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a(n) = Sum_{k=1..n} k^(n - floor(n/k)).
2

%I #17 Jun 09 2021 10:37:15

%S 1,3,14,96,971,12015,184286,3283598,67676125,1572527901,40843114146,

%T 1170338862814,36718016941445,1251213685475261,46033362584427670,

%U 1818364700307111794,76762441669319061911,3448793841153099408185,164309637864524321789042

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

%H Seiichi Manyama, <a href="/A345106/b345106.txt">Table of n, a(n) for n = 1..387</a>

%F G.f.: Sum_{k>=1} k^(k-1)*x^k * (1 - (k*x)^k)/((1 - k^(k-1)*x^k) * (1 - k*x)).

%t a[n_] := Sum[k^(n - Floor[n/k]), {k, 1, n}]; Array[a, 20] (* _Amiram Eldar_, Jun 08 2021 *)

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

%o (PARI) my(N=20, x='x+O('x^N)); Vec(sum(k=1, N, k^(k-1)*x^k*(1-(k*x)^k)/((1-k^(k-1)*x^k)*(1-k*x))))

%Y Cf. A076015, A344551, A345107.

%K nonn

%O 1,2

%A _Seiichi Manyama_, Jun 08 2021