OFFSET
1,3
COMMENTS
The number of all permutations of elements of sets {1..k}, k <= n, is b(n) = Sum_{k=0..n} k! while the number of all permutations of elements of all subsets of set {1,2..n} is c(n) = Sum_{k=0..n} binomial(n,k)!. So the required probability (in a sample space) is b(n)/c(n), n >= 1 (after reduction of the fractions).
Apparently a(n) = A014288(n) for n > 2. - Georg Fischer, Oct 23 2018
LINKS
Amiram Eldar, Table of n, a(n) for n = 1..30
MATHEMATICA
a[n_] := Numerator[Sum[k!, {k, 0, n}]/Sum[Binomial[n, k]!, {k, 0, n}]]; Array[a, 25] (* Amiram Eldar, Sep 21 2019 *)
PROG
(PARI) a(n) = numerator(sum(k=0, n, k!)/sum(k=0, n, binomial(n, k)!)); \\ Michel Marcus, Oct 12 2017
CROSSREFS
KEYWORD
nonn,frac
AUTHOR
Vladimir Shevelev, Oct 09 2017
EXTENSIONS
More terms from Peter J. C. Moses, Oct 09 2017
STATUS
approved