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A331501
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Decimal expansion of exp(3/4).
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2
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2, 1, 1, 7, 0, 0, 0, 0, 1, 6, 6, 1, 2, 6, 7, 4, 6, 6, 8, 5, 4, 5, 3, 6, 9, 8, 1, 9, 8, 3, 7, 0, 9, 5, 6, 1, 0, 1, 3, 4, 4, 9, 1, 5, 8, 4, 7, 0, 2, 4, 0, 3, 4, 2, 1, 7, 7, 9, 1, 3, 3, 0, 3, 0, 8, 1, 0, 9, 8, 4, 5, 3, 3, 3, 6, 4, 0, 1, 2, 8, 2, 0, 0, 0, 2, 7, 9, 1, 5, 6, 0, 2, 6, 6, 6, 1, 5, 7, 9, 8, 2, 1, 8, 8, 8
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OFFSET
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1,1
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COMMENTS
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Considering graph evolutions (see the Flajolet link) with 2n vertices initially isolated, the probability of the occurrence of an acyclic graph at the critical point n in the uniform model, will be denoted by P(n). In the case of the permutation model, the respective probability will be denoted by Pp(n).
Pp(n) / P(n) ~ exp(3/4) since Pp(n) = A302112(n) / A331505(2n) = A302112(n) / C(C(2n,2), n), and P(n) = A302112(n) * n! * 2^n / (2n)^(2n), Pp(n) / P(n) = (2n)^(2n) / (C(C(2n,2), n) * n! * 2^n), and lim_{n->oo} Pp(n) / P(n) = exp(3/4).
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LINKS
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Leonard Giugiuc and Dan Stefan Marinescu, Problem 4257, Crux Mathematicorum, Vol. 43, No. 6 (2017), pp. 263 and 265; Solution to Problem 4257, ibid., Vol. 44, No. 6 (2018), pp. 268-270.
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FORMULA
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Equals lim_{n->oo} Pp(n) / P(n) = lim_{n->oo} (2*n)^(2*n) / (binomial(binomial(2n,2), n) * n! * 2^n).
Equals lim_{n->oo} sqrt(n)/A000178(n)^(1/(n*(n+1))) (Giugiuc and Marinescu, 2017). - Amiram Eldar, Apr 12 2022
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EXAMPLE
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2.1170000166126746685453698198370956101344915847024...
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MAPLE
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evalf(exp(3/4), 134);
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MATHEMATICA
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RealDigits[Exp[3/4], 10, 100][[1]] (* Amiram Eldar, Apr 12 2022 *)
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CROSSREFS
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KEYWORD
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AUTHOR
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STATUS
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approved
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