As the title said, I'm very interested how many variants to choose $n$ signs from all $2^n$ variants when expression lead to zero. I tried to get recurrent formula but nothing happened.

Asked

Viewed
276 times

gp, usingFedor Petrov's hint:for(n=0,40,if((n+1)%4<2,print([n,polcoeff(prod(m=1,n,1+x^m),(n^2+n)/4)])))] $\endgroup$ – Noam D. Elkies Sep 26 '17 at 23:09infiniteproduct $\prod_{m=1}^\infty (1+x^m)$ is a modular function. The finite product $\prod_{m=1}^n (1+x^m)$ is not, and truncating the product at the $n$th term reduces the $x^{(n^2+n)/4}$ coefficient once $n>3$, so the Rademacher formula gives only an upper bound for the enumeration thatMr. Newmanasked for. $\endgroup$ – Noam D. Elkies Sep 27 '17 at 5:05for(n=1,20,print([n,polcoeff(prod(m=1,2*n,1+x^(2*m-1)),2*n^2)])), leads to oeis.org/A292476 and thence to oeis.org/A156700 but there's not nearly as much information there as there was on your original question. $\endgroup$ – Noam D. Elkies Sep 27 '17 at 18:19