A351242 a(n) = n^3 * Sum_{p|n, p prime} 1/p^3.

0, 1, 1, 8, 1, 35, 1, 64, 27, 133, 1, 280, 1, 351, 152, 512, 1, 945, 1, 1064, 370, 1339, 1, 2240, 125, 2205, 729, 2808, 1, 4591, 1, 4096, 1358, 4921, 468, 7560, 1, 6867, 2224, 8512, 1, 12221, 1, 10712, 4104, 12175, 1, 17920, 343, 16625, 4940, 17640, 1, 25515, 1456, 22464

Offset: 1

Author

Wesley Ivan Hurt, Feb 05 2022

Keywords

• nonn
• easy

Comments

Dirichlet convolution of A010051(n) and n^3. - Wesley Ivan Hurt, Jul 15 2025

Examples

			a(6) = 35; a(6) = 6^3 * Sum_{p|6, p prime} 1/p^3 = 216 * (1/2^3 + 1/3^3) = 35.
		

Links

• Seiichi Manyama, Table of n, a(n) for n = 1..10000

Crossrefs

Sequences of the form n^k * Sum_{p|n, p prime} 1/p^k for k = 0..10: A001221 (k=0), A069359 (k=1), A322078 (k=2), this sequence (k=3), A351244 (k=4), A351245 (k=5), A351246 (k=6), A351247 (k=7), A351248 (k=8), A351249 (k=9), A351262 (k=10).

Cf. A000040, A001221, A010051, A059376, A085964.

Programs

Mathematica

a[n_]:= n^3 * Sum[1/p^3,{p,Select[Divisors[n],PrimeQ]}]; Array[a,56] (* Stefano Spezia, Jul 14 2025 *)

PARI

a(n) = my(f = factor(n)); sum(i = 1, #f~, (n/f[i,1])^3) \\ David A. Corneth, Jul 14 2025

Formula

a(A000040(n)) = 1.

G.f.: Sum_{k>=1} x^prime(k) * (1 + 4*x^prime(k) + x^(2*prime(k))) / (1 - x^prime(k))^4. - Ilya Gutkovskiy, Feb 05 2022

Dirichlet g.f. = zeta(s-3)*primezeta(s). This follows because Sum_{n>=1} a(n)/n^s = Sum_{n>=1} (n^3/n^s) Sum_{p|n} 1/p^3. Since n = p*j, rewrite the sum as Sum_{p} Sum_{j>=1} 1/(p^3*(p*j)^(s-3)) = Sum_{p} 1/p^s Sum_{j>=1} 1/j^(s-3) = zeta(s-3)*primezeta(s). The result generalizes to higher powers of p in a(n). - Michael Shamos, Mar 01 2023

Sum_{k=1..n} a(k) ~ A085964 * n^4/4. - Vaclav Kotesovec, Mar 03 2023

a(n) = Sum_{d|n} A059376(d)*A001221(n/d). - Ridouane Oudra, Jul 14 2025

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

a(n) = Sum_{d|n} c(d) * (n/d)^3, where c = A010051.

a(p^k) = p^(3*k-3) for p prime and k>=1. (End)