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%I #17 Sep 11 2024 22:57:13
%S 1,1,2,4,18,106,816,4292,59698,594178,9066286,87498566,1784642080,
%T 20988667064,380829128200,4301687654136,167344151387170,
%U 2948286694377154,81332961594822202,1301097749397343978,48612398553534689114,904790963165201870170,26316129785192975106006,464241023562098660374014,24858620479726716329900336,556565016155501619684118816,20303230470838234228146518916,424323532462258172880428842252
%N a(n) = [x^(2^n+n-1)] (x + x^2 + x^4 + x^8 + ... + x^(2^n))^n / n for n>=1.
%F a(n) = A135068(n)/n for n>=1.
%t f[x_, n_] := (1/n)*(Sum[x^(2^k), {k, 0, n}])^n; Table[Coefficient[f[x, n], x^(2^n + n - 1)] , {n, 1, 10}] (* _G. C. Greubel_, Sep 22 2016 *)
%o (PARI) {a(n)=if(n<1,0,polcoeff(sum(j=0,n,x^(2^j)+O(x^(2^n+n)))^n,2^n+n-1)/n)}
%Y Cf. A135068, A135070, A135071.
%K nonn
%O 1,3
%A _Paul D. Hanna_, Nov 17 2007
%E a(15)-a(19) from _Alois P. Heinz_, Apr 29 2009
%E a(20)-a(22) from _Max Alekseyev_, Dec 03 2010
%E a(23)-a(28) from _Max Alekseyev_, Aug 31 2024