A365479 The sum of unitary divisors of the smallest square divisible by n.

1, 5, 10, 5, 26, 50, 50, 17, 10, 130, 122, 50, 170, 250, 260, 17, 290, 50, 362, 130, 500, 610, 530, 170, 26, 850, 82, 250, 842, 1300, 962, 65, 1220, 1450, 1300, 50, 1370, 1810, 1700, 442, 1682, 2500, 1850, 610, 260, 2650, 2210, 170, 50, 130, 2900, 850, 2810, 410

Offset: 1

Author

Amiram Eldar, Sep 05 2023

Keywords

• nonn
• easy
• mult

Comments

The number of unitary divisors of the smallest square divisible by n is the same as the number of unitary divisors of n, A034444(n).

Links

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

Crossrefs

Cf. A002117, A034444, A034448, A053143, A365346, A365480, A365481.

Programs

Mathematica

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

PARI

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

Python

from math import prod
from sympy import factorint
def A365479(n): return prod(p**(e+(e&1))+1 for p,e in factorint(n).items()) # Chai Wah Wu, Sep 05 2023

Formula

a(n) = A034448(A053143(n)).

Multiplicative with a(p^e) = p^(e + (e mod 2)) + 1.

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

Sum_{k=1..n} a(k) ~ c * n^3, where c = (Pi^2/45) * zeta(3) * Product_{p prime} (1 - 1/p^4 + 1/p^5 - 1/p^6) = 0.248414056414... .