It seems to me that all answers to this question will involve the symmetric group. Here is one example from the representation theory of $S_n$ over a field of characteristic $2$:$$\prod_{i=1}^\infty(1-x^i)^{-1}=\prod_{i=1}^\infty(1-x^{2i})^{-2}\sum_{n=0}^\infty x^{n(n+1)/2}.$$The left hand side is the generating function for partitions - the coefficient of $x^n$ is the number of partitions of $n$. The right hand side reflects the way the partitions of $n$ split into $2$-blocks of partitions, according to their $2$-cores. The $2$-cores have the form $[n,n-1,\dots,2,1]$ giving the sum on the extreme right. The other infinite product refers to pairs of partitions, and arises from the enumeration of partitions in a block using a $2$-abacus. Now this equality rearranges and simplifies to$$\prod_{i=1}^\infty\frac{(1-x^{2i})}{(1-x^{2i-1})}=\sum_{n=1}^\infty x^{n(n+1)/2}.$$The left hand side ranges over all partitions with all even parts distinct; a partition with $e$ distinct even parts contributes $(-1)^e$ to the sum.
So for $n$ not a triangular number, the number of partitions with an odd number of distinct even parts must equal the number of partitions with an even number of distinct even parts. This is not obvious to me, but it recalls Euler's famous pentagonal number theorem.
More generally, let $t>0$ and let $c_n$ be the number of $t$-cores of $n$. Then $C_t(x)=\sum_{n=0}^\infty c_nx^n$ is the generating function for $t$-cores. Apart from $t=2$, there does not seem to be any simple description of $C_t(x)$, although it is known that $c_n>0$ for all $n$ and $t>3$ a prime. Then the $t$-abacus idea gives$$\prod_{i=1}^\infty(1-x^i)^{-1}=\prod_{i=1}^\infty(1-x^{ti})^{-t}C_t(x).$$