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

0,2

Comments

Apply Riordan array (1/sqrt(1-4x), xc(x)) to A131056, c(x) the g.f. of A000108.

Apply Riordan array (c(x)/sqrt(1-4*x), x*c(x)^2) to A131055.

Hankel transform appears to be (-1)^n*A085046(n).

Coefficients T(2*n+1,n) of triangle A103450. [Emanuele Munarini, Jun 01 2012, corrected by Werner Schulte, Nov 27 2021]

Links

Vincenzo Librandi, Table of n, a(n) for n = 0..200

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)

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 *)

Prog

(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 */

Crossrefs

Sequence in context: A128746 A049675 A053154 * A127360 A116415 A026861

Adjacent sequences: A141219 A141220 A141221 * A141223 A141224 A141225

Keyword

easy,nonn

Author

Paul Barry, Jun 14 2008

Extensions

Name of the sequence corrected by Vaclav Kotesovec, Feb 13 2014

Status

approved