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Denominators of coefficients in the formal power series a(x) such that a(a(x)) = exp(x) - 1.
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%I #38 Nov 29 2021 10:00:14

%S 1,1,4,48,1,3840,92160,645120,3440640,30965760,14863564800,

%T 24222105600,7847962214400,40809403514880,5713316492083200,

%U 7617755322777600,5484783832399872000,5328075722902732800,1220613711064989696000

%N Denominators of coefficients in the formal power series a(x) such that a(a(x)) = exp(x) - 1.

%D R. P. Stanley, Enumerative Combinatorics, Cambridge, Vol. 2, 1999; see Problem 5.52c.

%H Alois P. Heinz, <a href="/A052105/b052105.txt">Table of n, a(n) for n = 0..120</a>

%H Dmitry Kruchinin and Vladimir Kruchinin, <a href="http://arxiv.org/abs/1302.1986">Method for solving an iterative functional equation A^{2^n}(x)=F(x)</a>, arXiv:1302.1986 [math.CO], 2013.

%F 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

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

%p T:= proc(n, k);

%p 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);

%p end proc;

%p a:= n -> denom(T(n, 1));

%p seq(a(n), n = 0..30); # _G. C. Greubel_, Apr 15 2021

%t (* First program *)

%t a[x_, n_] := Sum[c[k] x^k, {k, 0, n}] ;

%t f[x_, n_] := Series[Exp[x] - 1, {x, 0, n}] // Normal;

%t b[x_, n_] := Series[a[a[x, n], n], {x, 0, n}] // Normal;

%t eq[n_] := Thread[CoefficientList[f[x, n] - b[x, n], x] == 0] // Rest;

%t c[0] = 0; so[3] = Solve[eq[3], {c[1], c[2], c[3]}] // First;

%t so[n_] := so[n] = Solve[eq[n] /. Flatten[Table[so[k], {k, 3, n - 1}]], c[n]] // First

%t Array[c, 19, 0] /. Flatten[Table[so[k], {k, 3, 19}]] // Denominator

%t (* _Jean-François Alcover_, Jun 08 2011 *)

%t (* Second program *)

%t 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];

%t Table[Denominator[T[n, 1]], {n, 0, 30}] (* _G. C. Greubel_, Apr 15 2021 *)

%o (Sage)

%o @CachedFunction

%o def T(n,k):

%o if (k==n): return 1

%o 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

%o [denominator(T(n,1)) for n in (0..30)] # _G. C. Greubel_, Apr 15 2021

%Y Cf. A052104, A052122, A052123.

%K nonn,easy,nice,frac

%O 0,3

%A _N. J. A. Sloane_, Jan 22 2000