0,6

The number of degree-n permutations of order exactly p (where p is prime) satisfies a(n) =a(n-1)+(1+a(n-p))*(n-1)!/(n-p)! with a(n)=0 if p>n. Also a(n)=Sum_{j=1 to floor[n/p]}[n!/(j!*(n-p*j)!*(p^j))].

a(n) = a(n - 1) + (1 + a(n - 5))*(n - 1)(n - 2)(n - 3)(n - 4) = Sum_{j = 1 to floor[n/5]}[n!/(j!*(n - 5j)!*(5^j))].

Cf. A001471.

Sequence in context: A093699 A076835 A007900 this_sequence A054118 A001342 A029572

Adjacent sequences: A059590 A059591 A059592 this_sequence A059594 A059595 A059596

nonn

Henry Bottomley (se16(AT)btinternet.com), Jan 26 2001