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%I #22 Aug 31 2018 03:44:16
%S 1,1,5,49,243,14641,371293,253125,410338673,16983563041,1400846643,
%T 41426511213649,95367431640625,617673396283947,10260628712958602189,
%U 756943935220796320321,7474615974418932603,827909024473876953125,456487940826035155404146917,510798409623548623605717
%N Numerators of series related to triangular cacti.
%C From _L. Edson Jeffery_, Jan 09 2012: (Start)
%C The reference [Bergeron, et al.] lists the first few terms of the relevant series as S(x) = x + (1/2)*x^3 + (5/8)*x^5 + (49/48)*x^7 + (243/128)*x^9 + ..., from which the numerators were taken for this sequence and the denominators for A058928. This leads to the following
%C Conjecture: S(x) = Sum_{n>=0} ((2*n+1)^(n-1)/(n!*2^n))*x^(2*n+1) = (A052750(n)/A000165(n))*x^(2*n+1). Letting D_n be the set of divisors of n! and d_n = max(k in D_n : k | (2*n+1)^(n-1)), then a(n)=A052750(n)/d_n. (End)
%C The above conjecture is correct and follows from formula given in A034940 for the number of rooted labeled triangular cacti with 2n+1 nodes. - _Andrew Howroyd_, Aug 30 2018
%D F. Bergeron, G. Labelle and P. Leroux, Combinatorial Species and Tree-Like Structures, Camb. 1998, p. 307.
%H Andrew Howroyd, <a href="/A058927/b058927.txt">Table of n, a(n) for n = 0..200</a>
%F G.f.: A(x) satisfies A(x)=exp(x*A(x)^2). - _Vladimir Kruchinin_, Feb 09 2013
%F a(n) = numerator(A034940(n)/(2*n+1)!) = numerator((2*n+1)^(n-1)/(2^n*n!)). - _Andrew Howroyd_, Aug 30 2018
%o (PARI) a(n)={numerator((2*n+1)^(n-1)/(2^n*n!))} \\ _Andrew Howroyd_, Aug 30 2018
%Y Cf. A000165, A034940, A052750, A058928.
%K nonn,frac,easy
%O 0,3
%A _N. J. A. Sloane_, Jan 12 2001
%E More terms from Herman Jamke (hermanjamke(AT)fastmail.fm), Sep 25 2010
%E Terms a(12) and beyond from _Andrew Howroyd_, Aug 30 2018
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