A179952 Add 1 to all the divisors of n. a(n) = number of perfect squares in the set.

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

Michael De Vlieger, Table of n, a(n) for n = 1..10000

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

Adjacent sequences: A179949 A179950 A179951 * A179953 A179954 A179955

Keyword

nonn

Author

Jeff Burch, Aug 03 2010

Status

approved