A184011 Coefficients of the formal power series of a half-iterate of exp(x)-1 (rescaled).

0, 1, 2, 2, 0, 8, -56, 32, 10176, -215808, -78784, 150990912, -3405688576, -139041794560, 10385778676736, 130003936220160, -43016304236761088, 526545841919713280, 266085261164348628992, -12347306589339686547456

Offset: 0

Author

Gottfried Helms, Feb 13 2011

Keywords

• sign

Comments

Consider the formal power series for the real half-iterate of exp(x)-1 = Sum_{k>=0} c_k*x^k with c_1 = +1 then a(k) = c_k*k!*4^{k-1} and all a(k) seem to be integers.

For the general technique of finding the half-iterate of power series, see for instance the Comtet reference.

Examples

			f(x) = x + 1/4*x^2 + 1/48*x^3 + 1/3840*x^5 - 7/92160*x^6 + 1/645120*x^7 + O(x^8)
so   c_3  = 1/48
and  a(3) = c_3 * 4^2*3! = 16*6/48 = 2
		

References

• Comtet, L; Advanced Combinatorics (1974 edition), D. Reidel Publishing Company, Dordrecht - Holland, pp. 147-148.

Links

• Gottfried Helms, Coefficients for fractional iterates exp(x)-1
• Dmitry Kruchinin and Vladimir Kruchinin, Method for solving an iterative functional equation $A^{2^n}(x)=F(x)$, arXiv:1302.1986 [math.CO], 2013.

Crossrefs

Cf. A052122, A052123.

Programs

Mathematica

max = 19; f[x_] := Sum[c[k]*x^k, {k, 0, max}]; c[0] = 0; c[1] = 1; coes = CoefficientList[ Series[f[f[x]] - Exp[x] - 1, {x, 0, max}], x]; sol = Solve[Thread[coes == 0] // Rest] // First; Table[c[n]*4^(n-1)*n!, {n, 0, max}] /. sol (* Jean-François Alcover, Feb 11 2013 *)

PARI

{a(n)=local(A=x+x^2,B=x);for(i=1,n,B=serreverse(A+x*O(x^n));A=(A+exp(B)-1)/2);4^(n-1)*n!*polcoeff(A,n)} \\ Paul D. Hanna

PARI

{trisqrt(m) = local(tmp, rs=rows(m), cs=cols(m), c);
\\ computes sqrt of lower triangular matrix with unit-diagonal
   tmp=matid(#m);
   for(d=1,rs-1,
        for(r=d+1,rs,
              c=r-d;
              tmp[r,c]=(m[r,c]-sum(k=c+1,r-1,tmp[r,k]*tmp[k,c]))
                        /(tmp[c,c]+tmp[r,r])
           );
      );
return(tmp);}
ff = exp(x)-1
Mff = matrix(6,6,r,c,polcoeff(ff^(c-1),(r-1))) \\ create Bell-matrix for ff
Mf =  trisqrt ( Mff )  \\ = Mff^(1/2) is Bellmatrix for f
f = Ser(Mf[,2])  \\ coefficients of power series for half-iterate of exp(x)-1 from second column in Mf

Formula

G.f. f(x) where f(f(x)) = exp(x)-1 with f'(0)=1.

T(n,m) = if n=m then 1 else (stirling2(n,m)*m!/n!-sum(i=m+1..n-1, T(n,i)*T(i,m)))/2; a(n) = 4^(n-1)*n!*T(n,1). - Vladimir Kruchinin, Nov 09 2011

E.g.f. A(x), satisfies A(A(x))=(exp(4*x)-1)/4, T(n,m)=1/2*(4^(n-m)*stirling2(n,m)-sum(i=m+1..n-1, T(n,i)*T(i,m))), T(n,n)=1, a(n)=T(n,1), a(0)=0. - Dmitry Kruchinin, Dec 04 2012

a(n) = A052122(n) * 2^(2*n - 2 - A052123(n)). - Andrey Zabolotskiy, Aug 22 2022