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A331502 Decimal expansion of exp(4/9). 1
1, 5, 5, 9, 6, 2, 3, 4, 9, 7, 6, 0, 6, 7, 8, 0, 7, 1, 5, 5, 3, 3, 7, 0, 9, 2, 8, 6, 9, 7, 9, 4, 7, 1, 1, 8, 6, 3, 9, 4, 8, 2, 4, 0, 1, 1, 4, 2, 2, 1, 4, 2, 3, 5, 4, 3, 9, 0, 2, 7, 8, 4, 3, 1, 5, 4, 3, 5, 6, 3, 8, 5, 0, 1, 3, 3, 1, 0, 6, 3, 2, 6, 4, 2, 7, 5, 8, 1, 6, 1, 2, 4, 9, 2, 9, 9, 4, 0, 1, 5, 4, 2, 9, 1, 6, 9 (list; constant; graph; refs; listen; history; text; internal format)
OFFSET
1,2
COMMENTS
Considering graph evolutions (see the Flajolet link) with 3n vertices initially isolated, the probability of the occurrence of an acyclic graph at the point n, (n = 1/3 * 3n), in the uniform model, will be denoted by P13(n). In the case of the permutation model, the respective probability will be denoted by Pp13(n).
Pp13(n) / P13(n) ~ exp(4/9) since Pp13(n) = f(n) / C(N,n), where f(n) is the number of labeled forests with 3n nodes and n edges, and C(N,n), N = 3n *(3n-1)/2 (see the Lucatero link) is the number of labeled graphs with 3n nodes and n edges.
Because P13(n) = f(n)* n!* 2^n / (3n)^(2n), Pp13(n) / P13(n) = (3n)^(2n) / (C(N,n)* n! *2^n), and Lim_{n->oo} Pp13(n) / P13(n) = exp(4/9).
LINKS
P. Flajolet, D. E. Knuth, and B. Pittel, The first cycles in an evolving graph, Discrete Mathematics, 75(1-3):167-215, 1989.
FORMULA
Equals Lim_{n->oo} Pp13(n) / P13(n) = Lim_{n->oo} (3*n)^(2*n) / (binomial((3*n *(3*n-1)/2), n) * n! * 2^n).
EXAMPLE
1.55962349760678071553370928697947118639482401142214...
MAPLE
evalf(exp(4/9), 134);
MATHEMATICA
RealDigits[Exp[4/9], 10, 120][[1]] (* Harvey P. Dale, Jun 05 2023 *)
PROG
(PARI) exp(4/9)
CROSSREFS
Sequence in context: A046600 A344327 A030798 * A021951 A206772 A200679
KEYWORD
nonn,cons
AUTHOR
Washington Bomfim, Mar 04 2020
STATUS
approved

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