A098536 Expansion of 1/((1-x)^3 - 9*x^4)^(1/3).
1, 1, 1, 1, 4, 13, 31, 61, 124, 295, 757, 1873, 4402, 10237, 24421, 59701, 146455, 356308, 862810, 2096632, 5127391, 12583513, 30886735, 75775729, 186054142, 457662265, 1127659903, 2781162079, 6862930768, 16945704721, 41876228125, 103573882633, 256359901387, 634900971064
Offset: 0
Links
- G. C. Greubel, Table of n, a(n) for n = 0..1000
Programs
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Magma
Q:=Rationals(); R
:=PowerSeriesRing(Q,30); Coefficients(R!(1/((1-x)^3-9*x^4)^(1/3))); // G. C. Greubel, Jan 17 2018 -
Maple
with(FormalPowerSeries): # requires Maple 2022 re:= subs(n=n-1, FindRE(1/((1-x)^3 - 9*x^4)^(1/3), x, a(n))); # re = Mathar's recurrence f:= gfun:-rectoproc({re, a(0)=1, a(1)=1, a(2)=1, a(3)=1, a(4)=4}, a(n), remember): map(f, [$0..30]); # Georg Fischer, Oct 23 2022
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Mathematica
CoefficientList[Series[1/((1-x)^3-9*x^4)^(1/3),{x,0,40}],x] (* Harvey P. Dale, May 11 2011 *)
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PARI
x='x+O('x^30); Vec(1/((1-x)^3-9*x^4)^(1/3)) \\ G. C. Greubel, Jan 17 2018
Formula
Recurrence: n*a(n) +(-3*n+2)*a(n-1) +(3*n-4)*a(n-2) +(-n+2)*a(n-3) + 3*(-3*n+8)*a(n-4)=0. - R. J. Mathar, Nov 10 2014
a(n) = Sum_{k=0..floor(n/4)} (-9)^k * binomial(-1/3,k) * binomial(n-k,n-4*k). - Seiichi Manyama, Oct 04 2024
Comments