A322664 a(n) = n^2 * Sum_{p^k|n} Sum_{j=1..k} 1/p^(2*j), where p are primes dividing n with multiplicity k.

0, 1, 1, 5, 1, 13, 1, 21, 10, 29, 1, 61, 1, 53, 34, 85, 1, 121, 1, 141, 58, 125, 1, 253, 26, 173, 91, 261, 1, 361, 1, 341, 130, 293, 74, 565, 1, 365, 178, 589, 1, 673, 1, 621, 331, 533, 1, 1021, 50, 729, 298, 861, 1, 1093, 146, 1093, 370, 845, 1, 1669, 1, 965

Offset: 1

Author

Daniel Suteu, Dec 22 2018

Keywords

• nonn

Comments

The generalized formula is f(n,m) = n^m * Sum_{p^k|n} Sum_{j=1..k} 1/p^(m*j), where f(n,0) = A001222(n) and f(n,1) = A095112(n).

From Ridouane Oudra, Jul 21 2025: (Start)

a(n) is the sum of (n/d)^2 over all prime powers d which divide n.

Using the previous generalized formula we have :

f(n,m) = Sum_{d|n, d is a prime power} (n/d)^m.

f(n,m) = Sum_{d|n} bigomega(d)*J_m(n/d), where J_m is the m-th Jordan totient function. (End)

Examples

			The prime factorization of 24 is 2^3 * 3, so a(24) = 24^2 * (1/2^2 + 1/2^(2*2) + 1/2^(2*3) + 1/3^2) = 253.
		

Links

• Antti Karttunen, Table of n, a(n) for n = 1..20000

Crossrefs

Cf. A001222, A095112, A286229, A007434.

Programs

PARI

a(n) = my(f=factor(n)); sum(k=1, #f~, sum(j=1, f[k,2], n^2 / f[k,1]^(2*j)));

Formula

Sum_{k=1..n} a(k) ~ A286229 * A000330(n).

a(n) = Sum_{d|n} bigomega(d)*J_2(n/d), where J_2 = A007434. - Ridouane Oudra, Jul 21 2025