Does there exist some method for finding an analytic expression for the coefficient of $z\_1^kz\_2^kz\_3^k$ in:


$$[(1+z\_1)(1+z\_2)(1+z\_3)(1+z\_1z\_2)(1+z\_1z\_3)(1+z\_2z\_3)(1+z\_1z\_2z\_3)]^{k}$$


or is it hopeless?


I can't think of any other method than trying to expand each factor.


Background: the above polynomial is the generating function for a system of linear equations in binary values (see [this question](https://math.stackexchange.com/q/4413256/573047)).


For the simplest case of the coefficient of $z\_1^kz\_2^k$ in $[(1+z\_1)(1+z\_2)(1+z\_1z\_2)]^k$ I found the formula $\sum\_{j=0}^k \binom{k}{j}^3$ at [OEIS A000172](https://oeis.org/A000172).



Analytic expression for the coefficient of a multivariate polynomial
I was coming to the same conclusion that Brendan McKay posted in the comments at about the same time: the efficient way to calculate this is the direct approach $$\sum\_{r,s,t,u} \binom{k}{r} \binom{k}{s} \binom{k}{t} \binom{k}{u} \binom{k}{k-r-s-u} \binom{k}{k-r-t-u} \binom{k}{k-s-t-u}$$ where the sum is over the support implicit in the binomial coefficients. (The way I conceptualise this sum is that if you expand $[(1+z\_1z\_2)(1+z\_1z\_3)(1+z\_2z\_3)]^k$ first as $\sum\_{r,s,t} \binom{k}{r} \binom{k}{s} \binom{k}{t} (z\_1 z\_2)^r (z\_1 z\_3)^s (z\_2 z\_3)^t$ then you must take individual terms $z\_1$, $z\_2$, $z\_3$ from the $(1+z\_i)^k$ to balance them before you consider $(1+z\_1 z\_2 z\_3)^k$.)


I've used the first 300 terms to do a brute-force search for a D-finite recurrence without finding one, so if there is a recurrence then either it's non-linear, it has non-polynomial coefficients, or it's enormous.


