

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 ncycles?


0



1, 1, 1, 2, 4, 6, 11, 20, 36, 65, 118, 215, 389, 727, 1366, 2565, 4849, 9123, 17168, 32629
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OFFSET

1,4


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.


LINKS

Table of n, a(n) for n=1..20.


EXAMPLE

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 5cycles, 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).


PROG

(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;


CROSSREFS

Cf. A000048, A056303.
Sequence in context: A026385 A254532 A199926 * A017993 A049870 A093970
Adjacent sequences: A068327 A068328 A068329 * A068331 A068332 A068333


KEYWORD

nonn,nice,more


AUTHOR

Simon P. Norton (simon(AT)dpmms.cam.ac.uk), Feb 27 2002


STATUS

approved



