left(x\right)+4x^{10}{\mathrm{P}}_{n-5}^{-}\left(x\right).\hfill \end{array}$$

Under the change of variable $\boxed{x\to y^2-1}$, it can be verified that both ${\mathcal{P}}_{+}(y^2-1)$ and ${\mathcal{P}}_{-}(y^2-1)$ become polynomials:

$$\begin{array}{cc}\hfill {\mathcal{P}}_{+}(y^2-1)& ={\displaystyle \frac{{(-1)}^n}{2}}(1+\mathbf{i}y-2y^2){(y^2+\mathbf{i}y)}^{n-3}\hfill \\ & \times (1-9\mathbf{i}y-8y^2+34\mathbf{i}{y}^3+18y^4-41\mathbf{i}{y}^5-9y^6+15\mathbf{i}{y}^7-\mathbf{i}{y}^9),\hfill \\ \hfill {\mathcal{P}}_{-}(y^2-1)& ={\displaystyle \frac{{(-1)}^n}{2}}(2y^2+\mathbf{i}y-1){(y^2-\mathbf{i}y)}^{n-3}\hfill \\ & \times (1+9\mathbf{i}y-8y^2-34\mathbf{i}{y}^3+18y^4+41\mathbf{i}{y}^5-9y^6-15\mathbf{i}{y}^7+\mathbf{i}{y}^9).\hfill \end{array}$$

Observe further that ${\mathcal{P}}_{+}(y^2-1)-{\mathcal{P}}_{-}(y^2-1)$ is, in fact, a polynomial in $y^2$ since it is not only a polynomial but also an even function in y. By sending back $\boxed{y\to \sqrt{1+x}}$, we conclude that ${\mathcal{P}}_{+}\left(x\right)-{\mathcal{P}}_{-}\left(x\right)$ is a polynomial in x of degree “$n+2$”. This confirms the recurrence relation for ${\mathsf{\Lambda }}_d\left(n\right)$ stated in Proposition 4:

$${\Delta }_d\left(n\right)=\left[x^{2n+1}\right]\left\{{\mathcal{P}}_{+}\left(x\right)-{\mathcal{P}}_{-}\left(x\right)\right\}=0\mathrm{when}n>4.$$

5.3. Proof of Theorem 4(d)

According to the relation between the recurrent sequence and the generating function, there exists a polynomial $\rho \left(y\right)$ of a degree less than 5 such that

$$\sum _{n=0}^{\infty }{\mathsf{\Lambda }}_d\left(n\right)y^n={\displaystyle \frac{\rho \left(y\right)}{1-2y+11y^2-25y^3+37y^4+4y^5}}={\displaystyle \frac{\rho \left(y\right)}{(1-3y+4y^2)(1+y+10y^2+y^3)}}.$$

Finally, by computing the initial values

$${\left\{{\mathsf{\Lambda }}_d\left(n\right)\right\}}_{n=0}^4=\left\{1,8,9,-45,-26\right\}$$

and then resolving the linear equations concerning the coefficient of $\rho \left(y\right)$, we can determine explicitly

$$\rho \left(y\right)=1+6y+4y^2.$$

This completes the proof of Theorem 4(d).□

6. Conclusions and Further Problems

By means of the generating function method, we have succeeded in constructing recursively recurrence relations and generating functions for 16 triple sums of binomial circular products. By carrying out the same process, it is not difficult to establish the counterpart results corresponding to double sums, such as

$$\begin{array}{cccc}& {\mathrm{U}}_n=\sum _{i=0}^n\sum _{j=0}^n\left({\displaystyle \genfrac{}{}{0.0pt}{}{n-i}{2j}}\right)\left({\displaystyle \genfrac{}{}{0.0pt}{}{n-j}{2i}}\right):\hfill & & \sum _{n=0}^{\infty }{\mathrm{U}}_n{y}^n={\displaystyle \frac{1-2y-y^2}{(1-y)(1-2y-3y^2-y^3)}},\hfill \\ & {\mathrm{V}}_n=\sum _{i=0}^n\sum _{j=0}^n\left({\displaystyle \genfrac{}{}{0.0pt}{}{n-i}{2j}}\right)\left({\displaystyle \genfrac{}{}{0.0pt}{}{n-j}{2i}}\right){(-1)}^i:\hfill & & \sum _{n=0}^{\infty }{\mathrm{V}}_n{y}^n={\displaystyle \frac{{(1-y)}^3}{1-4y+6y^2-3y^3+3y^4+y^5}},\hfill \\ & {\mathrm{W}}_n=\sum _{i=0}^n\sum _{j=0}^n\left({\displaystyle \genfrac{}{}{0.0pt}{}{n-i}{2j}}\right)\left({\displaystyle \genfrac{}{}{0.0pt}{}{n-j}{2i}}\right){(-1)}^{i+j}:\hfill & \hfill & \sum _{n=0}^{\infty }{\mathrm{W}}_n{y}^n={\displaystyle \frac{1-2y+3y^2}{(1-y)(1-2y+5y^2-y^3)}}.\hfill \end{array}$$

From these rational generating functions, it is natural to ask what would happen to quadruplicate and multiple sums of circular binomial products. Unfortunately, the recursive construction process devised in this paper does not work, even though numerical evidences suggest that the corresponding quadruplicate sums satisfy linear recurrence relations. Particularly for the quadruplicate positive sums

$${\mathsf{\Xi }}_n:=\sum _{i=0}^n\sum _{j=0}^n\sum _{k=0}^n\sum _{\ell =0}^n\left({\displaystyle \genfrac{}{}{0.0pt}{}{n-i}{2j}}\right)\left({\displaystyle \genfrac{}{}{0.0pt}{}{n-j}{2k}}\right)\left({\displaystyle \genfrac{}{}{0.0pt}{}{n-k}{2\ell }}\right)\left({\displaystyle \genfrac{}{}{0.0pt}{}{n-\ell }{2i}}\right)$$

with the initial values

$$\{{\mathsf{\Xi }}_n{\}}_{0\le n\le 7}=\left\{1,1,7,56,512,4817,45600,432289\right\},$$

Mathematica detects a recurrence relation of order 7

$${\mathsf{\Xi }}_n\stackrel{?}{=}12{\mathsf{\Xi }}_{n-1}-24{\mathsf{\Xi }}_{n-2}+12{\mathsf{\Xi }}_{n-4}+4{\mathsf{\Xi }}_{n-5}-4{\mathsf{\Xi }}_{n-6}+{\mathsf{\Xi }}_{n-7}$$

and the following generating function:

$$\sum _{n=0}^{\infty }{\mathsf{\Xi }}_n{y}^n\stackrel{?}{=}{\displaystyle \frac{1-11y+19y^2-4y^3-4y^4+y^5}{(1-y)(1-y-2y^2-y^3)(1-10y+5y^2-y^3)}}.$$

Therefore, alternative approaches will be indispensable in order to handle problems of related multiple sums.