A369759 -id:A369759 - cp's OEIS frontend

A369757 The number of divisors of the smallest cubefull exponentially odd number that is divisible by n.

1, 4, 4, 4, 4, 16, 4, 4, 4, 16, 4, 16, 4, 16, 16, 6, 4, 16, 4, 16, 16, 16, 4, 16, 4, 16, 4, 16, 4, 64, 4, 6, 16, 16, 16, 16, 4, 16, 16, 16, 4, 64, 4, 16, 16, 16, 4, 24, 4, 16, 16, 16, 4, 16, 16, 16, 16, 16, 4, 64, 4, 16, 16, 8, 16, 64, 4, 16, 16, 64, 4, 16, 4

Offset: 1

Amiram Eldar, Jan 31 2024

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

Cf. A000005, A335988, A356192, A365348, A369758, A369759.

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

a(n) = vecprod(apply(x -> if(x%2, max(x, 3) + 1, x + 2), factor(n)[, 2]));

a(n) = A000005(A356192(n)).
Multiplicative with a(p^e) = max(e,3) + 1 if e is odd, and e+2 if e is even.
a(n) >= A000005(n), with equality if and only if n is cubefull exponentially odd number (A335988).
Dirichlet g.f.: zeta(s) * zeta(2*s) * Product_{p prime} (1 + 3/p^s - 1/p^(2*s) - 3/p^(3*s) + 2/p^(4*s)).
From Vaclav Kotesovec, Feb 02 2024: (Start)
Dirichlet g.f.: zeta(s)^4 * Product_{p prime} (1 + (7*p^(2*s) + 2*p^(3*s) - 6*p^(4*s) - 7*p^s + 2) / ((p^s+1)*p^(5*s))).
Sum_{k=1..n} a(k) = c * n*log(n)^3/6 + O(n*log(n)^2), where c = Product_{p prime} (1 - (6*p^4 - 2*p^3 - 7*p^2 + 7*p - 2) / ((p+1)*p^5)) = 0.124604542136592401049820049658828040278... (End)

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

Original entry on oeis.org

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

Amiram Eldar, Jan 31 2024

First differs from A369720 at n = 16.

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

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

Cf. A013662, A013664.

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]

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));}

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

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

cp's OEIS Frontend

A369757 The number of divisors of the smallest cubefull exponentially odd number that is divisible by n.

Views

Author

Keywords

Links

Crossrefs

Programs

Mathematica

PARI

Formula