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%I #20 May 03 2024 12:12:38
%S 0,1,2,2,2,2,0,4,6,-58,100,1052,-5924,-21972,322020,332392,-21168682,
%T 29068598,1724404180,-7070346036,-172304798980,1290100381724,
%U 20728501384592,-247269172883976,-2936888518668676,53037176259027580,477640220538178184
%N G.f. satisfies A(A(x)) = F(x), where F(x) is the g.f. for A001787(n) = n*2^(n-1).
%C A(x) is sometimes called a functional square root, or half-iterate of F(x).
%H Seiichi Manyama, <a href="/A309509/b309509.txt">Table of n, a(n) for n = 0..488</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, 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.
%H Wikipedia, <a href="https://en.wikipedia.org/wiki/Functional_square_root">Functional square root</a>.
%F Define the sequence b(n,k) as follows. If n<k, b(n,k) = 0, else if n=k, b(n,k) = 1, otherwise b(n,k) = 1/2 * (2^(n-k) * binomial(n+k-1,2*k-1) - Sum_{m=k+1..n-1} b(n,m) * b(m,k)). a(n) = b(n,1). - _Seiichi Manyama_, May 03 2024
%t half[q_] := half[q] = Module[{h}, h[0] = 0; h[1] = 1; h[n_Integer] := h[n] = Module[{c}, c[m_Integer /; m < n] := h[m]; c[n] /. Solve[q[n] == Sum[k! c[k] BellY[n, k, Table[m! c[m], {m, n - k + 1}]], {k, n}]/n!, c[n]][[1]]]; h]; a[n_Integer] := a[n] = half[Function[k, k 2^(k-1)]][n]; Table[a[n], {n, 0, 26}]
%Y Cf. A001787, A027436, A030274, A030275, A184011.
%K sign
%O 0,3
%A _Vladimir Reshetnikov_, Aug 05 2019
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