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A351397
Sum of the exponents in the prime factorizations of the prime power divisors of n.
1
0, 1, 1, 3, 1, 2, 1, 6, 3, 2, 1, 4, 1, 2, 2, 10, 1, 4, 1, 4, 2, 2, 1, 7, 3, 2, 6, 4, 1, 3, 1, 15, 2, 2, 2, 6, 1, 2, 2, 7, 1, 3, 1, 4, 4, 2, 1, 11, 3, 4, 2, 4, 1, 7, 2, 7, 2, 2, 1, 5, 1, 2, 4, 21, 2, 3, 1, 4, 2, 3, 1, 9, 1, 2, 4, 4, 2, 3, 1, 11, 10, 2, 1, 5, 2, 2, 2, 7, 1, 5, 2, 4, 2
OFFSET
1,4
COMMENTS
a(n) is the sum of all the k's in the divisors of n of the form p^k, where p is prime and k>=1.
LINKS
FORMULA
a(n) = Sum_{d|n} Omega(d) * [omega(d) = 1].
Additive with a(p^e) = e*(e+1)/2. - Amiram Eldar, Feb 10 2022
Sum_{k=1..n} a(k) ~ n * (log(log(n)) + B + C), where B is Mertens's constant (A077761) and C = Sum_{k>=2} (k * P(k)) = 2.14822166379843041578..., where P(s) is the prime zeta function. - Amiram Eldar, Oct 05 2023
From Ridouane Oudra, Aug 16 2024: (Start)
a(n) = (A001222(n) + A090885(n))/2 ;
a(n) = Sum_{d|n} A100995(d). (End)
EXAMPLE
a(8) = 6; The prime power divisors of 8 are 2,4,8 with prime factorizations 2^1,2^2,2^3 and the sum of the exponents in their prime factorizations is 1+2+3 = 6.
a(20) = 4; The prime power divisors of 20 are 2,4,5 with prime factorizations 2^1,2^2,5^1 and the sum of the exponents in each of their prime factorizations is 1+2+1 = 5.
MATHEMATICA
f[p_, e_] := e*(e + 1)/2; a[1] = 0; a[n_] := Plus @@ f @@@ FactorInteger[n]; Array[a, 100] (* Amiram Eldar, Feb 10 2022 *)
PROG
(PARI) a(n) = sumdiv(n, d, my(x); if (x=isprimepower(d), x)); \\ Michel Marcus, Feb 10 2022
CROSSREFS
Cf. A001221 (omega), A001222 (Omega), A077761, A246655.
Sequence in context: A010123 A039620 A008296 * A140185 A229341 A372245
KEYWORD
nonn,easy
AUTHOR
Wesley Ivan Hurt, Feb 09 2022
STATUS
approved