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A309509
G.f. satisfies A(A(x)) = F(x), where F(x) is the g.f. for A001787(n) = n*2^(n-1).
6
0, 1, 2, 2, 2, 2, 0, 4, 6, -58, 100, 1052, -5924, -21972, 322020, 332392, -21168682, 29068598, 1724404180, -7070346036, -172304798980, 1290100381724, 20728501384592, -247269172883976, -2936888518668676, 53037176259027580, 477640220538178184
OFFSET
0,3
COMMENTS
A(x) is sometimes called a functional square root, or half-iterate of F(x).
LINKS
Dmitry Kruchinin, Vladimir Kruchinin, Method for solving an iterative functional equation $A^{2^n}(x)=F(x)$, arXiv:1302.1986 [math.CO], 2013.
FORMULA
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
MATHEMATICA
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}]
CROSSREFS
KEYWORD
sign
AUTHOR
STATUS
approved