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a(n) = Sum_{d|n} max(d, n/d)^5.
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%I #32 Jan 13 2025 01:32:44

%S 1,64,486,2080,6250,16038,33614,67584,118341,206250,322102,515264,

%T 742586,1109262,1525000,2163712,2839714,3912786,4952198,6606250,

%U 8201816,10629366,12872686,16504000,19534375,24505338,28815912,35529998,41022298,50334302

%N a(n) = Sum_{d|n} max(d, n/d)^5.

%C If p is a prime, then 2*p^5 belongs to this sequence. Conjecture: The converse is true. - _Alexandra Hercilia Pereira Silva_, Oct 04 2022

%H Seiichi Manyama, <a href="/A297844/b297844.txt">Table of n, a(n) for n = 1..10000</a>

%F a(n) + A297795(n) = 2*A001160(n).

%F Sum_{k=1..n} a(k) ~ (zeta(6)/3) * n^6. - _Amiram Eldar_, Jan 12 2025

%t f[n_] := Block[{d = Divisors@ n}, Plus @@ (Max[#, n/#]^5 & /@ d)]; Array[f, 32] (* _Robert G. Wilson v_, Jan 07 2018 *)

%o (PARI) {a(n) = sumdiv(n, d, max(d, n/d)^5)}

%Y Sum_{d|n} max(d, n/d)^k: A117003 (k=1), A297841 (k=2), A297842 (k=3), A297843 (k=4), this sequence (k=5).

%Y Cf. A001160, A013664, A297795.

%K nonn

%O 1,2

%A _Seiichi Manyama_, Jan 07 2018