A226770 - OEIS

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.

%I #25 Jul 05 2013 03:02:56

%S 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,

%T 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,

%U 0,57,0,59,12,29,11,47,0,65,14,43,0,69,0,71,12,39

%N 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.

%C 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.

%C If n = p^2 - 1, where p^2 + p - 1 is prime (A053184), then a(n) = 1.

%C What one can say about other values of a(n)?

%H Peter J. C. Moses, <a href="/A226770/b226770.txt">Table of n, a(n) for n = 1..2000</a>

%e 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.

%t 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 *)

%K nonn

%O 1,5

%A _Vladimir Shevelev_, Jun 17 2013