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Exponential (or binomial) half-convolution of the sequence A000108 (Catalan) with itself.
2

%I #17 Oct 22 2022 16:19:27

%S 1,1,4,11,58,212,1304,5567,37734,178148,1284124,6501420,48758648,

%T 259775440,2000594288,11080668871,86930955662,496461841956,

%U 3947716126292,23113333523180,185660199980696,1109722749130576,8983793097101144,54645629076275356,445109373450545608,2748480598104423952

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

%C 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.

%H Vincenzo Librandi, <a href="/A203577/b203577.txt">Table of n, a(n) for n = 0..200</a>

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

%F 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.

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

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

%F 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

%F a(n) ~ 2^(3*n+2)/(Pi*n^3) * (1 + (1+(-1)^n)/sqrt(2*Pi*n)). - _Vaclav Kotesovec_, Feb 25 2014

%e With Catalan = A000108 = {1, 1, 2, 5, 14, 42, ...}

%e a(4) = 1*1*14 + 4*1*5 + 6*2*2 = 58.

%e a(5) = 1*1*42 + 5*1*14 + 10*2*5 = 212.

%t 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 *)

%o (PARI) hat(b,n) = sum(k=0,n\2,binomial(n,k)*b(k)*b(n-k))

%o A203577(n)=hat(A000108,n) \\ where A000108(n)=(2*n)!/n!/(n+1)! \\ - M. F. Hasler, Jan 13 2012

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

%K nonn,easy

%O 0,3

%A _Wolfdieter Lang_, Jan 13 2012