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A332423
If n = Product (p_j^k_j) then a(n) = Sum ((-1)^(k_j + 1) * k_j).
2
0, 1, 1, -2, 1, 2, 1, 3, -2, 2, 1, -1, 1, 2, 2, -4, 1, -1, 1, -1, 2, 2, 1, 4, -2, 2, 3, -1, 1, 3, 1, 5, 2, 2, 2, -4, 1, 2, 2, 4, 1, 3, 1, -1, -1, 2, 1, -3, -2, -1, 2, -1, 1, 4, 2, 4, 2, 2, 1, 0, 1, 2, -1, -6, 2, 3, 1, -1, 2, 3, 1, 1, 1, 2, -1, -1, 2, 3, 1, -3
OFFSET
1,4
COMMENTS
Sum of odd exponents in prime factorization of n minus the sum of even exponents in prime factorization of n.
FORMULA
From Amiram Eldar, Oct 09 2023: (Start)
Additive with a(p^e) = (-1)^(e+1) * e.
a(n) = A350387(n) - A350386(n).
Sum_{k=1..n} a(k) ~ n * (log(log(n)) + B - C), where B is Mertens's constant (A077761) and C = Sum_{p prime} (3*p+1)/(p*(p+1)^2) = 0.81918453457738985491 ... . (End)
EXAMPLE
a(2700) = a(2^2 * 3^3 * 5^2) = -2 + 3 - 2 = -1.
MATHEMATICA
a[n_] := Plus @@ ((-1)^(#[[2]] + 1) #[[2]] & /@ FactorInteger[n]); a[1] = 0; Table[a[n], {n, 1, 80}]
PROG
(PARI) a(n) = vecsum(apply(x -> (-1)^(x+1) * x, factor(n)[, 2])); \\ Amiram Eldar, Oct 09 2023
KEYWORD
sign,easy
AUTHOR
Ilya Gutkovskiy, Feb 12 2020
STATUS
approved