A369758 The sum of divisors of the smallest cubefull exponentially odd number that is divisible by n.

1, 15, 40, 15, 156, 600, 400, 15, 40, 2340, 1464, 600, 2380, 6000, 6240, 63, 5220, 600, 7240, 2340, 16000, 21960, 12720, 600, 156, 35700, 40, 6000, 25260, 93600, 30784, 63, 58560, 78300, 62400, 600, 52060, 108600, 95200, 2340, 70644, 240000, 81400, 21960, 6240

Offset: 1

Author

Amiram Eldar, Jan 31 2024

Keywords

• nonn
• easy
• mult

Comments

First differs from A369720 at n = 16.

Links

• Amiram Eldar, Table of n, a(n) for n = 1..10000

Crossrefs

Cf. A000203, A335988, A356192, A365349, A369720, A369757, A369759.

Cf. A013662, A013664.

Programs

Mathematica

f[p_, e_] := (p^If[OddQ[e], Max[e, 3] + 1, e + 2] - 1)/(p-1); a[1] = 1; a[n_] := Times @@ f @@@ FactorInteger[n]; Array[a, 100]

PARI

a(n) = {my(f = factor(n)); prod(i = 1, #f~, (f[i,1]^if(f[i,2]%2, max(f[i,2], 3) + 1, f[i,2] + 2) - 1)/(f[i,1] - 1));}

Python

from math import prod
from sympy import factorint
def A369758(n): return prod((p**((3 if e==1 else e)+1+(e&1^1))-1)//(p-1) for p,e in factorint(n).items()) # Chai Wah Wu, Feb 03 2024

Formula

a(n) = A000203(A356192(n)).

Multiplicative with a(p) = p^3 + p^2 + p + 1, a(p^e) = (p^(e+1)-1)/(p-1) for an odd e >= 3, and a(p^e) = (p^(e+2)-1)/(p-1) for an even e.

a(n) >= A000203(n), with equality if and only if n is cubefull exponentially odd number (A335988).

Dirichlet g.f.: zeta(s) * zeta(2*s-2) * Product_{p prime} (1 + 1/p^(s-3) + 1/p^(s-2) + 1/p^(s-1) - 1/p^(2*s-2) - 1/p^(3*s-5) - 1/p^(3*s-4) - 1/p^(3*s-3) + 1/p^(4*s-5) + 1/p^(4*s-4)).

Sum_{k=1..n} a(k) ~ c * n^4 / 4, where c = zeta(4) * zeta(6) * Product_{p prime} (1 - 1/p^4 - 1/p^6 + 1/p^10 + 1/p^11 - 1/p^13) = 1.00040193512214077945... .

Equivalently, c = Product_{p prime} (1 + 1/(p^3*(p^4 - 1)*(p^4 + p^2 + 1))). - Vaclav Kotesovec, Feb 02 2024