OFFSET
0,3
COMMENTS
From L. Edson Jeffery, Jan 09 2012: (Start)
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
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)
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
REFERENCES
F. Bergeron, G. Labelle and P. Leroux, Combinatorial Species and Tree-Like Structures, Camb. 1998, p. 307.
LINKS
Andrew Howroyd, Table of n, a(n) for n = 0..200
FORMULA
G.f.: A(x) satisfies A(x)=exp(x*A(x)^2). - Vladimir Kruchinin, Feb 09 2013
a(n) = numerator(A034940(n)/(2*n+1)!) = numerator((2*n+1)^(n-1)/(2^n*n!)). - Andrew Howroyd, Aug 30 2018
PROG
(PARI) a(n)={numerator((2*n+1)^(n-1)/(2^n*n!))} \\ Andrew Howroyd, Aug 30 2018
CROSSREFS
KEYWORD
nonn,frac,easy
AUTHOR
N. J. A. Sloane, Jan 12 2001
EXTENSIONS
More terms from Herman Jamke (hermanjamke(AT)fastmail.fm), Sep 25 2010
Terms a(12) and beyond from Andrew Howroyd, Aug 30 2018
STATUS
approved