OFFSET
0,3
REFERENCES
R. P. Stanley, Enumerative Combinatorics, Cambridge, Vol. 2, 1999; see Problem 5.52c.
LINKS
Alois P. Heinz, Table of n, a(n) for n = 0..120
Dmitry Kruchinin and Vladimir Kruchinin, Method for solving an iterative functional equation A^{2^n}(x)=F(x), arXiv:1302.1986 [math.CO], 2013.
FORMULA
a(n) = denominator(T(n,1)) where T(n, m) = if n=m then 1, otherwise ( StirlingS2(n, m)*m!/n! - Sum_{i=m+1..n-1} T(n, i) * T(i, m)))/2. - G. C. Greubel, Apr 15 2021
EXAMPLE
a(x) = x + x^2/4 + x^3/48 + x^5/3840 - 7*x^6/92160 + x^7/645120 + ...
MAPLE
T:= proc(n, k);
T(n, k):= `if`(n=k, 1, (Stirling2(n, k)*k!/n! - add(T(n, j)*T(j, k), j = k+1..n-1))/2);
end proc;
a:= n -> denom(T(n, 1));
seq(a(n), n = 0..30); # G. C. Greubel, Apr 15 2021
MATHEMATICA
(* First program *)
a[x_, n_] := Sum[c[k] x^k, {k, 0, n}] ;
f[x_, n_] := Series[Exp[x] - 1, {x, 0, n}] // Normal;
b[x_, n_] := Series[a[a[x, n], n], {x, 0, n}] // Normal;
eq[n_] := Thread[CoefficientList[f[x, n] - b[x, n], x] == 0] // Rest;
c[0] = 0; so[3] = Solve[eq[3], {c[1], c[2], c[3]}] // First;
so[n_] := so[n] = Solve[eq[n] /. Flatten[Table[so[k], {k, 3, n - 1}]], c[n]] // First
Array[c, 19, 0] /. Flatten[Table[so[k], {k, 3, 19}]] // Denominator
(* Jean-François Alcover, Jun 08 2011 *)
(* Second program *)
T[n_, k_]:= T[n, k]= If[k==n, 1, (StirlingS2[n, k]*k!/n! - Sum[T[n, j]*T[j, k], {j, k+1, n-1}])/2];
Table[Denominator[T[n, 1]], {n, 0, 30}] (* G. C. Greubel, Apr 15 2021 *)
PROG
(Sage)
@CachedFunction
def T(n, k):
if (k==n): return 1
else: return ( (factorial(k)/factorial(n))*stirling_number2(n, k) - sum(T(n, j)*T(j, k) for j in (k+1..n-1)) )/2
[denominator(T(n, 1)) for n in (0..30)] # G. C. Greubel, Apr 15 2021
CROSSREFS
KEYWORD
nonn,easy,nice,frac
AUTHOR
N. J. A. Sloane, Jan 22 2000
STATUS
approved