A369307 The number of exponentially odd divisors d of n such that n/d is also exponentially odd.
1, 2, 2, 1, 2, 4, 2, 2, 1, 4, 2, 2, 2, 4, 4, 2, 2, 2, 2, 2, 4, 4, 2, 4, 1, 4, 2, 2, 2, 8, 2, 2, 4, 4, 4, 1, 2, 4, 4, 4, 2, 8, 2, 2, 2, 4, 2, 4, 1, 2, 4, 2, 2, 4, 4, 4, 4, 4, 2, 4, 2, 4, 2, 3, 4, 8, 2, 2, 4, 8, 2, 2, 2, 4, 2, 2, 4, 8, 2, 4, 2, 4, 2, 4, 4, 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_] := If[OddQ[e], 2, e/2]; a[1] = 1; a[n_] := Times @@ f @@@ FactorInteger[n]; Array[a, 100]
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PARI
a(n) = vecprod(apply(x -> if(x%2, 2, x/2), factor(n)[,2]));
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PARI
for(n=1, 100, print1(direuler(p=2, n, (1 - X^2 + X)^2/(1 - X^2)^2)[n], ", ")) \\ Vaclav Kotesovec, Jan 19 2024
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Python
from math import prod from sympy import factorint def A369307(n): return prod(2 if e&1 else e>>1 for e in factorint(n).values()) # Chai Wah Wu, Jan 19 2024
Formula
Multiplicative with a(p^e) = 2 is e is odd, and e/2 if e is even.
a(n) >= 1, with equality if and only if n is the square of a squarefree number (A062503).
Dirichlet g.f.: zeta(2*s)^2 * (Product_{p prime} (1 + 1/p^s - 1/p^(2*s)))^2.
From Vaclav Kotesovec, Jan 19 2024: (Start)
Dirichlet g.f.: zeta(s)^2 * Product_{p prime} (p^(2*s) + p^s - 1)^2 / ((p^s + 1)^2 * p^(2*s)).
Let f(s) = Product_{p prime} (p^(2*s) + p^s - 1)^2 / ((p^s + 1)^2 * 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 - (2*p^2 + 2*p - 1) / (p^2*(p+1)^2)) = 0.49623881454854881762168565097162197963340069996226074849602334089041678...,
f'(1) = f(1) * Sum_{p prime} 2*(2*p + 1) * log(p) / ((p+1)*(p^2 + p - 1)) = f(1) * 1.49674466685934940187617305887881799198585080518913793200171026177150513...
and gamma is the Euler-Mascheroni constant A001620. (End)
Comments