A209305 Exponential generating function A(x) is the unique solution of the Cauchy problem: A'(x) = exp(x+A(x)^2-1), A(0) = 1.
1, 1, 3, 17, 151, 1821, 27675, 507177, 10879247, 267329317, 7403007507, 228096010961, 7739098803559, 286704779800173, 11514881722287051, 498352218891144249, 23120905868853862655, 1144719051552552433525, 60241163501500355522499
Offset: 0
Keywords
Links
- G. C. Greubel, Table of n, a(n) for n = 0..375
Programs
-
Mathematica
(* Expansion of the generating series *) A[x_] := InverseErf[(2 Exp[x] - 2 + Exp[1] Sqrt[Pi] Erf[1])/(Exp[1] Sqrt[Pi])]; CoefficientList[Series[A[x], {x, 0,40}], x] Table[n!, {n, 0,40}] (* Recurrences - More efficient *) a[n_] := a[n] = a[n-1]+2Sum[Binomial[n-2,k]a[k]b[n-2-k],{k,0,n-2}]; a[1] = 1; a[0] = 1; b[n_] := Sum[Binomial[n, k]a[k + 1]a[n - k + 1], {k, 0, n}]; Table[a[n], {n, 0, 100}]
Formula
E.g.f.: A(x) = inverse_erf((2*exp(x)-2+exp(1)*sqrt(Pi)*erf(1)) / (exp(1)*sqrt(Pi))), where inverse_erf is the inverse of the error function.
a(n) = a(n-1) + 2*Sum_{k=0..n-2} C(n-2,k)*a(k)*b(n-2-k) for n>0, a(0) = 1, where b(n) = Sum_{k=0..n} C(n,k)*a(k+1)*a(n-k+1) (A209306).
Limit_{n->oo} (a(n)/n!)^(1/n) = 1 / log(1 + exp(1)*sqrt(Pi)*erfc(1)/2) = 3.112237454247335904679793089483819785... . - Vaclav Kotesovec, Mar 31 2017