A302352 - OEIS

%I #11 Apr 07 2018 03:43:42

%S 0,1,19,155,936,4884,23465,107107,472600,2036838,8631206,36119798,

%T 149724940,616104450,2520629685,10265200035,41650094640,168481778790,

%U 679847488650,2737640364810,11005139655744,44176226269728,177114113623194,709364594864910,2838599638596176,11350436081373340

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

%C 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.

%H Seiichi Manyama, <a href="/A302352/b302352.txt">Table of n, a(n) for n = 0..1000</a>

%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/BiquadraticNumber.html">Biquadratic Number</a>

%F a(n) = [x^n] x*(1 + 11*x + 11*x^2 + x^3)/(1 - x)^(n+6).

%F 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)).

%F a(n) ~ 75*2^(2*n+1)/sqrt(Pi*n).

%t Table[Sum[k^4 Binomial[2 n - k, n], {k, 0, n}], {n, 0, 25}]

%t Table[SeriesCoefficient[x (1 + 11 x + 11 x^2 + x^3)/(1 - x)^(n + 6), {x, 0, n}], {n, 0, 25}]

%t 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}]

%t 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]

%t 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]!

%o (PARI) a(n) = sum(k=0, n, k^4*binomial(2*n-k,n)); \\ _Michel Marcus_, Apr 07 2018

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

%K nonn

%O 0,3

%A _Ilya Gutkovskiy_, Apr 06 2018