OFFSET
1,2
COMMENTS
The infinitary abundancy index of a number k is A049417(k)/k.
LINKS
Amiram Eldar, Table of n, a(n) for n = 1..10000
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).
EXAMPLE
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.
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]
PROG
(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)))); }
CROSSREFS
KEYWORD
nonn,easy,frac
AUTHOR
Amiram Eldar, Jul 20 2024
STATUS
approved