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%I #29 Aug 15 2021 15:08:59
%S 1,2,10,80,872,11928,195072,3702080,80065792,1950808000,53016791360,
%T 1587229842688,51619520360960,1808576831681536,68562454975587328,
%U 2830905156661645312,124395772159835529216,5504660984739184156672,250011277837808237105152,14799530615476409472303104
%N E.g.f. A(x) satisfies A(A(x)) = (1/4)*log(1/(1-4*x)).
%C a(23) = -4050933314339181211663673622528 is the first negative term. - _Vladimir Reshetnikov_, Aug 15 2021
%D Comtet, L; Advanced Combinatorics (1974 edition), D. Reidel Publishing Company, Dordrecht - Holland, pp. 147-148.
%H Vladimir Reshetnikov, <a href="/A220112/b220112.txt">Table of n, a(n) for n = 1..281</a>
%H Gottfried Helms, <a href="http://go.helms-net.de/math/tetdocs/CoefficientsForUTetration.htm">Coefficients for fractional iterates exp(x)-1</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) = 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.
%p A := proc(n, m) option remember; if n = m then 1 else
%p 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
%t 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 *)
%o (Maxima)
%o 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));
%o makelist((T(n,1)),n,1,10);
%Y Cf. A052122, A052123, A184011.
%K sign
%O 1,2
%A _Dmitry Kruchinin_, Dec 05 2012
%E More terms from _Vladimir Reshetnikov_, Aug 15 2021
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