OFFSET
1,3
LINKS
Vincenzo Librandi, Table of n, a(n) for n = 1..1000
Paul Barry, Centered polygon numbers, heptagons and nonagons, and the Robbins numbers, arXiv:2104.01644 [math.CO], 2021.
FORMULA
a(n) = Sum_{k=1..n} ((-1)^(n-k)*2^(n-k)*binomial(n, k)*binomial(3*k, k-1))/n.
G.f.: A(z) satisfies z*A(z)^3 + 3z*A(z)^2 + z*A(z) - A(z) + z = 0.
Recurrence: 2*n*(2*n+1)*a(n) = (n+2)*(3*n-1)*a(n-1) + 4*(n-2)*(15*n-13)*a(n-2) + 76*(n-3)*(n-2)*a(n-3). - Vaclav Kotesovec, Oct 08 2012
a(n) ~ 19^(n+1/2)/(3*sqrt(Pi)*n^(3/2)*2^(2*n+2)). - Vaclav Kotesovec, Oct 08 2012
a(n) = (-1)^(n+1)*2^(n-1)*hypergeom([4/3, 5/3, 1-n], [2, 5/2], 27/8). - Peter Luschny, Aug 03 2017
MAPLE
a := n -> (-1)^(n + 1)*2^(n - 1)*hypergeom([4/3, 5/3, 1 - n], [2, 5/2], 27/8):
seq(simplify(a(n)), n=1..24); # Peter Luschny, Aug 03 2017
MATHEMATICA
Table[Sum[(-1)^(n-k)*2^(n-k)*Binomial[n, k]*Binomial[3*k, k-1], {k, 1, n}]/n, {n, 1, 25}] (* Vaclav Kotesovec, Oct 08 2012 *)
PROG
(PARI) a(n) = sum(k=1, n, (-1)^(n-k)*2^(n-k)*binomial(n, k)*binomial(3*k, k-1))/n; \\ Andrew Howroyd, Nov 12 2017
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
STATUS
approved