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

1,5

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

Sequence in context: A247015 A241972 A357823 * A185782 A304027 A187886

Adjacent sequences: A226767 A226768 A226769 * A226771 A226772 A226773

Keyword

nonn

Author

Vladimir Shevelev, Jun 17 2013

Status

approved