A374539 The sum of the squares of the infinitary divisors of n.

1, 5, 10, 17, 26, 50, 50, 85, 82, 130, 122, 170, 170, 250, 260, 257, 290, 410, 362, 442, 500, 610, 530, 850, 626, 850, 820, 850, 842, 1300, 962, 1285, 1220, 1450, 1300, 1394, 1370, 1810, 1700, 2210, 1682, 2500, 1850, 2074, 2132, 2650, 2210, 2570, 2402, 3130, 2900

Offset: 1

Author

Amiram Eldar, Jul 11 2024

Keywords

• nonn
• easy
• mult

Comments

Also the sum of the infinitary divisors of n^2.

Links

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

Crossrefs

Cf. A049417, A050376, A077609.

Similar sequences: A001157, A034676, A050999, A067558, A076577, A351265, A351307, A351647.

Programs

Mathematica

f[p_, e_] := p^(2^Position[Reverse@IntegerDigits[e, 2], ?(# == 1 &)]); a[1] = 1; a[n] := Times @@ (Flatten@(f @@@ FactorInteger[n]) + 1); Array[a, 100]

PARI

a(n) = {my(f = factor(n), b); prod(i = 1, #f~, b = binary(2 * f[i, 2]); prod(k=1, #b, if(b[k], 1+f[i, 1]^(2^(#b-k)), 1)));}

Python

from math import prod
from sympy import factorint
def A374539(n): return prod(p**(1<Chai Wah Wu, Jul 11 2024

Formula

a(n) = A049417(n^2).

a(n) <= A001157(n), with equality if and only if n is in A036537.

Multiplicative with a(p^e) = Product{k>=1, e_k=1} (p^(2^(k+1)) + 1), where e = Sum_{k} e_k * 2^k is the binary representation of e, i.e., e_k is bit k of e.

Sum_{k=1..n} a(k) ~ c * n^3 / 3, where c = Product_{P} (1 + 1/(P^2*(P+1))) = 1.14142906130350119631..., and P are numbers of the form p^(2^k) where p is prime and k >= 0 (A050376).