A135070 a(n) = [x^(2^n+n-2)] (x + x^2 + x^4 + x^8 + ... + x^(2^n))^n for n>=1.
1, 1, 3, 18, 70, 600, 4956, 52528, 358128, 6654600, 79967800, 1453049400, 16239408120, 392541718660, 5252687631660, 108961629396480, 1395025456201408, 62831427044385384, 1223872353413404344, 37391632408971430120, 655014723078024641640, 27055523795138159291124, 547691411941958414420092, 17365164578604322437671664, 332211955074827711097949200, 19385342415197943809053622700, 466687147661484232610990714436, 18326221432646410203582181439808
Offset: 1
Keywords
Programs
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Mathematica
f[x_, n_] := (Sum[x^(2^k), {k, 0, n}])^n; Table[Coefficient[f[x, n], x^(2^n + n - 2)] , {n, 2, 10}] (* G. C. Greubel, Sep 22 2016 *) -
PARI
{a(n)=if(n<2,0,polcoeff(sum(j=0,n,x^(2^j)+O(x^(2^n+n)))^n,2^n+n-2))}
Formula
n(n-1)/2 divides a(n) for n>=2: A135071(n) = a(n)/[n(n-1)/2] for n>=2.
Extensions
a(15)-a(19) from Alois P. Heinz, Apr 29 2009
a(1) prepended and a(20)-a(28) added by Max Alekseyev, Aug 31 2024