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A088857
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Let Product[1+Sum[b(i,j) x^(i*j),{i,1,Infinity}],{j,1,Infinity}]=1+Sum[c(n) x^n,{n,1,Infinity}], where b(i,j) is plus or minus one and c(n) is plus or minus one or zero. Furthermore, let b(1,1)=1 (for definiteness). Then, for a given n, a(n) is the number of ways in which the coefficients b(i,j) i<=n, j<=n can be chosen.
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%I #3 Mar 31 2012 20:51:39
%S 1,2,6,30,80,634,1424,5392,10677,38052,39051,815616,1421550
%N Let Product[1+Sum[b(i,j) x^(i*j),{i,1,Infinity}],{j,1,Infinity}]=1+Sum[c(n) x^n,{n,1,Infinity}], where b(i,j) is plus or minus one and c(n) is plus or minus one or zero. Furthermore, let b(1,1)=1 (for definiteness). Then, for a given n, a(n) is the number of ways in which the coefficients b(i,j) i<=n, j<=n can be chosen.
%C In all of these series, the coefficients b(n) will be congruent, modulo 2, to the number of unrestricted partitions of n. Not much is known about the parity of the partition function, although much is known if the modulus 2 is replaced by other numbers such as 5 or 7. As for the sequence proposed here, it is not even known if it is finite or infinite. I have found coefficients satisfying the given conditions for n as large as 70.
%D G. E. Andrews, The Theory of Partitions.
%e a(2)=2 because (1+x+x^2)(1-x^2)=1+x+0x^2+... and (1+x-x^2)(1+x^2)=1+x+0x^2+... But, (1+x+x^2)(1+x^2)=1+x+2x^2+... and (1+x-x^2)(1-x^2)=1+x-2x^2+...
%K hard,nonn
%O 1,2
%A _David S. Newman_, Nov 29 2003
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