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a(n) = n^4 * Sum_{p|n, p prime} 1/p^4.
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%I #27 Mar 09 2023 20:51:23

%S 0,1,1,16,1,97,1,256,81,641,1,1552,1,2417,706,4096,1,7857,1,10256,

%T 2482,14657,1,24832,625,28577,6561,38672,1,61921,1,65536,14722,83537,

%U 3026,125712,1,130337,28642,164096,1,234193,1,234512,57186,279857,1,397312,2401,400625,83602

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

%H Seiichi Manyama, <a href="/A351244/b351244.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 + 11*x^prime(k) + 11*x^(2*prime(k)) + x^(3*prime(k))) / (1 - x^prime(k))^5. - _Ilya Gutkovskiy_, Feb 05 2022

%F 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

%F Sum_{k=1..n} a(k) ~ A085965 * n^5/5. - _Vaclav Kotesovec_, Mar 03 2023

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

%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), 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).

%Y Cf. A000040, A085965.

%K nonn

%O 1,4

%A _Wesley Ivan Hurt_, Feb 05 2022