Conjectural congruences for numbers related to Littlewood-Richardson coefficients

Question

For $n \geq 0$, let $a_n$ be the square of the Euclidean length of the vector of Littlewood-Richardson coefficients of $\sum_{\lambda \vdash n} s_\lambda^2$, where $s_\lambda$ are the symmetric Schur functions and the sum runs over all partitions $\lambda$ of $n$. These numbers can also be described via the generating function $\sum_{n \geq 0} a_n x^n = \prod_{i \geq 1} (1 - 4x^i)^{-1/2}$. This sequence appears in the OEIS (https://oeis.org/A067855) and begins 1, 2, 8, 26, 94, 326, 1196, 4358, $\dots$.

For $n > 1$, it appears empirically that $a_n$ is congruent to 0 (resp. 2) mod 4 when the exponent in the highest power of 2 dividing $n-1$ is odd (resp. even). Is this true?

(Additionally, it appears that $a_{4k}-a_{4k-2}$ is always 0 mod 8, while $a_{2^{i+1} k}-a_{2^{i+1} k- 2^i}$ is always 4 mod 8 when $i \geq 2$. There are similar but more complicated patterns modulo higher powers of 2. For instance, $a_{16k}-a_{16k-2}$ is 0 (resp. 32) mod 64 according to whether $k$ is 0, 4, or 5 (resp. 1, 2, or 3) mod 6.)

Answer 1

This is true, if you replace $n-1$ in "the exponent in the highest power of 2 dividing $n−1$" to $n$.

First of all, we study the coefficients of the series $(1-4t)^{-1/2}=\sum C_nt^n$ modulo 4. We have $C_n=(-4)^n{-1/2\choose n}=2^n\frac{(2n-1)!!}{n!}$. We have for 2-adic valuation: $$\nu_2(n!)=\sum_{k=1}^\infty \lfloor n/2^k\rfloor<n,$$ thus all $C_n$ for $n>1$ are even. And 4 does not divide $C_n$ if and only if $\nu_2(n!)=n-1$. I claim that this holds if and only if $n$ is a power of 2. This may be proved by induction (with base $n=1$): if $n>1$ is odd, then $\nu_2(n!)=\nu_2((n-1)!)<n-1$; if $n$ is even, then $\nu_2(n!)=n/2+\nu_2((n/2)!)$, this equals to $n-1$ if and only if $\nu_2((n/2)!)=n/2-1$, that holds by induction proposition if and only if $n/2$, or, equivalently, $n$ is a power of 2. Therefore, modulo 4 we have $$(1-4t)^{-1/2}=1+2(t+t^2+t^4+\ldots).$$ Substituting $t=x^i$ and multiplying, we get (again modulo 4) $$ \sum a_nx^n=\prod_{i=1}^\infty \left(1+2(x^i+x^{2i}+x^{4i}+\ldots)\right)\\= 1+\sum_{i\geqslant 1,k\geqslant 0} 2x^{i2^k}. $$ For $n=2^rs$ with odd $s$ the equation $n=i2^k$ has exactly $r+1$ solutions, thus the result.

Other your observations, modulo 8 and 16, look be provable by the same strategy (since we may control when $n!$ is not divisible by $2^{n-2}$ etc.), but more involved technically. I suggest on the first stage multiplying $(1-4x)(1-4x^2)...=1-4y$, write $y=x/(1-x)+4z$, and only after that raise to a power $-1/2$.