A351242 - OEIS

%I #26 Mar 09 2023 20:52:40

%S 0,1,1,8,1,35,1,64,27,133,1,280,1,351,152,512,1,945,1,1064,370,1339,1,

%T 2240,125,2205,729,2808,1,4591,1,4096,1358,4921,468,7560,1,6867,2224,

%U 8512,1,12221,1,10712,4104,12175,1,17920,343,16625,4940,17640,1,25515,1456,22464

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

%H Seiichi Manyama, <a href="/A351242/b351242.txt">Table of n, a(n) for n = 1..10000</a>

%F a(A000040(n)) = 1.

%F 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

%F 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

%F Sum_{k=1..n} a(k) ~ A085964 * n^4/4. - _Vaclav Kotesovec_, Mar 03 2023

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

%Y 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).

%Y Cf. A000040, A085964.

%K nonn

%O 1,4

%A _Wesley Ivan Hurt_, Feb 05 2022