A203577 Exponential (or binomial) half-convolution of the sequence A000108 (Catalan) with itself.

1, 1, 4, 11, 58, 212, 1304, 5567, 37734, 178148, 1284124, 6501420, 48758648, 259775440, 2000594288, 11080668871, 86930955662, 496461841956, 3947716126292, 23113333523180, 185660199980696, 1109722749130576, 8983793097101144, 54645629076275356, 445109373450545608, 2748480598104423952

Offset: 0

Author

Wolfdieter Lang, Jan 13 2012

Keywords

• nonn
• easy

Comments

For the definition of the exponential (also known as binomial) half-convolution of a sequence with itself see A203576, where also the rule for the e.g.f. is given.

Examples

			With Catalan = A000108 = {1, 1, 2, 5, 14, 42, ...}
a(4) = 1*1*14 + 4*1*5 + 6*2*2 = 58.
a(5) = 1*1*42 + 5*1*14 + 10*2*5 = 212.
		

Links

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

Crossrefs

Cf. A203576, A000108, A014330 (exponential convolution).

Programs

Mathematica

a[n_] := Sum[ Binomial[n, k]*CatalanNumber[k]*CatalanNumber[n - k], {k, 0, n/2}]; Table[a[n], {n, 0, 25}] (* Jean-François Alcover, Jun 21 2013 *)

PARI

hat(b,n) = sum(k=0,n\2,binomial(n,k)*b(k)*b(n-k))
A203577(n)=hat(A000108,n)  \\ where A000108(n)=(2*n)!/n!/(n+1)! \\ - M. F. Hasler, Jan 13 2012

Formula

a(n) = Sum_{k=0..floor(n/2)} binomial(n,k)*Catalan(k)*Catalan(n-k), n >= 0.

E.g.f.: (C(x)^2 + C2(x^2))/2 with the e.g.f. C(x) of A000108, and the e.g.f. C2(x) := Sum_{n>=0} Catalan(n)^2*x^n/(n!)^2 of the scaled Catalan squares. See a comment above.

C(x) = hypergeom([1/2],[2],4*x) (see A000108 for the version involving BesselI functions), and

C2(x) = hypergeom([1/2,1/2],[1,2,2],16*x).

Recurrence: n*(n+1)^2 * (n+2)^2 * (3*n^6 - 39*n^5 + 166*n^4 - 322*n^3 + 316*n^2 - 153*n + 27)*a(n) = 12*(n-1)*n*(n+1)^2 * (3*n^7 - 34*n^6 + 113*n^5 - 121*n^4 - 19*n^3 + 68*n^2 + 17*n - 18)*a(n-1) + 32*(3*n^11 - 45*n^10 + 220*n^9 - 448*n^8 + 173*n^7 + 920*n^6 - 1696*n^5 + 842*n^4 + 580*n^3 - 846*n^2 + 360*n - 54)*a(n-2) - 768*(n-2)^3 * n *(3*n^7 - 34*n^6 + 113*n^5 - 121*n^4 - 19*n^3 + 68*n^2 + 17*n - 18)*a(n-3) + 2048*(n-3)^3 * (n-2)^2 * (3*n^6 - 21*n^5 + 16*n^4 + 12*n^3 + n^2 - 2)*a(n-4). - Vaclav Kotesovec, Feb 25 2014

a(n) ~ 2^(3*n+2)/(Pi*n^3) * (1 + (1+(-1)^n)/sqrt(2*Pi*n)). - Vaclav Kotesovec, Feb 25 2014