A322078 a(n) = n^2 * Sum_{p|n, p prime} 1/p^2.

0, 1, 1, 4, 1, 13, 1, 16, 9, 29, 1, 52, 1, 53, 34, 64, 1, 117, 1, 116, 58, 125, 1, 208, 25, 173, 81, 212, 1, 361, 1, 256, 130, 293, 74, 468, 1, 365, 178, 464, 1, 673, 1, 500, 306, 533, 1, 832, 49, 725, 298, 692, 1, 1053, 146, 848, 370, 845, 1, 1444, 1, 965, 522

Offset: 1

Author

Daniel Suteu, Nov 25 2018

Keywords

• nonn

Comments

Generalized formula is f(n,m) = n^m * Sum_{p|n} 1/p^m, where f(n,0) = A001221(n) and f(n,1) = A069359(n).

Dirichlet convolution of A010051(n) and n^2. - Wesley Ivan Hurt, Jul 15 2025

Examples

			a(40) = 464 because the prime factors of 40 are 2 and 5, so we have 40^2 * (1/2^2 + 1/5^2) = 464.
		

Links

• Daniel Suteu, Table of n, a(n) for n = 1..10000

Crossrefs

Cf. A000040, A000330, A001221, A007434, A010051, A069359, A085541.

Sequences of the form n^k * Sum_{p|n, p prime} 1/p^k for k = 0..10: A001221 (k=0), A069359 (k=1), this sequence (k=2), A351242 (k=3), A351244 (k=4), A351245 (k=5), A351246 (k=6), A351247 (k=7), A351248 (k=8), A351249 (k=9), A351262 (k=10).

Programs

Magma

[0] cat [n^2*&+[1/p^2:p in PrimeDivisors(n)]:n in [2..70]]; // Marius A. Burtea, Oct 10 2019

Maple

a:= n-> n^2*add(1/i[1]^2, i=ifactors(n)[2]):
seq(a(n), n=1..70);  # Alois P. Heinz, Oct 11 2019

Mathematica

f[p_, e_] := 1/p^2; a[n_] := If[n==1, 0, n^2*Plus@@f@@@FactorInteger[n]]; Array[a, 60] (* Amiram Eldar, Nov 26 2018 *)

PARI

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

Formula

Sum_{k=1..n} a(k) ~ A085541 * A000330(n).

G.f.: Sum_{k>=1} x^prime(k) * (1 + x^prime(k)) / (1 - x^prime(k))^3. - Ilya Gutkovskiy, Oct 10 2019

a(n) = 1 <=> n is prime. _Alois P. Heinz, Oct 11 2019

Dirichlet g.f.: zeta(s-2)*primezeta(s). This follows because Sum_{n>=1} a(n)/n^s = Sum_{n>=1} (n^2/n^s) Sum_{p|n} 1/p^2. Since n = p*j, rewrite the sum as Sum_{p} Sum_{j>=1} 1/(p^2*(p*j)^(s-2)) = Sum_{p} 1/p^s Sum_{j>=1} 1/j^(s-2) = zeta(s-2)*primezeta(s). The result generalizes to higher powers of p. - Michael Shamos, Mar 02 2023

a(n) = Sum_{d|n} A007434(d)*A001221(n/d). - Ridouane Oudra, Jul 13 2025

From Wesley Ivan Hurt, Jul 15 2025: (Start)

a(n) = Sum_{d|n} c(d) * (n/d)^2, where c = A010051.

a(p^k) = p^(2*k-2) for p prime and k>=1. (End)