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A065704
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Number of squares or twice squares dividing n.
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4
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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
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OFFSET
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1,2
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LINKS
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FORMULA
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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.
Asymptotic mean: Limit_{m->oo} (1/m) * Sum_{k=1..m} a(k) = Pi^2/4 (A091476). - Amiram Eldar, Sep 25 2022
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EXAMPLE
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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.
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MATHEMATICA
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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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PROG
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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
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CROSSREFS
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KEYWORD
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mult,nonn
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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