A203576 Exponential (or binomial) half-convolution of A000984 (central binomial) with itself.

1, 2, 14, 56, 446, 2152, 18248, 97120, 848254, 4796552, 42454664, 250140640, 2226532712, 13516860320, 120553738144, 748819997056, 6679690686334, 42254745008840, 376638926040392, 2418457241945056, 21530200591563496, 139992790135717792, 1244418656720926624, 8178446389043428736

Offset: 0

Author

Wolfdieter Lang, Jan 13 2012

Keywords

• nonn
• easy

Comments

In general the exponential (also known as binomial) half-convolution of a sequence {b(n), n>=0} with itself is defined by

bhat(n) := Sum_{k=0..floor(n/2)} binomial(n,k)*b(k)*b(n-k), n>=0.

The e.g.f. of the sequence {bhat(n)} is Bhat(x) = ((B(x))^2 + B2(x^2))/2, with the e.g.f. B(x) of {b(n), n>=0} and the e.g.f. B2(x) := Sum_((b(n)^2/n!)*x^n/n!, n>=0) of the scaled squares. The proof runs along the same line as the one given for the ordinary half-convolution in a comment on A201204. In fact, bhat(n)/n! is the ordinary half-convolution of the sequence {b(n)/n!, n>=0} with itself.

Here b(n) = A000984(n) = binomial(2*n,n), n>=0, B(x) = exp(2*x)*BesselI(0,2*x) (see the Abramowitz-Stegun reference and link under A008277 for BesselI, p. 375, eq. 9.6.10) and B2(x) = hypergeometric([1/2,1/2],[1,1,1],16*x).

Examples

			With cbi = {1, 2, 6, 20, 70, 252, ...}
a(4) = 1*70 + 4*2*20 + 6*6^2 = 446,
a(5) = 1*252 + 5*2*70 + 10*6*20 = 2152.
		

Links

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

Crossrefs

Cf. A000984, A081085 (exponential convolution).

Programs

Mathematica

cbi[n_] := Binomial[2*n, n]; a[n_] := Sum[Binomial[n, k]*cbi[k]*cbi[n - k], {k, 0, Floor[n/2]}]; Table[a[n], {n, 0, 23}] (* Jean-François Alcover, Jul 09 2013 *)

PARI

A203576(n)=sum(k=0,n\2,binomial(n,k)*A000984(k)*A000984(n-k))  \\ M. F. Hasler, Jan 13 2012

Formula

a(n) = Sum_{k=0..floor(n/2)} binomial(n,k)*cbi(k)*cbi(n-k) n>=0, with cbi(n)=A000984(n).

E.g.f.: (exp(4*x)*BesselI(0, 2*x)^2 + hypergeom([1/2,1/2], [1,1,1],(4*x)^2))/2. See comment above.

Recurrence: (n-1)^2 * n^3 * (3*n^5 - 40*n^4 + 200*n^3 - 476*n^2 + 544*n - 241)*a(n) = 4*(n-1)^3 * (9*n^7 - 126*n^6 + 699*n^5 - 1997*n^4 + 3165*n^3 - 2770*n^2 + 1239*n - 228)*a(n-1) + 32*(3*n^10 - 55*n^9 + 420*n^8 - 1786*n^7 + 4731*n^6 - 8232*n^5 + 9630*n^4 - 7580*n^3 + 3900*n^2 - 1194*n + 162)*a(n-2) - 256*(n-2)^3 * (9*n^7 - 126*n^6 + 699*n^5 - 1997*n^4 + 3165*n^3 - 2770*n^2 + 1239*n - 228)*a(n-3) + 2048*(n-3)^3 * (n-2)^2 * (3*n^5 - 25*n^4 + 70*n^3 - 86*n^2 + 47*n - 10)*a(n-4). - Vaclav Kotesovec, Feb 25 2014

a(n) ~ 8^n / (Pi*n) * (1 + (1+(-1)^n)/sqrt(2*Pi*n)). - Vaclav Kotesovec, Feb 25 2014