A351245 a(n) = n^5 * Sum_{p|n, p prime} 1/p^5.

0, 1, 1, 32, 1, 275, 1, 1024, 243, 3157, 1, 8800, 1, 16839, 3368, 32768, 1, 66825, 1, 101024, 17050, 161083, 1, 281600, 3125, 371325, 59049, 538848, 1, 867151, 1, 1048576, 161294, 1419889, 19932, 2138400, 1, 2476131, 371536, 3232768, 1, 4629701, 1, 5154656, 818424, 6436375, 1

Offset: 1

Author

Wesley Ivan Hurt, Feb 05 2022

Keywords

• nonn

Comments

Dirichlet convolution of A010051(n) and n^5. - Wesley Ivan Hurt, Jul 15 2025

Examples

			a(6) = 275; a(6) = 6^5 * Sum_{p|6, p prime} 1/p^5 = 7776 * (1/2^5 + 1/3^5) = 275.
		

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), A351242 (k=3), A351244 (k=4), this sequence (k=5), A351246 (k=6), A351247 (k=7), A351248 (k=8), A351249 (k=9), A351262 (k=10).

Cf. A000040, A001221, A010051, A059378, A085966.

Programs

Mathematica

Array[#^5*DivisorSum[#, 1/#^5 &, PrimeQ] &, 47] (* Stefano Spezia, Jul 15 2025 *)

PARI

a(n) = my(f = factor(n)); sum(i = 1, #f~, (n/f[i,1])^5) \\ David A. Corneth, Jul 15 2025

Formula

a(A000040(n)) = 1.

Dirichlet g.f.: zeta(s-5)*primezeta(s). This follows because Sum_{n>=1} a(n)/n^s = Sum_{n>=1} (n^5/n^s) Sum_{p|n} 1/p^5. Since n = p*j, rewrite the sum as Sum_{p} Sum_{j>=1} 1/(p^5*(p*j)^(s-5)) = Sum_{p} 1/p^s Sum_{j>=1} 1/j^(s-5) = zeta(s-5)*primezeta(s). The result generalizes to higher powers of p. - Michael Shamos, Mar 03 2023

Sum_{k=1..n} a(k) ~ A085966 * n^6/6. - Vaclav Kotesovec, Mar 03 2023

a(n) = Sum_{d|n} A059378(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)^5, where c = A010051.

a(p^k) = p^(5*k-5) for p prime and k>=1. (End)