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.

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

Author

Vladimir Shevelev, Jun 17 2013

Keywords

• nonn

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)?

Examples

			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.
		

Links

• Peter J. C. Moses, Table of n, a(n) for n = 1..2000

Programs

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 *)