login
A226770
Let n+1 have proper divisors 1 < d_1,...,d_k < n+1; consider all proper divisors of n+d_1,...,n+d_k which did not appear earlier. Let them be d_{1,1}, d_{1,2},..., d_{k,1}, d_{k,2},..., d_{k,t}; then consider proper divisors of n+d_{1,1},...,n+d_{k,t} which did not appear earlier, repeat until no new divisor is introduced. a(n) is the total number of different divisors obtained.
6
0, 0, 1, 0, 3, 0, 5, 1, 5, 0, 9, 0, 11, 2, 3, 0, 15, 0, 17, 3, 11, 0, 21, 1, 19, 5, 17, 0, 27, 0, 29, 7, 19, 4, 23, 0, 35, 8, 23, 0, 39, 0, 41, 6, 23, 0, 45, 3, 41, 2, 31, 0, 51, 3, 39, 9, 35, 0, 57, 0, 59, 12, 29, 11, 47, 0, 65, 14, 43, 0, 69, 0, 71, 12, 39
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
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
Sequence in context: A247015 A241972 A357823 * A376625 A185782 A304027
KEYWORD
nonn
AUTHOR
Vladimir Shevelev, Jun 17 2013
STATUS
approved