A302352 a(n) = Sum_{k=0..n} k^4*binomial(2*n-k,n).

0, 1, 19, 155, 936, 4884, 23465, 107107, 472600, 2036838, 8631206, 36119798, 149724940, 616104450, 2520629685, 10265200035, 41650094640, 168481778790, 679847488650, 2737640364810, 11005139655744, 44176226269728, 177114113623194, 709364594864910, 2838599638596176, 11350436081373340

Offset: 0

Author

Ilya Gutkovskiy, Apr 06 2018

Keywords

• nonn

Comments

Main diagonal of iterated partial sums array of fourth powers (starting with the first partial sums). For nonnegative integers see A002054, for squares see A265612, for cubes see A293550.

Links

• Seiichi Manyama, Table of n, a(n) for n = 0..1000
• Eric Weisstein's World of Mathematics, Biquadratic Number

Crossrefs

Cf. A000538, A000583, A002054, A101089, A101090, A101091, A265612, A293550, A302353.

Programs

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]!

PARI

a(n) = sum(k=0, n, k^4*binomial(2*n-k,n)); \\ Michel Marcus, Apr 07 2018

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).