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A220112
E.g.f. A(x) satisfies A(A(x)) = (1/4)*log(1/(1-4*x)).
4
1, 2, 10, 80, 872, 11928, 195072, 3702080, 80065792, 1950808000, 53016791360, 1587229842688, 51619520360960, 1808576831681536, 68562454975587328, 2830905156661645312, 124395772159835529216, 5504660984739184156672, 250011277837808237105152, 14799530615476409472303104
OFFSET
1,2
COMMENTS
a(23) = -4050933314339181211663673622528 is the first negative term. - Vladimir Reshetnikov, Aug 15 2021
REFERENCES
Comtet, L; Advanced Combinatorics (1974 edition), D. Reidel Publishing Company, Dordrecht - Holland, pp. 147-148.
LINKS
Vladimir Reshetnikov, Table of n, a(n) for n = 1..281
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) = T(n,1), T(n,m) = (1/2)*(4^(n-m)*(-1)^(n-m)*Stirling1(n,m) - Sum_{i=m+1..n-1} T(n,i)*T(i,m)), T(n,n)=1.
MAPLE
A := proc(n, m) option remember; if n = m then 1 else
1/2*(4^(n-m)*(-1)^(n-m)*Stirling1(n, m) - add(A(n, k)*A(k, m), k =m+1..n-1)) fi end: a := n -> A(n, 1): seq(a(n), n = 1..23); # Peter Luschny, Aug 15 2021
MATHEMATICA
t[n_, m_] := t[n, m] = 1/2*(4^(n - m)*(-1)^(n - m)*StirlingS1[n, m] - Sum[t[n, i]*t[i, m], {i, m+1, n-1}]); t[n_, n_] = 1; Table[t[n, 1], {n, 1, 20}] (* Jean-François Alcover, Feb 22 2013 *)
PROG
(Maxima)
T(n, m):=if n=m then 1 else 1/2*(4^(n-m)*(-1)^(n-m)*stirling1(n, m)-sum(T(n, i)*T(i, m), i, m+1, n-1));
makelist((T(n, 1)), n, 1, 10);
CROSSREFS
KEYWORD
sign
AUTHOR
Dmitry Kruchinin, Dec 05 2012
EXTENSIONS
More terms from Vladimir Reshetnikov, Aug 15 2021
STATUS
approved