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%I #36 Jan 01 2024 23:47:50
%S 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,
%T 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,
%U 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
%N a(0)=a(2)=1, a(1)=2, a(n)=0 for n > 2.
%C Self-convolution of A019590. Up to a sign the convolutional inverse of the natural numbers sequence. - _Tanya Khovanova_, Jul 14 2007
%C 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
%H Paul Barry, <a href="https://arxiv.org/abs/2104.01644">Centered polygon numbers, heptagons and nonagons, and the Robbins numbers</a>, arXiv:2104.01644 [math.CO], 2021.
%F G.f.: 1 + 2*x + x^2.
%F a(n) = binomial(2n, n^2). - _Wesley Ivan Hurt_, Mar 08 2014
%p A130713:=n->binomial(2*n, n^2); seq(A130713(n), n=0..100); # _Wesley Ivan Hurt_, Mar 08 2014
%t Table[Binomial[2 n, n^2], {n, 0, 100}] (* _Wesley Ivan Hurt_, Mar 08 2014 *)
%K easy,nonn
%O 0,2
%A _Paul Curtz_ and _Tanya Khovanova_, Jul 01 2007
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