A141222 Expansion of -1/(2*x) + (2*x-1)^2/(2*x*(1-4x)^(3/2)).
1, 5, 22, 95, 406, 1722, 7260, 30459, 127270, 529958, 2200276, 9111830, 37650172, 155266100, 639191160, 2627302995, 10784089350, 44208873390, 181025067300, 740483276610, 3026059513620, 12355464845100
Offset: 0
Links
- Vincenzo Librandi, Table of n, a(n) for n = 0..200
Programs
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Mathematica
Table[((1+3*n+n^2)*Binomial[2*n, n])/(n+1),{n,0,20}] (* Vaclav Kotesovec, Feb 13 2014 *) CoefficientList[Series[-1/(2*x)+(2*x-1)^2/(2*x*(1-4x)^(3/2)),{x,0,20}],x] (* Vaclav Kotesovec, Feb 13 2014 *) a[n_] := (1 + 3 n + n^2) CatalanNumber[n]; Table[a[n], {n, 0, 21}] (* Peter Luschny, Nov 28 2021 *)
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Maxima
a(n):=sum(binomial(2*n,k)*binomial(n+1,2*n-k),k,0,n); makelist(a(n),n,0,40); /* Emanuele Munarini, Jun 01 2012 */
Formula
a(n) = Sum_{k=0..n} (1 + (k+1)*2^(k-1) - 0^k/2)*C(2n-k,n-k); a(n) = Sum_{k=0..n} C(2n,k)*C(n+1,2n-k).
Equals the Narayana transform (A001263) of integer squares. - Gary W. Adamson, Jul 29 2011
Conjecture: (n+1)*a(n) + 2*(-3*n-1)*a(n-1) + 4*(2*n-3)*a(n-2) = 0. - R. J. Mathar, Nov 24 2012
From Vaclav Kotesovec, Feb 13 2014: (Start)
G.f.: -1/(2*x) + (2*x-1)^2/(2*x*(1-4x)^(3/2)).
a(n) = (1 + 3*n + n^2) * C(2*n,n) / (n+1).
Recurrence: (n+1)*(n^2 + n - 1)*a(n) = 2*(2*n-1)*(n^2 + 3*n + 1)*a(n-1).
(End)
Extensions
Name of the sequence corrected by Vaclav Kotesovec, Feb 13 2014
Comments