OFFSET
1,2
COMMENTS
Suppose that p = [x(1),...,x(k)], is a partition of n, where x(1) <= x(2) <= ... <= x(k). If x(1) = x(k), there is only one cyclic permutation of p; otherwise, there are k of them.
LINKS
Alois P. Heinz, Table of n, a(n) for n = 1..400
FORMULA
a(n) = (d(n), f(2), f(3),..., f(n-1))*(1,2,3,...,n-1), where d(n) = (number of divisors of n) = (number of constant partitions of n), and f(k) = number of nonconstant partitions of n, for k = 2,3,...,n-1.
EXAMPLE
a(6) = (4,2,2,2,1)*(1,2,3,4,5) = 27, where * = convolution. The 5 components of (4,2,2,2,1) count these partitions: (6, 33, 222, 1111); (51, 42); (411, 321); (3111, 2211); (211111).
MATHEMATICA
Map[Total[Map[Length, Map[(# /. Table[x_, {Length[#]}] -> {x}) &, IntegerPartitions[#]]]] &, Range[40]] (* A236292 *)
(* Peter J. C. Moses, Jan 21 2014 *)
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Clark Kimberling, Jan 22 2014
STATUS
approved