A232193 Numerators of the expected value of the length of a random cycle in a random n-permutation.
1, 3, 23, 55, 1901, 4277, 198721, 16083, 14097247, 4325321, 2132509567, 4527766399, 13064406523627, 905730205, 13325653738373, 362555126427073, 14845854129333883, 57424625956493833, 333374427829017307697, 922050973293317, 236387355420350878139797
Offset: 1
Examples
Expectations for n=1,... are 1/1, 3/2, 23/12, 55/24, 1901/720, 4277/1440, 198721/60480, 16083/4480, ... = A232193/A232248 For n=3 there are 6 permutations. We have probability 1/6 of selecting (1)(2)(3) and the cycle size is 1. We have probability 3/6 of selecting a permutation with cycle type (1)(23) and (on average) the cycle length is 3/2. We have probability 2/6 of selecting a permutation of the form (123) and the cycle size is 3. 1/6*1 + 3/6*3/2 + 2/6*3 = 23/12.
Links
- Alois P. Heinz, Table of n, a(n) for n = 1..250
Crossrefs
Programs
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Maple
with(combinat): b:= proc(n, i) option remember; `if`(n=0, 1, `if`(i<1, 0, expand(add(multinomial(n, n-i*j, i$j)/j!*(i-1)!^j *b(n-i*j, i-1) *x^j, j=0..n/i)))) end: a:= n->numer((p->add(coeff(p, x, i)/i, i=1..n))(b(n$2))/(n-1)!): seq(a(n), n=1..30); # Alois P. Heinz, Nov 21 2013 # second Maple program: a:= n-> numer(add(abs(combinat[stirling1](n, i))/i, i=1..n)/(n-1)!): seq(a(n), n=1..30); # Alois P. Heinz, Nov 23 2013 -
Mathematica
Table[Numerator[Total[Map[Total[#]!/Product[#[[i]],{i,1,Length[#]}]/Apply[Times,Table[Count[#,k]!,{k,1,Max[#]}]]/(Total[#]-1)!/Length[#]&,Partitions[n]]]],{n,1,25}]
Formula
a(n) = Numerator( 1/(n-1)! * Sum_{i=1..n} A132393(n,i)/i ). - Alois P. Heinz, Nov 23 2013
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