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%I #17 Nov 15 2016 12:16:01
%S 2,2,3,2,3,2,4,3,4,2,3,2,4,4,4,2,3,2,5,3,4,2,4,3,4,3,4,2,3,2,4,3,4,4,
%T 5,2,4,3,4,2,3,2,4,3,4,2,4,3,5,3,4,2,5,4,6,3,4,2,3,2,4,4,6,4,4
%N a(n) = number of positive divisors of (n+1) that are each the average of a positive divisor of n and a positive divisor of (n+2).
%H G. C. Greubel, <a href="/A135615/b135615.txt">Table of n, a(n) for n = 1..2500</a>
%F a(n) <= A000005(n+1). - _Michel Marcus_, Oct 23 2016
%e The divisors of 5 are (1,5). The divisors of 6 are (1,2,3,6). And the divisors of 7 are (1,7). Looking at the divisors of 6, 1 is the average of 1 (from the divisors of 5) and 1 (from the divisors of 7). 3 is the average of 5 (from the divisors of 5) and 1 (from the divisors of 7). And 6 is the average of 5 (from the divisors of 5) and 7 (from the divisors of 7). 2 is not the average of any divisor of 5 and any divisor of 7. There are 3 divisors of 6 that are such averages, so a(5) = 3.
%t Table[Length[Intersection[Divisors[n + 1], Mean /@ Flatten[Outer[List, Divisors[n], Divisors[n + 2]], 1]]], {n, 1, 65}] (* Brad Chalfan (brad(AT)chalfan.net), Aug 31 2010 *)
%Y Cf. A000005 (number of divisors).
%K nonn
%O 1,1
%A _Leroy Quet_, Feb 28 2008
%E More terms from Brad Chalfan (brad(AT)chalfan.net), Aug 31 2010
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