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

1,5

Comments

Number of permutations with k1 1-cycles, k2 2-cycles, ...

Links

Gheorghe Coserea, Table of n, a(n) for n = 1..3000

Eric Weisstein's World of Mathematics, Permutation cycle

Formula

For n=p1^k1*p2^k2*... where 2=p1<p2<... are the sequence of all primes, a(n) = a([k1,k2,...]) = (k1+2*k2+...)!/((k1!*k2!*...)*(1^k1*2^k2*...)

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 *)

Prog

(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

Crossrefs

Cf. A000040.

Sequence in context: A275732 A200594 A319191 * A368547 A346560 A084459

Adjacent sequences: A124792 A124793 A124794 * A124796 A124797 A124798

Keyword

nonn

Author

Max Alekseyev, Nov 07 2006

Status

approved