A319273 -id:A319273 - cp's OEIS frontend

A332423 If n = Product (p_j^k_j) then a(n) = Sum ((-1)^(k_j + 1) * k_j).

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

Ilya Gutkovskiy, Feb 12 2020

Sum of odd exponents in prime factorization of n minus the sum of even exponents in prime factorization of n.

			a(2700) = a(2^2 * 3^3 * 5^2) = -2 + 3 - 2 = -1.

Cf. A001222, A037916, A067255, A071321, A077761, A195017, A316523, A319273, A332422, A350386, A350387.

a[n_] := Plus @@ ((-1)^(#[[2]] + 1) #[[2]] & /@ FactorInteger[n]); a[1] = 0; Table[a[n], {n, 1, 80}]

a(n) = vecsum(apply(x -> (-1)^(x+1) * x, factor(n)[, 2])); \\ Amiram Eldar, Oct 09 2023

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)

A326440 a(n) = 1 - tau(1) + tau(2) - tau(3) + ... + (-1)^n tau(n), where tau = A000005 is number of divisors.

Original entry on oeis.org

1, 0, 2, 0, 3, 1, 5, 3, 7, 4, 8, 6, 12, 10, 14, 10, 15, 13, 19, 17, 23, 19, 23, 21, 29, 26, 30, 26, 32, 30, 38, 36, 42, 38, 42, 38, 47, 45, 49, 45, 53, 51, 59, 57, 63, 57, 61, 59, 69, 66, 72, 68, 74, 72, 80, 76, 84, 80, 84, 82, 94, 92, 96, 90, 97, 93, 101, 99

Offset: 0

Gus Wiseman, Jul 06 2019

nonn

Is this sequence nonnegative?

As tau(n) is odd when n is a square, there are alternating strings of even and odd integers with change of parity for each n square. Indeed, between m^2 and (m+1)^2-1, there is a string of 2m+1 even terms if m is odd, or a string of 2m+1 odd terms if m is even. - Bernard Schott, Jul 10 2019

			The first 6 terms of A000005 are 1, 2, 2, 3, 2, 4, so a(6) = 1 - 1 + 2 - 2 + 3 - 2 + 4 = 5.

Michel Marcus, Table of n, a(n) for n = 0..5000

Cf. A000005, A001222, A008683, A054519, A071321, A195017, A268387, A307704, A316523, A316524, A319273.

[1] cat [1+(&+[(-1)^(k)*#Divisors(k):k in [1..n]]):n in [1..70]]; // Marius A. Burtea, Jul 10 2019

Accumulate[Table[If[k==0,1,(-1)^k*DivisorSigma[0,k]],{k,0,30}]]

a(n) = 1 - sum(k=1, n, (-1)^(k+1)*numdiv(k)); \\ Michel Marcus, Jul 09 2019

a(n) = 1 + Sum_{k=1..n} (-1)^k A000005(k).
For n > 0, a(n) = 1 + A307704(n).
If p prime, a(p) = a(p-1) - 2. - Bernard Schott, Jul 10 2019

cp's OEIS Frontend

A332423 If n = Product (p_j^k_j) then a(n) = Sum ((-1)^(k_j + 1) * k_j).

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