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A374786 Numerator of the mean infinitary abundancy index of the infinitary divisors of n.

%I A374786 #17 Jul 20 2024 15:12:20
%S A374786 1,5,7,9,11,35,15,45,19,11,23,21,27,75,77,33,35,95,39,99,5,115,47,105,
%T A374786 51,135,133,135,59,77,63,165,161,175,33,19,75,195,63,99,83,25,87,207,
%U A374786 209,235,95,77,99,51,245,243,107,665,23,675,91,295,119,231,123,315
%N A374786 Numerator of the mean infinitary abundancy index of the infinitary divisors of n.
%C A374786 The infinitary abundancy index of a number k is A049417(k)/k.
%C A374786 The record values of a(n)/A374787(n) are attained at the terms of A037992.
%C A374786 The least number k such that a(k)/A374787(k) is larger than 2, 3, 4, ..., is A037992(6) = 7560, A037992(33) = 1370819010042780920891599455129161859473627856000, ... .
%H A374786 Amiram Eldar, <a href="/A374786/b374786.txt">Table of n, a(n) for n = 1..10000</a>
%F A374786 Let f(n) = a(n)/A374787(n). Then:
%F A374786 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 A374786 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 A374786 f(n) = (Sum_{d infinitary divisor of n} d*id(d)) / (n*id(n)).
%F A374786 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.
%F A374786 Lim sup_{n->oo} f(n) = oo (i.e., f(n) is unbounded).
%F A374786 f(n) <= A374777(n)/A374778(n) with equality if and only if n is squarefree (A005117).
%F A374786 f(n) >= A374783(n)/A374784(n) with equality if and only if n is in A138302.
%e A374786 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.
%t A374786 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]
%o A374786 (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))));}
%Y A374786 Cf. A005117, A037445, A037992, A049417, A050376, A077609, A138302, A327574, A374787 (denominators).
%Y A374786 Similar sequences: A374777/A374778, A374783/A374784.
%K A374786 nonn,easy,frac
%O A374786 1,2
%A A374786 _Amiram Eldar_, Jul 20 2024