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A232193
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Numerators of the expected value of the length of a random cycle in a random n-permutation.
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4
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1, 3, 23, 55, 1901, 4277, 198721, 16083, 14097247, 4325321, 2132509567, 4527766399, 13064406523627, 905730205, 13325653738373, 362555126427073, 14845854129333883, 57424625956493833, 333374427829017307697, 922050973293317, 236387355420350878139797
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OFFSET
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1,2
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COMMENTS
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In this experiment we randomly select (uniform distribution) an n-permutation and then randomly select one of the cycles from that permutation. Cf. A102928/A175441 which gives the expected cycle length when we simply randomly select a cycle.
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LINKS
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FORMULA
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EXAMPLE
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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.
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MAPLE
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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)!):
# second Maple program:
a:= n-> numer(add(abs(combinat[stirling1](n, i))/i, i=1..n)/(n-1)!):
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MATHEMATICA
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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}]
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CROSSREFS
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Cf. A028417(n)/n! the expected value of the length of the shortest cycle in a random n-permutation.
Cf. A028418(n)/n! the expected value of the length of the longest cycle in a random n-permutation.
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KEYWORD
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nonn,frac
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AUTHOR
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STATUS
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approved
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