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A374777
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Numerator of the mean abundancy index of the divisors of n.
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9
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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
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OFFSET
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1,2
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COMMENTS
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First differs from A318491 at n = 27.
The abundancy index of a number k is sigma(k)/k = A017665(k)/A017666(k).
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LINKS
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FORMULA
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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).
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).
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EXAMPLE
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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.
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MATHEMATICA
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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]
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PROG
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(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))); }
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CROSSREFS
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Cf. A000005, A000203, A013661, A017665, A017666, A038040, A060640, A245214, A318491, A318492, A374778 (denominators), A374779, A374780, A374781.
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KEYWORD
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nonn,easy,frac
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AUTHOR
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STATUS
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approved
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