A331502 Decimal expansion of exp(4/9).

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

Offset: 1

Author

Washington Bomfim, Mar 04 2020

Keywords

• nonn
• cons

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).

Examples

			1.55962349760678071553370928697947118639482401142214...
		

Links

• P. Flajolet, D. E. Knuth, and B. Pittel, The first cycles in an evolving graph, Discrete Mathematics, 75(1-3):167-215, 1989.
• Carlos R. Lucatero, Combinatorial Enumeration of Graphs.
• Index entries for transcendental numbers

Crossrefs

Cf. A001113, A019774, A092042, A302112, A331500, A331501, A331505.

Programs

Maple

evalf(exp(4/9), 134);

Mathematica

RealDigits[Exp[4/9],10,120][[1]] (* Harvey P. Dale, Jun 05 2023 *)

PARI

exp(4/9)

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).