OFFSET
1,6
COMMENTS
a(n) depends on exponents of prime power factorization of n only; moreover, it is invariant with respect to permutations of them. An equivalent multiset formulation of the problem: for a given finite multiset A, we should, beginning with A, to get all submultisets of A, if, by every step, we remove or join 1 element and such that, joining to the last submultiset one element, we again obtain A. How many ways to do this?
Via Seqfan Discussion List (Aug 07 2010), Alois P. Heinz proved that every subsequence of the form a(p), a(p*q), a(p*q*r), ..., where p, q, r, ... are distinct primes, coincides with A003042. - Vladimir Shevelev, Nov 07 2014
LINKS
Vladimir Shevelev, Combinatorial minors of matrix functions and their applications, arXiv:1105.3154 [math.CO], 2011-2014.
Vladimir Shevelev, Combinatorial minors of matrix functions and their applications, Zesz. Nauk. PS., Mat. Stosow., Zeszyt 4, pp. 5-16. (2014).
FORMULA
a(p)=1, and, for k>=2, a(p^k)=0; a(p*q)=a(p^2*q)=a(p^3*q)=2; a(p^2*q^2)=0; a(p*q*r)=12, etc. (here p,q,r are distinct primes).
EXAMPLE
If n=p*q, then we have exactly two required chains: p*q, p, 1, q and p*q, q, 1, p. Thus a(6)=a(10)=a(14)=...=2.
CROSSREFS
KEYWORD
nonn
AUTHOR
Vladimir Shevelev, Aug 07 2010
EXTENSIONS
Corrected and extended by Alois P. Heinz from a(48) via Seqfan Discussion List (Aug 07 2010)
STATUS
approved