OFFSET
0,3
COMMENTS
LINKS
Seiichi Manyama, Table of n, a(n) for n = 0..1000
Eric Weisstein's World of Mathematics, Biquadratic Number
FORMULA
a(n) = [x^n] x*(1 + 11*x + 11*x^2 + x^3)/(1 - x)^(n+6).
a(n) = 2^(2*n+1)*n*(75*n^3 + 52*n^2 - 3*n - 4)*Gamma(n+3/2)/(sqrt(Pi)*Gamma(n+6)).
a(n) ~ 75*2^(2*n+1)/sqrt(Pi*n).
MATHEMATICA
Table[Sum[k^4 Binomial[2 n - k, n], {k, 0, n}], {n, 0, 25}]
Table[SeriesCoefficient[x (1 + 11 x + 11 x^2 + x^3)/(1 - x)^(n + 6), {x, 0, n}], {n, 0, 25}]
Table[2^(2 n + 1) n (75 n^3 + 52 n^2 - 3 n - 4) Gamma[n + 3/2]/(Sqrt[Pi] Gamma[n + 6]), {n, 0, 25}]
CoefficientList[Series[(24 - 180 x + 410 x^2 - 285 x^3 + 31 x^4 + Sqrt[1 - 4 x] (-24 + 132 x - 194 x^2 + 65 x^3 - x^4))/(2 Sqrt[1 - 4 x] x^5), {x, 0, 25}], x]
CoefficientList[Series[E^(2 x) (-576 + 360 x - 244 x^2 + 75 x^3) BesselI[0, 2 x]/x^3 + E^(2 x) (576 - 360 x + 532 x^2 - 255 x^3 + 75 x^4) BesselI[1, 2 x]/x^4, {x, 0, 25}], x]* Range[0, 25]!
PROG
(PARI) a(n) = sum(k=0, n, k^4*binomial(2*n-k, n)); \\ Michel Marcus, Apr 07 2018
CROSSREFS
KEYWORD
nonn
AUTHOR
Ilya Gutkovskiy, Apr 06 2018
STATUS
approved