%N Add 1 to all the divisors of n. a(n)=number of perfect squares in the set.
%C a(24)=3 because the divisors of 24 are 1,2,3,4,6,8,12,24. Adding one to each gives 2,3,4,5,7,9,13,25 and of those 4,9 and 25 are perfect squares.
%C Number of k>=2 such that both k-1 and k+1 divide n. - _Joerg Arndt_, Jan 06 2015
%H Michael De Vlieger, <a href="/A179952/b179952.txt">Table of n, a(n) for n = 1..10000</a>
%F G.f.: sum(n>=2, x^(n^2-1) / (1 - x^(n^2-1)) ). - _Joerg Arndt_, Jan 06 2015
%p N:= 1000: # to get a(1) to a(N)
%p A:= Vector(N):
%p for n from 2 to floor(sqrt(N+1)) do
%p for k from 1 to floor(N/(n^2-1)) do
%p A[k*(n^2-1)]:= A[k*(n^2-1)]+1
%p convert(A,list); # _Robert Israel_, Jan 06 2015
%t a179952[n_] := Count[Sqrt[Divisors[#] + 1], _Integer] & /@ Range@n; a179952 (* _Michael De Vlieger_, Jan 06 2015 *)
%o (PARI) a(n) = sumdiv(n, d, issquare(d+1)); \\ _Michel Marcus_, Jan 06 2015
%A _Jeff Burch_, Aug 03 2010