|
|
A351242
|
|
a(n) = n^3 * Sum_{p|n, p prime} 1/p^3.
|
|
11
|
|
|
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
(list;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
OFFSET
|
1,4
|
|
LINKS
|
|
|
FORMULA
|
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
|
|
EXAMPLE
|
a(6) = 35; a(6) = 6^3 * Sum_{p|6, p prime} 1/p^3 = 216 * (1/2^3 + 1/3^3) = 35.
|
|
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).
|
|
KEYWORD
|
nonn
|
|
AUTHOR
|
|
|
STATUS
|
approved
|
|
|
|