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
Keywords
Examples
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.
Links
- Michael De Vlieger, Table of n, a(n) for n = 1..10000
Programs
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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 *)
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PARI
a(n) = sumdiv(n, d, issquare(d+1)); \\ Michel Marcus, Jan 06 2015
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
Comments