A068330 Consider all sublists of [(2,1),(3,2,1),(4,3,2,1),...,(n,...,4,3,2,1)] and multiply these permutations in that order. How many of the products are n-cycles?

1, 1, 1, 2, 4, 6, 11, 20, 36, 65, 118, 215, 389, 727, 1366, 2565, 4849, 9123, 17168, 32629, 62121, 118353, 226603, 434512, 833776, 1605642, 3101121, 5993545, 11593548, 22443167, 43459975, 84209877, 163359383, 317230531, 616506533, 1199200964, 2334860706, 4549377408, 8870694723

Offset: 1

Author

Simon P. Norton, Feb 27 2002

Keywords

• nonn
• nice

Comments

If we take the inverse permutations to the above, or, equivalently, multiply them in the reverse order, we get another description of the sequences A000048 or A056303 with the first term omitted in each case.

Examples

			a[5] (the output of the program below in which a is the list of the first n terms of the sequence) is 4 because that is the number of products of sublists of [(2,1),(3,2,1),(4,3,2,1),(5,4,3,2,1)] which are 5-cycles, namely (5,4,3,2,1) itself, (3,2,1)*(5,4,3,2,1) = (5,4,3,1,2), (2,1)*(4,3,2,1)*(5,4,3,2,1) = (5,4,2,3,1) and (2,1)*(3,2,1)*(4,3,2,1)*(5,4,3,2,1) = (5,4,2,1,3).
		

Links

• Sean A. Irvine, Java program (github)

Crossrefs

Cf. A000048, A056303.

Programs

GAP

a := []; p := (); perms := [p]; for i in [1..n] do pp := perms*p; pp1 := Filtered(pp,m -> CycleLength(m,[1..i],1) = i); a[i] := Length(pp1); perms := Union(perms, pp); p := p*(i,i+1); od;

Extensions

a(21)-a(33) from Sean A. Irvine, Feb 10 2024

a(34)-a(39) from Bert Dobbelaere, May 29 2025