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A300057
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Coefficient of z^(3*n) in the expansion of (1 + 9*z + 9*z^2 + z^3)^(2*n).
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2
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1, 164, 47148, 15454820, 5361965980, 1919987703504, 701459496193236, 259867456921970040, 97260263038893462300, 36686877800581349096240, 13924013746979490475444528, 5311128944356277793155688612, 2034235241375650519750351973188
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OFFSET
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0,2
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LINKS
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FORMULA
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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).
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
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MATHEMATICA
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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 *)
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PROG
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(PARI) a(n) = polcoeff((1 + 9*z + 9*z^2 + z^3)^(2*n), 3*n); \\ Michel Marcus, Mar 06 2018
(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
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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