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a(n) = Sum_{k=1..n} n^(floor(n/k) - 1).
4

%I #17 Jun 07 2021 09:09:20

%S 1,3,11,70,633,7821,117709,2097684,43047545,1000010125,25937439391,

%T 743008621422,23298085496173,793714780786669,29192926036832363,

%U 1152921504875352376,48661191876077295937,2185911559749718388655,104127350297928227579629

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

%H Seiichi Manyama, <a href="/A345030/b345030.txt">Table of n, a(n) for n = 1..387</a>

%F a(n) = [x^n] (1/(1 - x)) * Sum_{k>=1} x^k * (1 - x^k)/(1 - n*x^k).

%t a[n_] := Sum[n^(Floor[n/k] - 1), {k, 1, n}]; Array[a, 20] (* _Amiram Eldar_, Jun 06 2021 *)

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

%Y Diagonal of A345032.

%Y Cf. A023037, A332533, A344551, A345036.

%K nonn

%O 1,2

%A _Seiichi Manyama_, Jun 06 2021