A188678 Alternate partial sums of binomial(3*n,n)/(2*n+1).

1, 0, 3, 9, 46, 227, 1201, 6551, 36712, 209963, 1220752, 7193888, 42873220, 257957352, 1564809168, 9559946496, 58768808463, 363261736872, 2256369305793, 14076552984507, 88163556913188, 554148894304557, 3494365949734563

Offset: 0

Author

Emanuele Munarini, Apr 08 2011

Keywords

• nonn
• easy

Links

• Seiichi Manyama, Table of n, a(n) for n = 0..1000 (terms 0..100 from Vincenzo Librandi)

Crossrefs

Cf. A005809, A001764, A188676, A104859, A188679, A188680, A188681, A188682, A188683, A188684, A188685, A188686, A188687.

Programs

Mathematica

Table[Sum[Binomial[3k,k](-1)^(n-k)/(2k+1),{k,0,n}],{n,0,20}]

Maxima

makelist(sum(binomial(3*k,k)*(-1)^(n-k)/(2*k+1),k,0,n),n,0,20);

Formula

a(n) = Sum_{k=0..n} binomial(3*k,k)*(-1)^(n-k)/(2*k+1).

Recurrence: 2*(2*n^2+9*n+10)*a(n+2)-(23*n^2+63*n+40)*a(n+1)-3*(9*n^2+27*n+20)*a(n)=0.

G.f.: 2*sin((1/3)*arcsin(3*sqrt(3*x)/2))/((1+x)*sqrt(3*x)).

a(n) ~ 3^(3*n+3+1/2)/(31*sqrt(Pi)*n^(3/2)*2^(2*n+2)). - Vaclav Kotesovec, Aug 06 2013

G.f. A(x) satisfies: A(x) = 1 / (1 + x) + x * (1 + x)^2 * A(x)^3. - Ilya Gutkovskiy, Jul 25 2021