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A365171
The number of divisors d of n such that gcd(d, n/d) is a square.
6
1, 2, 2, 2, 2, 4, 2, 2, 2, 4, 2, 4, 2, 4, 4, 3, 2, 4, 2, 4, 4, 4, 2, 4, 2, 4, 2, 4, 2, 8, 2, 4, 4, 4, 4, 4, 2, 4, 4, 4, 2, 8, 2, 4, 4, 4, 2, 6, 2, 4, 4, 4, 2, 4, 4, 4, 4, 4, 2, 8, 2, 4, 4, 4, 4, 8, 2, 4, 4, 8, 2, 4, 2, 4, 4, 4, 4, 8, 2, 6, 3, 4, 2, 8, 4, 4, 4
OFFSET
1,2
COMMENTS
The sum of these divisors is A365172(n).
FORMULA
Multiplicative with a(p^e) = floor((e + 3)/4) + floor((e + 4)/4) = A004524(e+3).
a(n) <= A000005(n), with equality if and only if n is squarefree (A005117).
a(n) >= A034444(n), with equality if and only if n is not a biquadrateful number (A046101).
a(n) == 1 (mod 2) if and only if n is a fourth power (A000583).
From Vaclav Kotesovec, Jan 20 2024: (Start)
Dirichlet g.f.: zeta(s)^2 * Product_{p prime} (1 - 1/(p^(2*s) + 1)).
Let f(s) = Product_{p prime} (1 - 1/(p^(2*s) + 1)).
Sum_{k=1..n} a(k) ~ f(1) * n * (log(n) + 2*gamma - 1 + f'(1)/f(1)), where
f(1) = Product_{p prime} (1 - 1/(p^2 + 1)) = Pi^2/15 = A182448,
f'(1) = f(1) * Sum_{p prime} 2*log(p) / (p^2 + 1) = f(1) * 0.8852429263675811068149340172820329246145172848406469350087715037483367369...
and gamma is the Euler-Mascheroni constant A001620. (End)
MATHEMATICA
f[p_, e_] := Floor[(e + 3)/4] + Floor[(e + 4)/4]; a[1] = 1; a[n_] := Times @@ f @@@ FactorInteger[n]; Array[a, 100]
PROG
(PARI) a(n) = vecprod(apply(x -> (x+3)\4 + (x+4)\4, factor(n)[, 2]));
(PARI) for(n=1, 100, print1(direuler(p=2, n, 1/((1 - X)^2 * (1 + X^2)))[n], ", ")) \\ Vaclav Kotesovec, Jan 20 2024
KEYWORD
nonn,easy,mult
AUTHOR
Amiram Eldar, Aug 25 2023
STATUS
approved