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A179952 Add 1 to all the divisors of n. a(n)=number of perfect squares in the set. 1
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 (list; graph; refs; listen; history; text; internal format)
OFFSET

1,15

COMMENTS

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.

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

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

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Last modified January 20 23:20 EST 2019. Contains 319343 sequences. (Running on oeis4.)