A351244 a(n) = n^4 * Sum_{p|n, p prime} 1/p^4.

0, 1, 1, 16, 1, 97, 1, 256, 81, 641, 1, 1552, 1, 2417, 706, 4096, 1, 7857, 1, 10256, 2482, 14657, 1, 24832, 625, 28577, 6561, 38672, 1, 61921, 1, 65536, 14722, 83537, 3026, 125712, 1, 130337, 28642, 164096, 1, 234193, 1, 234512, 57186, 279857, 1, 397312, 2401, 400625, 83602

Offset

1,4

Comments

Dirichlet convolution of A010051(n) and n^4. - Wesley Ivan Hurt, Jul 15 2025

Links

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

Formula

a(A000040(n)) = 1.

G.f.: Sum_{k>=1} x^prime(k) * (1 + 11*x^prime(k) + 11*x^(2*prime(k)) + x^(3*prime(k))) / (1 - x^prime(k))^5. - Ilya Gutkovskiy, Feb 05 2022

Dirichlet g.f.: zeta(s-4)*primezeta(s). This follows because Sum_{n>=1} a(n)/n^s = Sum_{n>=1} (n^4/n^s) Sum_{p|n} 1/p^4. Since n = p*j, rewrite the sum as Sum_{p} Sum_{j>=1} 1/(p^4*(p*j)^(s-4)) = Sum_{p} 1/p^s Sum_{j>=1} 1/j^(s-4) = zeta(s-4)*primezeta(s). The result generalizes to higher powers of p. - Michael Shamos, Mar 02 2023

Sum_{k=1..n} a(k) ~ A085965 * n^5/5. - Vaclav Kotesovec, Mar 03 2023

a(n) = Sum_{d|n} A059377(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)^4, where c = A010051.

a(p^k) = p^(4*k-4) for p prime and k>=1. (End)

Example

a(6) = 97; a(6) = 6^4 * Sum_{p|6, p prime} 1/p^4 = 1296 * (1/2^4 + 1/3^4) = 97.

Mathematica

a[n_]:= n^4 * Sum[1/p^4, {p, Select[Divisors[n], PrimeQ]}]; Array[a, 51] (* Stefano Spezia, Jul 14 2025 *)

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), this sequence (k=4), A351245 (k=5), A351246 (k=6), A351247 (k=7), A351248 (k=8), A351249 (k=9), A351262 (k=10).

Cf. A000040, A001221, A010051, A059377, A085965.

Sequence in context: A036179 A309132 A099923 * A105671 A145828 A369913

Adjacent sequences: A351241 A351242 A351243 * A351245 A351246 A351247

Keyword

nonn

Author

Wesley Ivan Hurt, Feb 05 2022

Status

approved