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a(n) = Sum_{k=1..n} floor((2*n-k)/k).
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%I #23 Dec 23 2020 04:04:51

%S 0,1,4,8,12,17,23,27,34,40,46,52,60,65,73,81,87,93,104,108,118,126,

%T 132,140,150,157,165,173,183,189,201,205,216,226,232,242,254,258,268,

%U 278,288,295,307,313,323,335,343,349,363,369,382,390,398,408,420,428,440,448,456,464,482

%N a(n) = Sum_{k=1..n} floor((2*n-k)/k).

%F From _Vaclav Kotesovec_, Dec 23 2020: (Start)

%F For n>0, a(n) = 2*A006218(n) + A075989(n) - n.

%F a(n) ~ 2*n * (log(2*n) + 2*gamma - 2), where gamma is the Euler-Mascheroni constant A001620. (End)

%t Table[Sum[Floor[(2 n - i)/i], {i, n}], {n, 0, 60}]

%o (PARI) a(n) = sum(k=1, n, (2*n-k)\k); \\ _Michel Marcus_, Dec 22 2020

%Y Cf. A002541, A153485, A279847, A338991, A339370.

%K nonn,easy

%O 0,3

%A _Wesley Ivan Hurt_, Dec 22 2020