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a(n) = Sum_{k=1..n} floor((n/k)^n).
3

%I #19 Sep 14 2021 02:23:44

%S 1,5,31,276,3238,47463,830415,16845619,388198577,10009945747,

%T 285452668383,8918294580680,302912273410475,11112687415252836,

%U 437907284782655738,18447025981637731050,827246579683710818081,39346558272075085340201,1978423430905859200399397

%N a(n) = Sum_{k=1..n} floor((n/k)^n).

%H Seiichi Manyama, <a href="/A347416/b347416.txt">Table of n, a(n) for n = 1..386</a>

%F a(n) ~ n^n. - _Vaclav Kotesovec_, Sep 14 2021

%e a(3) = [(3/1)^3] + [(3/2)^3] + [(3/3)^3] = 27 + 3 + 1 = 31.

%t a[n_] := Sum[Floor[(n/k)^n], {k, 1, n}]; Array[a, 20] (* _Amiram Eldar_, Aug 31 2021 *)

%o (PARI) a(n) = sum(k=1, n, n^n\k^n);

%Y Cf. A006218, A153818, A332469, A344675, A347415.

%K nonn

%O 1,2

%A _Seiichi Manyama_, Aug 31 2021