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A162510
Dirichlet inverse of A076479.
9
1, 1, 1, 2, 1, 1, 1, 4, 2, 1, 1, 2, 1, 1, 1, 8, 1, 2, 1, 2, 1, 1, 1, 4, 2, 1, 4, 2, 1, 1, 1, 16, 1, 1, 1, 4, 1, 1, 1, 4, 1, 1, 1, 2, 2, 1, 1, 8, 2, 2, 1, 2, 1, 4, 1, 4, 1, 1, 1, 2, 1, 1, 2, 32, 1, 1, 1, 2, 1, 1, 1, 8, 1, 1, 2, 2, 1, 1, 1, 8, 8, 1, 1, 2, 1, 1, 1, 4, 1, 2, 1, 2, 1, 1, 1, 16, 1, 2, 2, 4, 1, 1, 1, 4, 1
OFFSET
1,4
COMMENTS
Apart from signs, this sequence is identical to A162512.
FORMULA
Multiplicative with a(p^e) = 2^(e-1) for any prime p and any positive exponent e.
a(n) = n/2 when n is a power of 2 (A000079).
a(n) = 1 when n is a squarefree number (A005117).
a(n) = 2^A046660(n) = A061142(n)/A034444(n). - R. J. Mathar, Nov 02 2016
a(n) = Sum_{d|n} mu(d) * 2^A001222(n/d). - Daniel Suteu, May 21 2020
a(1) = 1; a(n) = -Sum_{d|n, d < n} (-1)^omega(n/d) * a(d). - Ilya Gutkovskiy, Mar 10 2021
Dirichlet g.f.: (1/zeta(s)) * Product_{p prime} (1/(1 - 2/p^s)). - Amiram Eldar, Sep 16 2023
Sum_{k=1..n} 1/a(k) = c * n + o(n), where c = Product_{p prime} (1 - 1/(p*(2*p-1))) = 0.74030830284678515949... (Jakimczuk, 2024, Theorem 2.4, p. 16). - Amiram Eldar, Mar 08 2024
From Vaclav Kotesovec, Mar 08 2024: (Start)
Dirichlet g.f.: zeta(s) * (1 + 1/(2^s*(2^s - 2))) * f(s), where f(s) = Product_{p prime, p>2} (1 + 1/(p^s*(p^s - 2))).
Sum_{k=1..n} a(k) ~ (f(1)*n / (4*log(2))) * (log(n) - 1 + gamma + 5*log(2)/2 + f'(1)/f(1)), where
f(1) = Product_{p prime, p>2} (1 + 1/(p*(p-2))) = A167864 = 1.51478012813749125771853381230067247330485921179389884042843306025133959...,
f'(1) = f(1) * Sum_{p prime, p>2} (-2*log(p)/((p-1)*(p-2))) = -2*f(1)*A347195 = -2.6035805486753944250682818932032862770113061830543948257159113584026980...
and gamma is the Euler-Mascheroni constant A001620. (End)
MAPLE
A162510 := proc(n)
local a, f;
a := 1;
for f in ifactors(n)[2] do
a := a*2^(op(2, f)-1) ;
end do:
return a;
end proc: # R. J. Mathar, May 20 2017
MATHEMATICA
a[n_] := 2^(PrimeOmega[n] - PrimeNu[n]); Array[a, 100] (* Jean-François Alcover, Apr 24 2017, after R. J. Mathar *)
PROG
(PARI) a(n)=my(f=factor(n)[, 2]); 2^(vecsum(f)-#f) \\ Charles R Greathouse IV, Nov 02 2016
(Python)
from sympy import factorint
from operator import mul
def a(n):
f=factorint(n)
return 1 if n==1 else reduce(mul, [2**(f[i] - 1) for i in f]) # Indranil Ghosh, May 20 2017
KEYWORD
easy,mult,nonn
AUTHOR
Gerard P. Michon, Jul 05 2009
STATUS
approved