A130713 a(0)=a(2)=1, a(1)=2, a(n)=0 for n > 2.

1, 2, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0

Offset

0,2

Comments

Self-convolution of A019590. Up to a sign the convolutional inverse of the natural numbers sequence. - Tanya Khovanova, Jul 14 2007

Iterated partial sums give the chain A130713 -> A113311 -> A008574 -> A001844 -> A005900 -> A006325 -> A033455 -> A259181, up to index. The k-th term of the n-th partial sums is (n^2-7n+14 + 4k(k+n-4))(k+n-4)!/(k-1)!/(n-1)!, for k > 3-n. Iterating partial sums in reverse (n-th differences with n zeros prepended) gives row (n+3) of A182533, modulo signs and trailing zeros. - Travis Scott, Feb 19 2023

Links

Table of n, a(n) for n=0..104.

Paul Barry, Centered polygon numbers, heptagons and nonagons, and the Robbins numbers, arXiv:2104.01644 [math.CO], 2021.

Formula

G.f.: 1 + 2*x + x^2.

a(n) = binomial(2n, n^2). - Wesley Ivan Hurt, Mar 08 2014

Maple

A130713:=n->binomial(2*n, n^2); seq(A130713(n), n=0..100); # Wesley Ivan Hurt, Mar 08 2014

Mathematica

Table[Binomial[2 n, n^2], {n, 0, 100}] (* Wesley Ivan Hurt, Mar 08 2014 *)

Crossrefs

Sequence in context: A230093 A033322 A329679 * A236619 A355627 A300828

Adjacent sequences: A130710 A130711 A130712 * A130714 A130715 A130716

Keyword

easy,nonn

Author

Paul Curtz and Tanya Khovanova, Jul 01 2007

Status

approved