A026944 E.g.f. is inverse function to y -> integral from 0 to y of exp(-s^2).

1, 2, 28, 1016, 69904, 7796768, 1282366912, 291885678464, 87844207042816, 33775227494400512, 16152024497964817408, 9402833148376976193536, 6546848699382209957269504, 5372168190357763804164005888, 5130820073307731596716765724672, 5642704273822755928641583754215424

Offset: 1

Author

F. Chapoton, Mar 22 2000

Keywords

• nonn

Comments

The generating function is odd, so this list contains only the nonzero coefficients in the Taylor expansion.

Limit_{n->oo} (a(n)/(n!)^2)^(1/n) = 16/Pi. - Vaclav Kotesovec, Nov 19 2014

Links

• Vaclav Kotesovec, Table of n, a(n) for n = 1..220
• D. Dominici, Asymptotic analysis of the derivatives of the inverse error function, arXiv:math/0607230 [math.CA], 2006-2007.
• D. Dominici, Nested derivatives: A simple method for computing series expansions of inverse functions, arXiv:math/0501052 [math.CA], 2005.

Crossrefs

Cf. A002067.

Programs

Mathematica

MakeTable[n_] := Select[CoefficientList[Series[InverseErf[2x/Sqrt[Pi]],{x,0,2n+1}],x] Table[k!, {k,0,2n+1}], # != 0 &]; MakeTable[15] (* Emanuele Munarini, Dec 17 2012 *)
nmax=20; c = ConstantArray[0,nmax]; c[[1]]=1; Do[c[[k+1]] = Sum[c[[m+1]]*c[[k-m]]/(m+1)/(2*m+1),{m,0,k-1}],{k,1,nmax-1}]; A026944=c*(2*Range[0,nmax-1])! (* Vaclav Kotesovec, Feb 25 2014 *)

Maxima

f(n):=n!/2*coeff(taylor(2*inverse_erf(2*x/sqrt(%pi)),x,0,n),x,n); makelist(f(2*n+1),n,0,12); /* Emanuele Munarini, Dec 17 2012 */

PARI

v=Vec(serlaplace(serreverse(intformal(exp(-x^2)))));
vector(#v\2,n,v[2*n-1])  /* show terms */
/* Demonstration of Kruchinin's differential equation: */
default(seriesprecision,55); /* that many terms */
A=serreverse(intformal(exp(-x^2))); /* e.g.f. */
deriv(A)-exp(A^2)  /* gives O(x^57), i.e., zero up to order */

Formula

Nonzero constant terms of the polynomials P_{2n-1} in t defined by P_1=1, P_{n+1} = P'n+2*n*t*P_n.

E.g.f.: (1/2*sqrt(Pi)*erf)^{-1}(x).

a(n) = A002067(n-1) * 2^(n-1).

E.g.f. A(x) satisfies the differential equation A'(x) = exp(A(x)^2). - Vladimir Kruchinin, Jan 22 2011

Let D denote the operator g(x) -> d/dx(exp(x^2)*g(x)). Then a(n) = D^(2*n-2)(1) evaluated at x = 0. See [Dominici, Example 11]. - Peter Bala, Sep 08 2011