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%I #22 Feb 26 2024 11:46:40
%S 1,1,1,5,6,22,40,108,192,536,1072,2528,5344,12288,26624,61312,129024,
%T 286720,646656,1389568,3028992,6717440,14708736,31604736,69763072,
%U 150110208,329809920,714473472,1546649600,3324772352,7332954112,15626403840,33840693248,73194799104,158456610816
%N Expansion of Product_{n>=1} (1 - 2^(n-1)*x^n) * (1 + 2^(n-1)*x^n)^2.
%C Compare to Product_{n>=1} (1 - 2^n*x^n) * (1 + 2^n*x^n)^2 = Sum_{n>=0} 2^(n*(n+1)/2) * x^(n*(n+1)/2).
%H Paul D. Hanna, <a href="/A370337/b370337.txt">Table of n, a(n) for n = 0..1035</a>
%F a(n) ~ c^(1/4) * 2^n * exp(sqrt(c*n)) / (3*sqrt(Pi)*n^(3/4)), where c = 2*log(2)^2 - Pi^2/3 - 8*polylog(2,-1/2) = Pi^2 + 6*log(2)^2 + 8*polylog(2,-2) = 1.258351549529119595933889966687474131697... - _Vaclav Kotesovec_, Feb 26 2024
%e G.f.: A(x) = 1 + x + x^2 + 5*x^3 + 6*x^4 + 22*x^5 + 40*x^6 + 108*x^7 + 192*x^8 + 536*x^9 + 1072*x^10 + 2528*x^11 + 5344*x^12 + ...
%e where A(x) is the series expansion of the infinite product given by
%e A(x) = (1 - x)*(1 + x)^2 * (1 - 2*x^2)*(1 + 2*x^2)^2 * (1 - 4*x^3)*(1 + 4*x^3)^2 * (1 - 8*x^4)*(1 + 8*x^4)^2 * ... * (1 - 2^(n-1)*x^n)*(1 + 2^(n-1)*x^n)^2 * ...
%e Compare A(x) to the series that results from a similar infinite product:
%e (1 - 2*x)*(1 + 2*x)^2 * (1 - 4*x^2)*(1 + 4*x^2)^2 * (1 - 8*x^3)*(1 + 8*x^3)^2 * (1 - 16*x^4)*(1 + 16*x^4)^2 * ... = 1 + 2*x + 8*x^3 + 64*x^6 + 1024*x^10 + 32768*x^15 + 2097152*x^21 + 268435456*x^28 + 68719476736*x^36 + ...
%t CoefficientList[Series[8*QPochhammer[1/2, 2*x] * QPochhammer[-1/2, 2*x]^2/9, {x, 0, 40}], x] (* _Vaclav Kotesovec_, Feb 26 2024 *)
%o (PARI) {a(n) = polcoeff( prod(k=1,n, (1 - 2^(k-1)*x^k) * (1 + 2^(k-1)*x^k)^2 +x*O(x^n)), n)}
%o for(n=0,40, print1(a(n),", "))
%Y Cf. A006125, A075900, A304961, A337299, A355234, A370338.
%K nonn
%O 0,4
%A _Paul D. Hanna_, Feb 26 2024