

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
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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 k1 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^21) / (1  x^(n^21)) ).  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^21)) do
A[k*(n^21)]:= A[k*(n^21)]+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



