

A181590


Least value of n such that P(n)  1/e < 10^(i), i=1,2,3... . P(n)=floor(n!/e + 1/2)/n! is the probability of a random permutation on n objects be a derangement.


2



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
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OFFSET

1,1


COMMENTS

Both P(n) and the probability that a rooted forest on [n] be a tree tend to 1/e when n rises to infinity. So the events random forest be a tree and random permutation be a derangement become equiprobable as n tends to infinity.
The probability P(n) approaches 1/e quite quickly as this sequence shows. See image clicking the first link.


LINKS



EXAMPLE

a(2) = 4, a(3) = 6, so for n in the interval 4...5, 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+k)1, k the least positive integer such that a(i+k) > a(i), this error is less than 10(ik+1).
For example, a(11) = a(12) = 14, k = 2 and if n is in the interval 14...14, if we use 1/e as the probability P, we make an error less than 10^(12).


MATHEMATICA

$MaxExtraPrecision = 100; f[n_] := Block[{k = 1}, While[ Abs[ Floor[(k!/E + 1/2)]/k!  1/E] > 1/10^n, k++ ]; k]; Array[f, 71] (* Robert G. Wilson v, Nov 05 2010 *)


CROSSREFS



KEYWORD

nonn


AUTHOR



EXTENSIONS



STATUS

approved



