This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.
%I A300057 #32 Jun 24 2018 12:06:56
%S A300057 1,164,47148,15454820,5361965980,1919987703504,701459496193236,
%T A300057 259867456921970040,97260263038893462300,36686877800581349096240,
%U A300057 13924013746979490475444528,5311128944356277793155688612,2034235241375650519750351973188
%N A300057 Coefficient of z^(3*n) in the expansion of (1 + 9*z + 9*z^2 + z^3)^(2*n).
%H A300057 Muniru A Asiru, <a href="/A300057/b300057.txt">Table of n, a(n) for n = 0..100</a>
%F A300057 a(n) = 1/(2*Pi)*Integral_{0..2*Pi}(12*cos^2(x)*sin(x) + 20*sin^3(x))^(2*n) dx.
%F A300057 a(n) = Sum_{k1=0..2*n} Sum_{k2=0..2*n} binomial(2*n,k1)*binomial(2*n,k2)*binomial(2*n,3*n-k1-k2)*((4-sqrt(15))^(2*n-k1))*((4+sqrt(15))^(2*n-k2)).
%F A300057 a(n) = (c1/c3)*a(n-1)+(c2/c3)*a(n-2); with a(0)=1; a(1)=164; and
%F A300057 c1=16*(n-1/2)*(-230+2259*n-3933*n^2+1863*n^3);
%F A300057 c2=1036800*(n-1)*(n-3/2)*(n-1/2)*(n-1/9);
%F A300057 c3=81*n*(n-2/3)*(n-1/3)*(n-10/9).
%F A300057 From _Wolfdieter Lang_, Apr 06 2018: (Start)
%F A300057 a(n) = 4^(2*n)*(2/Pi)*Integral_{0..Pi/2} sin(x)^(2*n)*(3 + 2*sin(x)^2)^(2*n) dx. With the binomial formula and integrals over even powers of sin(x) this becomes
%F A300057 a(n) = 6^(2*n)*Sum_{k=0..2*n} binomial(2*n, k)*binomial(2*(n+k), n+k)*(1/6)^k = 6^(2*n)*binomial(2*n, n)*hypergeometric([-2*n, n+1/2], [n+1], -2/3). (End)
%F A300057 a(n) ~ 2^(4*n) * 5^(2*n + 1/2) / (3*sqrt(Pi*n)). - _Vaclav Kotesovec_, Apr 18 2018
%t A300057 c1=16*(n-1/2)*(-230+2259*n-3933*n^2+1863*n^3);c2=1036800*(n-1)*(n-3/2)*(n-1/2)*(n-1/9);c3=81*n*(n-2/3)*(n-1/3)*(n-10/9);a[0]=1;a[1]=164;a[n0_]:=ReplaceAll[(c1/c3)*a[n0-1]+(c2/c3)*a[n0-2],n->n0]
%t A300057 b[NN_]:=Expand[Total[Flatten[#]]&/@Table[Binomial[2*n,k2]*Binomial[2*n,k1]*Binomial[2*n, 3*n-k1-k2]*(4 + Sqrt[15])^(2*n-k1)*(4-Sqrt[15])^(2*n-k2),{n,0,NN},{k1,0,2*n},{k2,0,2*n}]]
%t A300057 ({#,SameQ[Coefficient[(1+9*z+9*z^2+z^3)^(2*#),z,3*#]&/@Range[0,10],#],SameQ[a/@Range[0,10],#]}&@b[10])[[1]]
%t A300057 Table[SeriesCoefficient[(1 + 9*z + 9*z^2 + z^3)^(2*n), {z, 0, 3*n}], {n, 0, 15}] (* _Vaclav Kotesovec_, Apr 18 2018 *)
%o A300057 (PARI) a(n) = polcoeff((1 + 9*z + 9*z^2 + z^3)^(2*n), 3*n); \\ _Michel Marcus_, Mar 06 2018
%o A300057 (GAP) List([0..15],n->6^(2*n)Sum([0..2*n],k->Binomial(2*n,k)*Binomial(2*(n+k),n+k)*(1/6)^k)); # _Muniru A Asiru_, Apr 07 2018
%Y A300057 Cf. A300058, A295870, A005721.
%K A300057 nonn
%O A300057 0,2
%A A300057 _Bradley Klee_, Feb 23 2018