Fix a sequence $a_n={n+2\choose 2}$ of triangular numbers with the initial condition $a_0=1$,such that


given by

$F(x)=\frac{1}{(1-x)^3}=\sum_{n=0}^{\infty} a_n x^n\tag1$

Now if we consider the following generating function


How do we prove

$$G(x)\overset{\color{red}?}=\sum_{n=0}^{\infty} a_n(4x^{3n+2}+x^{(6n+(-1)^{n}-1)/4})$$

Using the simplified form of $(2)$,we have the relation


$$\sum_{n=0}^{\infty} a_n(4x^{3n+2}+x^{(6n+(-1)^{n}-1)/4})=\Big(\sum_{n=0}^{\infty} a_n x^n\Big)\Big(\sum_{n=0}^{\infty} (-1)^{n}x^{(6n-(-1)^n+1)/4}\tag3\Big)^2$$

which relates the number of partitions of $6n$ into two odd parts $\frac{(6n-(-1)^n+1)}{4}$ A007494 oeis to the number of partitions of $6n$ into two even parts $\frac{(6n+(-1)^n-1)}{4}$ A032766 oeis and the number of partitions of 6n into at most 2 parts $3n+1$.

Evidently there's an interesting combinatorial information about partitions of $6n$ encoded in the exponents of the identity

If we introduce the notation $P(6n)=\frac{(6n-(-1)^n+1)}{4}$,$Q(6n)=\frac{(6n+(-1)^n-1)}{4}$ and $R(6n)=3n+1$

the identity can be written succinctly as follows

$$\sum_{n=0}^{\infty} a_n(4x^{R(6n)+1}+x^{Q(6n)})=\Big(\sum_{n=0}^{\infty} a_n x^n\Big)\Big(\sum_{n=0}^{\infty} (-1)^{n}x^{P(6n)}\Big)^2$$

whereby the ff. algebraic equation is satisfied

$R(6n)-Q(6n)-P(6n)=1\tag4$ for natural number $n$

which is an equation of a triangular plane with equal sides $x-y-z=1$ on a 3D space

The connection between triangular numbers $a_n={n+2\choose 2}$(which count objects arranged in an equilateral triangle) as coefficients in the identity $(3)$ and the equation $R(6n)-Q(6n)-P(6n)=1$ for natural number $n$, of a triangular plane with equal sides $(4)$ is quite interesting, since both deal with equilateral triangles

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    $\begingroup$ @Nicco Simplified $G(x) = -\dfrac{(x+1)^2}{(x-1)^3(x^2+x+1)^2}$. $\endgroup$ – orlp Oct 13 '17 at 19:03
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    $\begingroup$ Multiply both things by $(1-x^3)^2(1-x)$ and check that the resulting polynomials are equal. $\endgroup$ – mercio Oct 13 '17 at 19:05

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