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

%I A203576 #25 Jan 27 2021 10:36:31
%S A203576 1,2,14,56,446,2152,18248,97120,848254,4796552,42454664,250140640,
%T A203576 2226532712,13516860320,120553738144,748819997056,6679690686334,
%U A203576 42254745008840,376638926040392,2418457241945056,21530200591563496,139992790135717792,1244418656720926624,8178446389043428736
%N A203576 Exponential (or binomial) half-convolution of A000984 (central binomial) with itself.
%C A203576 In general the exponential (also known as binomial) half-convolution of a sequence {b(n), n>=0} with itself is defined by
%C A203576 bhat(n) := Sum_{k=0..floor(n/2)} binomial(n,k)*b(k)*b(n-k), n>=0.
%C A203576 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.
%C A203576 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).
%H A203576 Vincenzo Librandi, <a href="/A203576/b203576.txt">Table of n, a(n) for n = 0..200</a>
%F A203576 a(n) = Sum_{k=0..floor(n/2)} binomial(n,k)*cbi(k)*cbi(n-k) n>=0, with cbi(n)=A000984(n).
%F A203576 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.
%F A203576 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
%F A203576 a(n) ~ 8^n / (Pi*n) * (1 + (1+(-1)^n)/sqrt(2*Pi*n)). - _Vaclav Kotesovec_, Feb 25 2014
%e A203576 With cbi = {1, 2, 6, 20, 70, 252, ...}
%e A203576 a(4) = 1*70 + 4*2*20 + 6*6^2 = 446,
%e A203576 a(5) = 1*252 + 5*2*70 + 10*6*20 = 2152.
%t A203576 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 *)
%o A203576 (PARI) A203576(n)=sum(k=0,n\2,binomial(n,k)*A000984(k)*A000984(n-k))  \\ _M. F. Hasler_, Jan 13 2012
%Y A203576 Cf. A000984, A081085 (exponential convolution).
%K A203576 nonn,easy
%O A203576 0,2
%A A203576 _Wolfdieter Lang_, Jan 13 2012