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

%I #23 Sep 14 2021 02:20:13

%S 1,3,6,11,18,31,48,76,118,184,279,426,641,966,1448,2163,3228,4805,

%T 7137,10586,15681,23198,34278,50606,74632,109987,161954,238312,350432,

%U 514999,756407,1110391,1629219,2389346,3502578,5132354,7517523,11007078,16110784,23573102,34480937,50420909

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

%H Seiichi Manyama, <a href="/A347415/b347415.txt">Table of n, a(n) for n = 1..5000</a>

%F a(n) ~ sqrt(2*Pi*n) * exp(exp(-1)*n - 1/2). - _Vaclav Kotesovec_, Sep 14 2021

%e a(3) = [3/1] + [(3/2)^2] + [(3/3)^3] = 3 + 2 + 1 = 6.

%t a[n_] := Sum[Floor[(n/k)^k], {k, 1, n}]; Array[a, 40] (* _Amiram Eldar_, Aug 31 2021 *)

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

%Y Cf. A006218, A062071, A153818, A344675, A345176, A347416.

%K nonn

%O 1,2

%A _Seiichi Manyama_, Aug 31 2021