A065704 Number of squares or twice squares dividing n.
1, 2, 1, 3, 1, 2, 1, 4, 2, 2, 1, 3, 1, 2, 1, 5, 1, 4, 1, 3, 1, 2, 1, 4, 2, 2, 2, 3, 1, 2, 1, 6, 1, 2, 1, 6, 1, 2, 1, 4, 1, 2, 1, 3, 2, 2, 1, 5, 2, 4, 1, 3, 1, 4, 1, 4, 1, 2, 1, 3, 1, 2, 2, 7, 1, 2, 1, 3, 1, 2, 1, 8, 1, 2, 2, 3, 1, 2, 1, 5, 3, 2, 1, 3, 1, 2, 1, 4, 1, 4, 1, 3, 1, 2, 1, 6, 1, 4, 2, 6, 1, 2, 1, 4, 1
Offset: 1
Examples
divisors(36) = {1, 2, 3, 4, 6, 9, 12, 18, 36}, thus a(36) = #{1, 2, 4, 9, 18, 36}=6. a(36) = 1/2*(tau(36)-((-1)^sigma(1)+(-1)^sigma(2)+(-1)^sigma(3)+(-1)^sigma(4)+(-1)^sigma(6)+(-1)^sigma(9)+(-1)^sigma(12)+(-1)^sigma(18)+(-1)^sigma(36))) = 1/2*(9-(-3)) = 6. a(36) = a(2^2*3^2) = a(2^2)*a(3^2) = (2+1)*(1+1) = 6.
Links
- Carl R. White, Table of n, a(n) for n = 1..10000
Programs
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Mathematica
f[n_] := Total[1 - (-1)^DivisorSigma[1, Divisors@n]]/2; Array[f, 105] (* Robert G. Wilson v, Jan 02 2013 *) f[p_, e_] := If[p == 2, e+1, Floor[e/2] + 1]; a[n_] := Times @@ f @@@ FactorInteger[n]; Array[a, 100] (* Amiram Eldar, Sep 25 2022 *)
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PARI
a(n) = {my(e = valuation(n, 2), o = n>>e, f = factor(o)); (e+1)*prod(i=1 , #f~, floor(f[i,2]/2)+1)}; \\ Amiram Eldar, Sep 25 2022
Formula
a(n) = (1/2)*Sum_{ d divides n } (1-(-1)^sigma(d)).
Multiplicative with a(2^e) = e+1 and a(p^e) = floor(e/2)+1 for an odd prime p.
a(n) = A005361(2*n) for n>0 (conjectured). - Werner Schulte, Jan 15 2018 [This is true only for numbers whose odd part (A000265) is cubefree (A004709). Therefore, the least counterexample is n=3^3=27: a(27) = 2 while A005361(2*27) = 3. - Amiram Eldar, Sep 25 2022]
Asymptotic mean: Limit_{m->oo} (1/m) * Sum_{k=1..m} a(k) = Pi^2/4 (A091476). - Amiram Eldar, Sep 25 2022
Extensions
More terms from David Wasserman, Sep 09 2002