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a(n) = Sum_{k=1..n} k^floor((n-k)/k).
7

%I #13 Jun 06 2021 09:00:15

%S 1,2,3,5,6,11,12,20,27,40,41,93,94,133,208,328,329,658,659,1217,1746,

%T 2269,2270,5768,6269,8330,12777,20253,20254,45253,45254,74390,113867,

%U 146652,161211,401275,401276,532367,886818,1412574,1412575,3053234,3053235,4889475,8396664

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

%H Seiichi Manyama, <a href="/A344551/b344551.txt">Table of n, a(n) for n = 1..5000</a>

%F a(n) ~ 3^((n - 3 - mod(n,3))/3). - _Vaclav Kotesovec_, May 28 2021

%F G.f.: (1/(1 - x)) * Sum_{k>=1} x^k * (1 - x^k)/(1 - k*x^k). - _Seiichi Manyama_, Jun 06 2021

%t Table[Sum[k^Floor[(n - k)/k], {k, n}], {n, 80}]

%o (PARI) a(n) = sum(k=1, n, k^(n\k-1)); \\ _Seiichi Manyama_, Jun 06 2021

%o (PARI) my(N=66, x='x+O('x^N)); Vec(sum(k=1, N, x^k*(1-x^k)/(1-k*x^k))/(1-x)) \\ _Seiichi Manyama_, Jun 06 2021

%Y Cf. A002541, A006218, A153485.

%K nonn

%O 1,2

%A _Wesley Ivan Hurt_, May 22 2021