A135069 a(n) = [x^(2^n+n-1)] (x + x^2 + x^4 + x^8 + ... + x^(2^n))^n / n for n>=1.
1, 1, 2, 4, 18, 106, 816, 4292, 59698, 594178, 9066286, 87498566, 1784642080, 20988667064, 380829128200, 4301687654136, 167344151387170, 2948286694377154, 81332961594822202, 1301097749397343978, 48612398553534689114, 904790963165201870170, 26316129785192975106006, 464241023562098660374014, 24858620479726716329900336, 556565016155501619684118816, 20303230470838234228146518916, 424323532462258172880428842252
Offset: 1
Keywords
Programs
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Mathematica
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 *)
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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)}
Formula
a(n) = A135068(n)/n for n>=1.
Extensions
a(15)-a(19) from Alois P. Heinz, Apr 29 2009
a(20)-a(22) from Max Alekseyev, Dec 03 2010
a(23)-a(28) from Max Alekseyev, Aug 31 2024