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

1,2

Comments

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

Links

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

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).

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]

Prog

(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<<i)+1 for p, e in factorint(n).items() for i, j in enumerate(bin(e)[-1:1:-1], 1) if j=='1') # Chai Wah Wu, Jul 11 2024

Crossrefs

Cf. A049417, A050376, A077609.

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

Sequence in context: A098749 A034676 A076598 * A306011 A080341 A271328

Adjacent sequences: A374536 A374537 A374538 * A374540 A374541 A374542

Keyword

nonn,easy,mult

Author

Amiram Eldar, Jul 11 2024

Status

approved