A374786 Numerator of the mean infinitary abundancy index of the infinitary divisors of n.
1, 5, 7, 9, 11, 35, 15, 45, 19, 11, 23, 21, 27, 75, 77, 33, 35, 95, 39, 99, 5, 115, 47, 105, 51, 135, 133, 135, 59, 77, 63, 165, 161, 175, 33, 19, 75, 195, 63, 99, 83, 25, 87, 207, 209, 235, 95, 77, 99, 51, 245, 243, 107, 665, 23, 675, 91, 295, 119, 231, 123, 315
Offset: 1
Examples
For n = 4, 4 has 2 infinitary divisors, 1 and 4. Their infinitary abundancy indices are isigma(1)/1 = 1 and isigma(4)/4 = 5/4, and their mean infinitary abundancy index is (1 + 5/4)/2 = 9/8. Therefore a(4) = numerator(9/8) = 9.
Links
- Amiram Eldar, Table of n, a(n) for n = 1..10000
Crossrefs
Programs
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Mathematica
f[p_, e_] := p^(2^(-1 + Position[Reverse@IntegerDigits[e, 2], ?(# == 1 &)])); a[1] = 1; a[n] := Numerator[Times @@ (1 + 1/(2*Flatten@ (f @@@ FactorInteger[n])))]; Array[a, 100]
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PARI
a(n) = {my(f = factor(n), b); numerator(prod(i = 1, #f~, b = binary(f[i, 2]); prod(k=1, #b, if(b[k], 1 + 1/(2*f[i, 1]^(2^(#b-k))), 1))));}
Formula
Let f(n) = a(n)/A374787(n). Then:
f(n) = (Sum_{d infinitary divisor of n} isigma(d)/d) / id(n), where isigma(n) is the sum of infinitary divisors of n (A049417), and id(n) is their number (A037445).
f(n) is multiplicative with f(p^e) = Product{k>=1, e_k=1} (1 + 1/(2*p^(2^(k+1)))), where e = Sum_{k} e_k * 2^k is the binary representation of e, i.e., e_k is bit k of e.
f(n) = (Sum_{d infinitary divisor of n} d*id(d)) / (n*id(n)).
Asymptotic mean: Limit_{m->oo} (1/m) * Sum_{k=1..m} f(k) = Product_{P} (1 + 1/(2*P*(P+1))) = 1.21407233718434377029..., where P are numbers of the form p^(2^k) where p is prime and k >= 0 (A050376). For comparison, the asymptotic mean of the infinitary abundancy index over all the positive integers is 1.461436... = 2 * A327574.
Lim sup_{n->oo} f(n) = oo (i.e., f(n) is unbounded).
Comments