A300058 a(n) = binomial(3*n,n)/(2*Pi)*Integral_{x=0..2*Pi} (12*cos^2(x)*sin(x) + 20*sin^3(x))^(2*n) dx.
1, 492, 707220, 1298204880, 2654173160100, 5765723073622512, 13021894087331233104, 30217387890886676251200, 71532102917478013611243300, 171944976047709681477985038000, 418347201888204996027087975427920
Offset: 0
Keywords
A300057 Coefficient of z^(3*n) in the expansion of (1 + 9*z + 9*z^2 + z^3)^(2*n).
1, 164, 47148, 15454820, 5361965980, 1919987703504, 701459496193236, 259867456921970040, 97260263038893462300, 36686877800581349096240, 13924013746979490475444528, 5311128944356277793155688612, 2034235241375650519750351973188
Offset: 0
Keywords
Links
- Muniru A Asiru, Table of n, a(n) for n = 0..100
Programs
-
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
-
Mathematica
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] 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}]] ({#,SameQ[Coefficient[(1+9*z+9*z^2+z^3)^(2*#),z,3*#]&/@Range[0,10],#],SameQ[a/@Range[0,10],#]}&@b[10])[[1]] Table[SeriesCoefficient[(1 + 9*z + 9*z^2 + z^3)^(2*n), {z, 0, 3*n}], {n, 0, 15}] (* Vaclav Kotesovec, Apr 18 2018 *) -
PARI
a(n) = polcoeff((1 + 9*z + 9*z^2 + z^3)^(2*n), 3*n); \\ Michel Marcus, Mar 06 2018
Formula
a(n) = 1/(2*Pi)*Integral_{0..2*Pi}(12*cos^2(x)*sin(x) + 20*sin^3(x))^(2*n) dx.
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)).
a(n) = (c1/c3)*a(n-1)+(c2/c3)*a(n-2); with a(0)=1; a(1)=164; and
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).
From Wolfdieter Lang, Apr 06 2018: (Start)
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
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)
a(n) ~ 2^(4*n) * 5^(2*n + 1/2) / (3*sqrt(Pi*n)). - Vaclav Kotesovec, Apr 18 2018
Comments
Links
Crossrefs
Programs
Maple
Mathematica
c1=12*(-230+2259*n-3933*n^2+1863*n^3);c2=5248800*(n-5/3)*(n-4/3)*(n-1/9);c3=9*n^2*(n-10/9);a[0]=1;a[1]=492;a[n0_]:=ReplaceAll[(c1/c3)*a[n0-1]+(c2/c3)*a[n0-2],n->n0]; b[NN_]:=Expand[Total[Flatten[#]]&/@Table[Binomial[3*n,n]*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}]]; ({#,SameQ[#,a/@Range[0,10]]}&@b[10])[[1]] Table[Binomial[3*n, n] * SeriesCoefficient[(1 + 9*z + 9*z^2 + z^3)^(2*n), {z, 0, 3*n}], {n, 0, 15}] (* Vaclav Kotesovec, Apr 18 2018 *)Formula