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A179952
Add 1 to all the divisors of n. a(n) = number of perfect squares in the set.
2
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, 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, 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
OFFSET
1,15
COMMENTS
Number of k>=2 such that both k-1 and k+1 divide n. - Joerg Arndt, Jan 06 2015
LINKS
FORMULA
G.f.: Sum_{n>=2} x^(n^2-1) / (1 - x^(n^2-1)). - Joerg Arndt, Jan 06 2015
Asymptotic mean: Limit_{m->oo} (1/m) * Sum_{k=1..m} a(k) = 3/4. - Amiram Eldar, Jan 19 2024
EXAMPLE
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.
MAPLE
N:= 1000: # to get a(1) to a(N)
A:= Vector(N):
for n from 2 to floor(sqrt(N+1)) do
for k from 1 to floor(N/(n^2-1)) do
A[k*(n^2-1)]:= A[k*(n^2-1)]+1
od
od;
convert(A, list); # Robert Israel, Jan 06 2015
MATHEMATICA
a179952[n_] := Count[Sqrt[Divisors[#] + 1], _Integer] & /@ Range@n; a179952[105] (* Michael De Vlieger, Jan 06 2015 *)
PROG
(PARI) a(n) = sumdiv(n, d, issquare(d+1)); \\ Michel Marcus, Jan 06 2015
CROSSREFS
Sequence in context: A250214 A073423 A219180 * A321930 A134023 A257931
KEYWORD
nonn
AUTHOR
Jeff Burch, Aug 03 2010
STATUS
approved