%I #9 Jul 20 2024 03:40:46
%S 1,5,7,17,11,35,15,49,34,11,23,119,27,75,77,129,35,85,39,187,5,115,47,
%T 343,86,135,71,85,59,77,63,107,161,175,33,289,75,195,63,539,83,25,87,
%U 391,187,235,95,301,54,43,245,153,107,355,23,105,91,295,119,1309,123,315
%N Numerator of the mean abundancy index of the divisors of n.
%C First differs from A318491 at n = 27.
%C The abundancy index of a number k is sigma(k)/k = A017665(k)/A017666(k).
%H Amiram Eldar, <a href="/A374777/b374777.txt">Table of n, a(n) for n = 1..10000</a>
%F Let f(n) = a(n)/A374778(n). Then:
%F f(n) = (Sum_{d|n} sigma(d)/d) / tau(n), where sigma(n) is the sum of divisors of n (A000203), and tau(n) is their number (A000005).
%F f(n) is multiplicative with f(p^e) = ((e+1)*p^2 - (e+2)*p + p^(-e))/((e+1)*(p-1)^2).
%F f(n) = A318491(n)/(A318492(n)*A000005(n)).
%F f(n) = (Sum_{d|n} d*tau(d)) / (n*tau(n)) = A060640(n)/A038040(n).
%F Dirichlet g.f. of f(n): zeta(s) * Product_{p prime} ((p/(p-1)^2) * ((p^s-1)*log((1-1/p^s)/(1-1/p^(s+1))) + p-1)).
%F Asymptotic mean: Limit_{m->oo} (1/m) * Sum_{k=1..m} f(k) = Product_{p prime} ((p/(p-1)) * (1 - log(1 + 1/p))) = 1.3334768464... . For comparison, the asymptotic mean of the abundancy index over all the positive integers is zeta(2) = 1.644934... (A013661).
%F Lim sup_{n->oo} f(n) = oo (i.e., f(n) is unbounded).
%e For n = 2, n has 2 divisors, 1 and 2. Their abundancy indices are sigma(1)/1 = 1 and sigma(2)/2 = 3/2, and their mean abundancy index is (1 + 3/2)/2 = 5/4. Therefore a(2) = numerator(5/4) = 5.
%t f[p_, e_] := ((e+1)*p^2 - (e+2)*p + p^(-e))/((e+1)*(p-1)^2); a[1] = 1; a[n_] := Numerator[Times @@ f @@@ FactorInteger[n]]; Array[a, 100]
%o (PARI) a(n) = {my(f = factor(n), p, e); numerator(prod(i = 1, #f~, p = f[i, 1]; e = f[i, 2]; ((e+1)*p^2 - (e+2)*p + p^(-e))/((e+1)*(p-1)^2)));}
%Y Cf. A000005, A000203, A013661, A017665, A017666, A038040, A060640, A245214, A318491, A318492, A374778 (denominators), A374779, A374780, A374781.
%K nonn,easy,frac
%O 1,2
%A _Amiram Eldar_, Jul 19 2024