A365173 The number of divisors d of n such that gcd(d, n/d) is an exponentially odd number (A268335).
1, 2, 2, 3, 2, 4, 2, 4, 3, 4, 2, 6, 2, 4, 4, 4, 2, 6, 2, 6, 4, 4, 2, 8, 3, 4, 4, 6, 2, 8, 2, 4, 4, 4, 4, 9, 2, 4, 4, 8, 2, 8, 2, 6, 6, 4, 2, 8, 3, 6, 4, 6, 2, 8, 4, 8, 4, 4, 2, 12, 2, 4, 6, 5, 4, 8, 2, 6, 4, 8, 2, 12, 2, 4, 6, 6, 4, 8, 2, 8, 4, 4, 2, 12, 4, 4
Offset: 1
Links
- Amiram Eldar, Table of n, a(n) for n = 1..10000
- Vaclav Kotesovec, Graph - the asymptotic ratio (100000 terms)
Programs
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Mathematica
f[p_, e_] := Floor[(e + 5)/4] + Floor[(e + 6)/4]; a[1] = 1; a[n_] := Times @@ f @@@ FactorInteger[n]; Array[a, 100]
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PARI
a(n) = vecprod(apply(x -> (x+5)\4 + (x+6)\4, factor(n)[, 2]));
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PARI
for(n=1, 100, print1(direuler(p=2, n, (1 + X^2 - X^4)/((1 - X)^2*(1 + X^2)))[n], ", ")) \\ Vaclav Kotesovec, Jan 20 2024
Formula
Multiplicative with a(p^e) = floor((e+5)/4) + floor((e+6)/4) = A004524(e+5).
a(n) == 1 (mod 2) if and only if n is a square of an exponentially odd number (i.e., a number whose prime factorization include only exponents e such that e == 2 (mod 4)).
From Vaclav Kotesovec, Jan 20 2024: (Start)
Dirichlet g.f.: zeta(s)^2 * Product_{p prime} (1 - 1/(p^(2*s)*(1 + p^(2*s)))).
Let f(s) = Product_{p prime} (1 - 1/(p^(2*s)*(1 + p^(2*s)))).
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 + p^2))) = 0.937494282731300250789438325050116436995101826036120273493270589183132928...,
f'(1) = f(1) * Sum_{p prime} (4*p^2 + 2) * log(p) / (p^6 + 2*p^4 - 1) = f(1) * 0.192452062257404507109731932640803706644036700262364333369815000973104583...
and gamma is the Euler-Mascheroni constant A001620. (End)
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