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a(n) is the numerator of A346009(n)/A346010(n) - A001221(n)/2.
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%I #10 Jul 02 2021 03:48:56

%S 0,0,0,1,0,0,0,1,1,0,0,1,0,0,0,3,0,1,0,1,0,0,0,1,1,0,1,1,0,0,0,1,0,0,

%T 0,1,0,0,0,1,0,0,0,1,1,0,0,3,1,1,0,1,0,1,0,1,0,0,0,1,0,0,1,5,0,0,0,1,

%U 0,0,0,5,0,0,1,1,0,0,0,3,3,0,0,1,0,0,0

%N a(n) is the numerator of A346009(n)/A346010(n) - A001221(n)/2.

%C a(n)/A346013 is the difference between two functions, the average number of distinct prime factors of the divisors of n and half the number of distinct prime factors of n.

%C Duncan (1961) proved that these two functions have the same average order, log(log(n))/2, and that their difference has a constant average order, or an asymptotic mean, c (A346011; see the Formula section).

%C The nonzero values occur exactly at the nonsquarefree numbers (A013929) and their asymptotic mean is c/(1-6/Pi^2) = 0.2439041253...

%H Amiram Eldar, <a href="/A346012/b346012.txt">Table of n, a(n) for n = 1..10000</a>

%H R. L. Duncan, <a href="https://www.jstor.org/stable/2311587">Note on the divisors of a number</a>, The American Mathematical Monthly, Vol. 68, No. 4 (1961), pp. 356-359.

%H Sébastien Gaboury, <a href="http://hdl.handle.net/20.500.11794/18899">Sur les convolutions de fonctions arithmétiques</a>, M.Sc. thesis, Laval University, Quebec, 2007.

%H Wikipedia, <a href="https://en.wikipedia.org/wiki/Average_order_of_an_arithmetic_function">Average order of an arithmetic function</a>.

%F Let f(n) = a(n)/A346013(n) be the sequence of fractions. Then:

%F f(n) depends only on the prime signature of n: If n = Product_{i} p_i^e_i, then f(n) = Sum_{i} (e_i - 1)/(2*(e_i + 1)).

%F f(n) = 0 if and only if n is squarefree (A005117), and f(n) > 0 otherwise.

%F f(n) = (Sum_{p prime, p^2|n} d(n/p^2))/(2*d(n)), where d(n) is the number of divisors of n (A000005).

%F Asymptotic mean: lim_{n->oo} (1/n) * Sum_{k=1..n} f(n) = 0.095628... (A346011).

%e The fractions begin with 0, 0, 0, 1/6, 0, 0, 0, 1/4, 1/6, 0, 0, 1/6, ....

%t f[p_, e_] := e/(e + 1); a[1] = 0; a[n_] := Numerator[Plus @@ f @@@ (fct = FactorInteger[n]) - Length[fct]/2]; Array[a, 100]

%Y Cf. A000005, A001221, A005117, A013929, A346009, A346010, A346011, A346013 (denominators).

%K nonn,frac

%O 1,16

%A _Amiram Eldar_, Jul 01 2021