A384554 -id:A384554 - cp's OEIS frontend

A384555 The largest infinitary divisor of n that is cubefree.

1, 2, 3, 4, 5, 6, 7, 4, 9, 10, 11, 12, 13, 14, 15, 1, 17, 18, 19, 20, 21, 22, 23, 12, 25, 26, 9, 28, 29, 30, 31, 2, 33, 34, 35, 36, 37, 38, 39, 20, 41, 42, 43, 44, 45, 46, 47, 3, 49, 50, 51, 52, 53, 18, 55, 28, 57, 58, 59, 60, 61, 62, 63, 4, 65, 66, 67, 68, 69

Offset: 1

Amiram Eldar, Jun 03 2025

The number of these divisors is A368883(n), and their sum is A384554(n).

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

Cf. A004709, A007948, A013666, A077609, A368883, A384554.

f[p_, e_] := Switch[Mod[e, 4], 0, 1, 1, p, 2, p^2, 3, p^2]; a[1] = 1; a[n_] := Times @@ f @@@ FactorInteger[n]; Array[a, 100]

a(n) = {my(f = factor(n)); prod(i = 1, #f~, p = f[i,1]; e = f[i,2]; [1, p, p^2, p^2][e%4+1]);}

Multiplicative with a(p^e) = 1 if e == 0 (mod 4), p if e == 1 (mod 4), p^2 if e == 2 or 3 (mod 4).
a(n) = n if and only if n is cubefree (A004709).
Dirichlet g.f.: zeta(4*s) * Product_{p prime} (1 + 1/p^(s-1) + 1/p^(2*s-2) + 1/p^(3*s-2)).
Sum_{k=1..n} a(k) = c * n^2 / 2, where c = zeta(8) * Product_{p prime} (1 - 1/p^3 + 1/p^4 - 1/p^5) = 0.87406992849637563411... .

A384556 The sum of the exponential divisors of n that are cubefree.

Original entry on oeis.org

1, 2, 3, 6, 5, 6, 7, 2, 12, 10, 11, 18, 13, 14, 15, 6, 17, 24, 19, 30, 21, 22, 23, 6, 30, 26, 3, 42, 29, 30, 31, 2, 33, 34, 35, 72, 37, 38, 39, 10, 41, 42, 43, 66, 60, 46, 47, 18, 56, 60, 51, 78, 53, 6, 55, 14, 57, 58, 59, 90, 61, 62, 84, 6, 65, 66, 67, 102, 69

Offset: 1

Amiram Eldar, Jun 03 2025

The number of these divisors is A056624(n), and the largest of them is A066990(n).

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

Cf. A004709, A013662, A051377, A056624, A066990, A073185, A322791, A384554.

f[p_, e_] := If[OddQ[e], p, p + p^2]; 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), f[i,1]^2));}

from math import prod
from sympy import factorint
def A384556(n): return prod(p*(1+p*(e&1^1)) for p, e in factorint(n).items()) # Chai Wah Wu, Jun 03 2025

Multiplicative with a(p^e) = p if e is odd, and p+p^2 is e is even.
a(n) <= A051377(n), with equality if and only if n is cubefree (A004709).
Dirichlet g.f.: zeta(2*s) * Product_{p prime} (1 + 1/p^(s-1) + 1/p^(2*s-2) + 1/p^(2*s-1) - 1/p^(2*s)).
Sum_{k=1..n} a(k) ~ c * n^2 / 2, where c = zeta(4) * Product_{p prime} (1 - 2/p^4 + 1/p^5) = 0.95692470821076622881...

cp's OEIS Frontend

A384555 The largest infinitary divisor of n that is cubefree.

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