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%I #20 Jan 19 2024 04:57:17
%S 0,0,1,0,0,1,0,1,1,0,0,1,0,0,2,1,0,1,0,0,1,0,0,3,0,0,1,0,0,2,0,1,1,0,
%T 1,1,0,0,1,1,0,1,0,0,2,0,0,4,0,0,1,0,0,1,0,1,1,0,0,2,0,0,2,1,0,1,0,0,
%U 1,1,0,3,0,0,2,0,0,1,0,2,1,0,0,1,0,0,1,1,0,2,0,0,1,0,0,4,0,0,2,0,0,1,0,1,3
%N Add 1 to all the divisors of n. a(n) = number of perfect squares in the set.
%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
%F Asymptotic mean: Limit_{m->oo} (1/m) * Sum_{k=1..m} a(k) = 3/4. - _Amiram Eldar_, Jan 19 2024
%e 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.
%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 od
%p od;
%p convert(A,list); # _Robert Israel_, Jan 06 2015
%t a179952[n_] := Count[Sqrt[Divisors[#] + 1], _Integer] & /@ Range@n; a179952[105] (* _Michael De Vlieger_, Jan 06 2015 *)
%o (PARI) a(n) = sumdiv(n, d, issquare(d+1)); \\ _Michel Marcus_, Jan 06 2015
%K nonn
%O 1,15
%A _Jeff Burch_, Aug 03 2010
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