A181590 -id:A181590 - cp's OEIS frontend

A181589 Least value of n such that P(n) - 1/e < 10^(-i), i=1,2,3... . P(n) = (n/(n+1))^(n-1) the probability of a random forest on n be a tree.

6, 56, 553, 5519, 55183, 551820, 5518192, 55181917, 551819162, 5518191618, 55181916176, 551819161758, 5518191617572, 55181916175717, 551819161757164, 5518191617571636, 55181916175716349, 551819161757163483

Offset: 1

Washington Bomfim, Oct 31 2010

nonn

The probability P(n) = A000169(n)/A000272(n+1). It is known that lim_{n->inf}p(n) = 1/e. (See Flajolet and Sedgewick link, pp 632, where we can find a function of the number of components k).
Both P(n) and the probability that a permutation on n objects be a derangement tend to 1/e when n rises to infinity. So the events a random forest be a tree and a random permutation be a derangement become equiprobable as n tends to infinity.
The probability P(n) approaches 1/e quite slowly as this sequence shows. See image clicking the first link.

			a(1) = 6, a(2) = 56, so for n in the interval 6...55 if we use 1/e as the probability P, we make an error less than 10^(-1). In general if n is in the interval a(i), ... , a(i+1)-1, this error is less than 10^(-i).

Flajolet and Sedgewick, Analytic Combinatorics
Wikipedia, Graph of probabilities
Wikipedia, Derangements

Cf. A000169, A000272, A068985, A000166, A181590.

A243293 Number of factorials < 10^n.

Original entry on oeis.org

3, 4, 6, 7, 8, 9, 10, 11, 12, 13, 14, 14, 15, 16, 17, 18, 18, 19, 20, 21, 21, 22, 23, 24, 24, 25, 26, 26, 27, 28, 29, 29, 30, 31, 31, 32, 33, 33, 34, 34, 35, 36, 36, 37, 38, 38, 39, 40, 40, 41, 41, 42, 43, 43, 44, 44, 45, 46, 46, 47, 47, 48, 49, 49, 50, 50, 51, 52, 52, 53, 53

Offset: 1

Derek Orr, Jun 02 2014

nonn

A181590(150) = 95 does not agree with a(150) = 96.
0! and 1! are the same number so a(n) counts it as 1 number.
a(n) is also the number of terms needed in the series Sum_{k=0..m} 1/k! to calculate exp(1) with a precision of at least n - 1 digits, i.e., exp(1) - Sum_{k=0..a(n)}1/k! < 10^(-n). - Martin Renner, Feb 18 2020

			There are 4 factorials < 10^2: 0! = 1! = 1, 2! = 2, 3! = 6, and 4! = 24. Thus a(2) = 4.

Giovanni Resta, Table of n, a(n) for n = 1..10000

Cf. A000142, A181590.

f=1; t=0; n=10; L={}; While[Length[L] < 100, t++; f*=t; While[f > n, AppendTo[ L, t-1]; n *= 10]]; L (* Giovanni Resta, Feb 19 2020 *)

a(n) = {my(tot=0); for(k=1,10^n,if(k!<10^n,tot++); if(k!>=10^n, break)); return(tot)}
n=1; while(n<100,print1(a(n),", "); n++)

cp's OEIS Frontend

A181589 Least value of n such that P(n) - 1/e < 10^(-i), i=1,2,3... . P(n) = (n/(n+1))^(n-1) the probability of a random forest on n be a tree.

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