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A365401
The number of divisors of the largest unitary divisor of n that is a square.
4
1, 1, 1, 3, 1, 1, 1, 1, 3, 1, 1, 3, 1, 1, 1, 5, 1, 3, 1, 3, 1, 1, 1, 1, 3, 1, 1, 3, 1, 1, 1, 1, 1, 1, 1, 9, 1, 1, 1, 1, 1, 1, 1, 3, 3, 1, 1, 5, 3, 3, 1, 3, 1, 1, 1, 1, 1, 1, 1, 3, 1, 1, 3, 7, 1, 1, 1, 3, 1, 1, 1, 3, 1, 1, 3, 3, 1, 1, 1, 5, 5, 1, 1, 3, 1, 1, 1
OFFSET
1,4
COMMENTS
First differs from A212181 at n = 32.
The sum of these divisors is A351568(n).
All the terms are odd.
FORMULA
a(n) = A000005(A350388(n)).
a(n) = A000005(n) / A365402(n).
a(n) <= A000005(n) with equality if and only if n is a square (A000290).
a(n) >= 1 with equality if and only if n is an exponentially odd number (A268335).
Multiplicative with a(p^e) = 1 if e is odd, and e+1 if e is even.
Dirichlet g.f.: zeta(2*s)^2 * Product_{p prime} (1 + 1/p^s + 1/p^(2*s) - 1/p^(3*s)).
From Vaclav Kotesovec, Sep 05 2023: (Start)
Dirichlet g.f.: zeta(s) * zeta(2*s)^2 * Product_{p prime} (1 - 2/p^(3*s) + 1/p^(4*s)).
Let f(s) = Product_{p prime} (1 - 2/p^(3*s) + 1/p^(4*s)).
Sum_{k=1..n} a(k) ~ f(1) * Pi^4 * n / 36 + sqrt(n) * zeta(1/2) * f(1/2)/2 * (log(n) + 4*gamma - 2 + zeta'(1/2)/zeta(1/2) + f'(1/2)/f(1/2)), where
f(1) = Product_{p prime} (1 - 2/p^3 + 1/p^4) = 0.7446954979060674204391238715944543281179691329049241118630718137015097502...,
f(1/2) = Product_{p prime} (1 - 2/p^(3/2) + 1/p^2) = 0.2312522106782016049013780988087017618011735848676872392115785564006277675...,
f'(1/2) = f(1/2) * Sum_{p prime} 2*(3*sqrt(p) - 2) * log(p) / (1 - 2*sqrt(p) + p^2) = f(1/2) * 6.937179176924511608542644054340717439502789953858512457656... and gamma is the Euler-Mascheroni constant A001620. (End)
MATHEMATICA
f[p_, e_] := If[OddQ[e], 1, e + 1]; a[1] = 1; a[n_] := Times @@ f @@@ FactorInteger[n]; Array[a, 100]
PROG
(PARI) a(n) = vecprod(apply(x -> if(x%2, 1, x+1), factor(n)[, 2]));
KEYWORD
nonn,easy,mult
AUTHOR
Amiram Eldar, Sep 03 2023
STATUS
approved