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A052104 Numerators of coefficients in function a(x) such that a(a(x)) = exp(x) - 1. 4
0, 1, 1, 1, 0, 1, -7, 1, 53, -281, -1231, 87379, -13303471, -54313201, 10142361989, 2821265977, -10502027401553, 1836446156249, 2952828271088741, -1004826382596003137, -7006246797736924249, 14607119841651449406947, 1868869263315549659372569 (list; graph; refs; listen; history; text; internal format)
OFFSET

0,7

REFERENCES

R. P. Stanley, Enumerative Combinatorics, Cambridge, Vol. 2, 1999; see Problem 6.52.

LINKS

Alois P. Heinz, Table of n, a(n) for n = 0..120

I. N. Baker, Zusammensetzungen ganzer Funktionen Math. Z. 69 (1) (1958) 121-163

Dmitry Kruchinin, Vladimir Kruchinin, Method for solving an iterative functional equation A^{2^n}(x)=F(x), arXiv:1302.1986

Mathoverflow, f(f(x))=exp(x)-1 and...

FORMULA

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) = numerator(T(n,1)). - Vladimir Kruchinin, Nov 08 2011

EXAMPLE

a(x) = x+1/4*x^2+1/48*x^3+1/3840*x^5-7/92160*x^6+1/645120*x^7+...

MAPLE

T:= proc(n, m) T(n, m):= `if`(n=m, 1, (Stirling2(n, m)*m!/n!-

       add(T(n, i)*T(i, m), i=m+1..n-1))/2)

    end:

a:= n-> numer(T(n, 1)):

seq(a(n), n=0..30);  # Alois P. Heinz, Feb 11 2013

MATHEMATICA

T[n_, n_] = 1; T[n_, m_] := T[n, m] = (StirlingS2[n, m]*m!/n! - Sum[T[n, i]*T[i, m], {i, m+1, n-1}])/2; Table[T[n, 1] // Numerator, {n, 0, 30}] (* Jean-Fran├žois Alcover, Mar 03 2014, after Alois P. Heinz *)

CROSSREFS

Cf. A052105, A052122, A052123.

Sequence in context: A027466 A218017 A075502 * A144450 A051339 A134141

Adjacent sequences:  A052101 A052102 A052103 * A052105 A052106 A052107

KEYWORD

sign,nice,easy,frac

AUTHOR

N. J. A. Sloane, Jan 22 2000

STATUS

approved

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Last modified February 21 14:29 EST 2018. Contains 299414 sequences. (Running on oeis4.)