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%I #10 Sep 23 2016 04:43:45
%S 1,3,18,70,600,4956,52528,358128,6654600,79967800,1453049400,
%T 16239408120,392541718660,5252687631660,108961629396480,
%U 1395025456201408,62831427044385384,1223872353413404344
%N a(n) = [x^(2^n+n-2)] (x + x^2 + x^4 + x^8 + ... + x^(2^n))^n for n>=2.
%F n(n-1)/2 divides a(n): A135071(n) = a(n)/[n(n-1)/2] for n>=2.
%t 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 *)
%o (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))}
%Y Cf. A135068, A135069, A135071.
%K nonn
%O 2,2
%A _Paul D. Hanna_, Nov 17 2007
%E a(15) - a(19) from _Alois P. Heinz_, Apr 29 2009
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