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

Author

Amiram Eldar, Jul 20 2024

Keywords

• nonn
• easy
• frac

Comments

The unitary abundancy index of a number k is A034448(k)/k = A332882(k)/A332883(k).

The record values of a(n)/A374784(n) are attained at the primorial numbers (A002110).

The least number k such that a(k)/A374784(k) is larger than 2, 3, 4, ..., is A002110(9) = 223092870, A002110(314) = 7.488... * 10^878, A002110(65599) = 5.373... * 10^356774, ... .

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

Cf. A002110, A005117, A034444, A034448, A077610, A306633, A332882, A332883, A343525, A374784 (denominators).

Similar sequences: A374777/A374778, A374786/A374787.

Programs

Mathematica

f[p_, e_] := 1 + 1/(2*p^e); a[1] = 1; a[n_] := Numerator[Times @@ f @@@ FactorInteger[n]]; Array[a, 100]

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).

f(n) = (Sum_{d|n, gcd(d, n/d) = 1} d*ud(d))/(n*ud(n)) = A343525(n)/(n*A034444(n)).

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).

f(n) <= A374777(n)/A374778(n) with equality if and only if n is squarefree (A005117).