A374777 Numerator of the mean abundancy index of the divisors of n.

1, 5, 7, 17, 11, 35, 15, 49, 34, 11, 23, 119, 27, 75, 77, 129, 35, 85, 39, 187, 5, 115, 47, 343, 86, 135, 71, 85, 59, 77, 63, 107, 161, 175, 33, 289, 75, 195, 63, 539, 83, 25, 87, 391, 187, 235, 95, 301, 54, 43, 245, 153, 107, 355, 23, 105, 91, 295, 119, 1309, 123, 315

Offset: 1

Author

Amiram Eldar, Jul 19 2024

Keywords

• nonn
• easy
• frac

Comments

First differs from A318491 at n = 27.

The abundancy index of a number k is sigma(k)/k = A017665(k)/A017666(k).

Examples

			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.
		

Links

• Amiram Eldar, Table of n, a(n) for n = 1..10000

Crossrefs

Cf. A000005, A000203, A013661, A017665, A017666, A038040, A060640, A245214, A318491, A318492, A374778 (denominators), A374779, A374780, A374781.

Programs

Mathematica

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]

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)));}

Formula

Let f(n) = a(n)/A374778(n). Then:

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(n) is multiplicative with f(p^e) = ((e+1)*p^2 - (e+2)*p + p^(-e))/((e+1)*(p-1)^2).

f(n) = A318491(n)/(A318492(n)*A000005(n)).

f(n) = (Sum_{d|n} d*tau(d)) / (n*tau(n)) = A060640(n)/A038040(n).

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)).

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).

Lim sup_{n->oo} f(n) = oo (i.e., f(n) is unbounded).