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%I #19 Oct 05 2023 04:07:51
%S 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,
%T 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,
%U 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
%N Sum of the exponents in the prime factorizations of the prime power divisors of n.
%C 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.
%H Amiram Eldar, <a href="/A351397/b351397.txt">Table of n, a(n) for n = 1..10000</a>
%F a(n) = Sum_{d|n} Omega(d) * [omega(d) = 1].
%F Additive with a(p^e) = e*(e+1)/2. - _Amiram Eldar_, Feb 10 2022
%F 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
%e 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.
%e 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.
%t f[p_, e_] := e*(e + 1)/2; a[1] = 0; a[n_] := Plus @@ f @@@ FactorInteger[n]; Array[a, 100] (* _Amiram Eldar_, Feb 10 2022 *)
%o (PARI) a(n) = sumdiv(n, d, my(x); if (x=isprimepower(d), x)); \\ _Michel Marcus_, Feb 10 2022
%Y Cf. A001221 (omega), A001222 (Omega), A077761, A246655.
%K nonn,easy
%O 1,4
%A _Wesley Ivan Hurt_, Feb 09 2022
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