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A184011 Coefficients of the formal power series of a half-iterate of exp(x)-1 (rescaled). 0
0, 1, 2, 2, 0, 8, -56, 32, 10176, -215808, -78784, 150990912, -3405688576, -139041794560, 10385778676736, 130003936220160, -43016304236761088, 526545841919713280, 266085261164348628992, -12347306589339686547456 (list; graph; refs; listen; history; text; internal format)
OFFSET

0,3

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 integer.

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

REFERENCES

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

LINKS

Table of n, a(n) for n=0..19.

Gottfried Helms, Coefficients for fractional iterates exp(x)-1

Dmitry Kruchinin, Vladimir Kruchinin, Method for solving an iterative functional equation $A^{2^n}(x)=F(x)$, arXiv:1302.1986 [math.CO], 2013.

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

EXAMPLE

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

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-Fran├žois Alcover, Feb 11 2013 *)

PROG

(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

CROSSREFS

Cf. A052122.

Sequence in context: A009187 A009803 A009615 * A079194 A179198 A117739

Adjacent sequences:  A184008 A184009 A184010 * A184012 A184013 A184014

KEYWORD

sign

AUTHOR

Gottfried Helms, Feb 13 2011

STATUS

approved

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Last modified September 26 05:08 EDT 2017. Contains 292502 sequences.