A124795 Number of permutations with given cycle structure, in the prime factorization order.
1, 1, 1, 1, 2, 3, 6, 1, 3, 8, 24, 6, 120, 30, 20, 1, 720, 15, 5040, 20, 90, 144, 40320, 10, 40, 840, 15, 90, 362880, 120, 3628800, 1, 504, 5760, 420, 45, 39916800, 45360, 3360, 40, 479001600, 630, 6227020800, 504, 210, 403200, 87178291200, 15, 1260, 280, 25920
Offset: 1
Keywords
Links
- Gheorghe Coserea, Table of n, a(n) for n = 1..3000
- Eric Weisstein's World of Mathematics, Permutation cycle
Crossrefs
Cf. A000040.
Programs
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Mathematica
a[1] = 1; a[n_] := (f1 = FactorInteger[n]; rr = Range[PrimePi[f1[[-1, 1]]]]; f2 = {Prime[#], 0}& /@ rr; ff = Union[f1, f2] //. {b___, {p_, 0}, {p_, k_}, c___} -> {b, {p, k}, c}; kk = ff[[All, 2]]; (kk.rr)!/Times @@ (kk!)/Times @@ (rr^kk)); Array[a, 100] (* Jean-François Alcover, Feb 02 2018 *) -
PARI
a(n) = { my(f=factor(n), fsz=matsize(f)[1], g=sum(k=1, fsz, primepi(f[k, 1]) * f[k, 2])!, h=prod(k=1, fsz, primepi(f[k, 1])^f[k, 2])); g/(prod(k=1, fsz, f[k, 2]!) * h); }; vector(51, n, a(n)) \\ Gheorghe Coserea, Feb 02 2018; edited by Max Alekseyev, Feb 05 2018
Formula
For n = p1^k1*p2^k2*... where 2=p1
Comments