A374783 Numerator of the mean unitary abundancy index of the unitary divisors of n.
1, 5, 7, 9, 11, 35, 15, 17, 19, 11, 23, 21, 27, 75, 77, 33, 35, 95, 39, 99, 5, 115, 47, 119, 51, 135, 55, 135, 59, 77, 63, 65, 161, 175, 33, 19, 75, 195, 63, 187, 83, 25, 87, 207, 209, 235, 95, 77, 99, 51, 245, 243, 107, 275, 23, 255, 91, 295, 119, 231, 123, 315
Offset: 1
Examples
For n = 4, 4 has 2 unitary divisors, 1 and 4. Their unitary abundancy indices are usigma(1)/1 = 1 and usigma(4)/4 = 5/4, and their mean unitary 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_] := 1 + 1/(2*p^e); a[1] = 1; a[n_] := Numerator[Times @@ f @@@ FactorInteger[n]]; Array[a, 100]
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PARI
a(n) = {my(f = factor(n)); numerator(prod(i = 1, #f~, 1 + 1/(2*f[i,1]^f[i,2])));}
Formula
Let f(n) = a(n)/A374784(n). Then:
f(n) = (Sum_{d|n, gcd(d, n/d) = 1} usigma(d)/d) / ud(n), where usigma(n) is the sum of unitary divisors of n (A034448), and ud(n) is their number (A034444).
f(n) is multiplicative with f(p^e) = 1 + 1/(2*p^e).
Dirichlet g.f. of f(n): zeta(s) * zeta(s+1) * Product_{p prime} (1 - 1/(2*p^(s+1)) - 1/(2*p^(2*s+1))).
Asymptotic mean: Limit_{m->oo} (1/m) * Sum_{k=1..m} f(k) = Product_{p prime} (1 + 1/(2*p*(p+1))) = 1.17443669198552182119... . For comparison, the asymptotic mean of the unitary abundancy index over all the positive integers is zeta(2)/zeta(3) = 1.368432... (A306633).
Lim sup_{n->oo} f(n) = oo (i.e., f(n) is unbounded).
Comments