%I #20 Aug 31 2024 23:29:06
%S 1,1,3,7,40,236,1876,9948,147880,1453960,22015900,208197540,
%T 4313645260,50025596492,908013578304,10257540119128,410662921858728,
%U 7157148265575464,196798065310375948,3119117728942974484,117123479632632724204,2164788189493906776364,62917262965957689991564,1107373183582759036993164,59647207431378288643241916,1329593013280581859290571836,48482067282133360326936987936
%N a(n) = [x^(2^n+n-2)] (x + x^2 + x^4 + x^8 + ... + x^(2^n))^n /(n*(n-1)/2) for n>=2.
%F a(n) = A135070(n) / (n(n-1)/2) for n>=2.
%t f[x_, n_] := (1/Binomial[n, 2])*(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)/(n*(n-1)/2))}
%Y Cf. A135068, A135069, A135070.
%K nonn
%O 2,3
%A _Paul D. Hanna_, Nov 17 2007
%E a(15)-a(19) from _Alois P. Heinz_, Apr 29 2009
%E a(20)-a(21) from _Max Alekseyev_, Jul 25 2009
%E a(22)-a(28) from _Max Alekseyev_, Aug 31 2024