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%I #18 Sep 29 2023 04:05:38
%S 0,1,1,2,1,2,1,2,2,2,1,3,1,2,2,3,1,3,1,3,2,2,1,3,2,2,2,3,1,3,1,2,2,2,
%T 2,4,1,2,2,3,1,3,1,3,3,2,1,4,2,3,2,3,1,3,2,3,2,2,1,4,1,2,3,4,2,3,1,3,
%U 2,3,1,4,1,2,3,3,2,3,1,4,3,2,1,4,2,2,2,3,1,4,2,3,2,2,2,3,1,3,3,4,1,3,1,3,3
%N a(n) is the number of prime powers (not including 1) that are (1+e)-divisors of n.
%C (1+e)-divisors are defined in A049599.
%C First differs from A106490 at n = 64.
%C The total number of prime powers (not including 1) that divide n is A001222(n).
%C If p|n and p^e is the highest power of p that divides n, then the powers of p that are (1+e)-divisors of n are of the form p^d where d|e.
%H Amiram Eldar, <a href="/A349281/b349281.txt">Table of n, a(n) for n = 1..10000</a>
%F Additive with a(p^e) = A000005(e).
%F a(n) <= A001222(n), with equality if and only if n is cubefree (A046099).
%F a(n) <= A049599(n)-1, with equality if and only if n is a prime power (including 1, A000961).
%F Sum_{k=1..n} a(n) ~ n * (log(log(n)) + B + C), where B is Mertens's constant (A077761) and C = Sum_{p prime} f(1/p) = 0.51780076119050171903..., where f(x) = -x + (1-x) * Sum_{k>=1} x^k/(1-x^k). - _Amiram Eldar_, Sep 29 2023
%e 8 has 3 (1+e)-divisors, 1, 2 and 8. Two of these divisors, 2 and 8 = 2^3 are prime powers. Therefore, a(8) = 2.
%t f[p_,e_] := DivisorSigma[0, e]; a[1] = 0; a[n_] := Plus @@ f @@@ FactorInteger[n]; Array[a,100]
%o (PARI) A349281(n) = vecsum(apply(e->numdiv(e),factor(n)[,2])); \\ _Antti Karttunen_, Nov 13 2021
%Y Cf. A000005, A000961, A001222, A046099, A049599, A077761, A106490, A246655, A349258.
%K nonn,easy
%O 1,4
%A _Amiram Eldar_, Nov 13 2021
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