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

1,4

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)

Links

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

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

Example

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.

Prog

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

Crossrefs

Cf. A001222, A095112, A286229, A007434.

Sequence in context: A393210 A094049 A286254 * A286457 A327766 A147348

Adjacent sequences: A322661 A322662 A322663 * A322665 A322666 A322667

Keyword

nonn

Author

Daniel Suteu, Dec 22 2018

Status

approved