OFFSET
0,2
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.
LINKS
Vincenzo Librandi, Table of n, a(n) for n = 0..200
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
EXAMPLE
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.
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 *)
PROG
(PARI) A203576(n)=sum(k=0, n\2, binomial(n, k)*A000984(k)*A000984(n-k)) \\ M. F. Hasler, Jan 13 2012
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Wolfdieter Lang, Jan 13 2012
STATUS
approved