OFFSET
1,5
COMMENTS
a(n) = 0, iff n = p - 1, where p is prime; we conjecture that a(p) = p - 2 and, more general, for odd prime p and k>=1, a(p^k) = p^k - p^(k-1) - 1.
If n = p^2 - 1, where p^2 + p - 1 is prime (A053184), then a(n) = 1.
What one can say about other values of a(n)?
LINKS
Peter J. C. Moses, Table of n, a(n) for n = 1..2000
EXAMPLE
Let n=9; the proper divisors >1 of n + 1 are 2,5; consider n + 2 = 11 and n + 5 = 14. These numbers give only one "new" proper divisor (>1) 7; the "new" proper divisors >1 of n + 7 = 16 are 4,8 and n + 4 = 13, n + 8 = 17 do not have proper divisors >1. The set of proper divisors of all considered sums is {2,5,7,4,8}. It contains 5 elements. Thus a(9) = 5.
MATHEMATICA
Table[(div=Most[Rest[Divisors[n+1]]]; If[div=={}, 0, Length[FixedPoint[ Union[Flatten[AppendTo[div, Map[Most[Rest[Divisors[n+#]]]&, #]]]]&, div]]]), {n, 50}] (* Peter J. C. Moses, Jun 17 2013 *)
CROSSREFS
KEYWORD
nonn
AUTHOR
Vladimir Shevelev, Jun 17 2013
STATUS
approved