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a(n) = Sum_{k=1..n} floor(1/{k*sqrt(6)}) where {x} := x - floor(x).
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%I #21 Sep 08 2022 08:44:48

%S 2,3,5,6,10,11,17,18,40,42,43,45,46,49,50,55,56,67,68,69,71,72,74,75,

%T 79,80,87,88,116,118,119,121,122,125,126,131,132,144,145,146,148,149,

%U 152,153,157,158,165,166,206,208,209,211,212,215,216,221,222,236,237,238,240,241,244

%N a(n) = Sum_{k=1..n} floor(1/{k*sqrt(6)}) where {x} := x - floor(x).

%H Clark Kimberling, <a href="/A024560/b024560.txt">Table of n, a(n) for n = 1..1000</a>

%t Table[Sum[Floor[1/FractionalPart[k*Sqrt[6]]], {k, 1, n}], {n, 1, 100}]

%t (* _Clark Kimberling_, Aug 16 2012 *)

%o (Magma) a:=Sqrt(6); [&+[Floor(1/(k*a-Floor(k*a))):k in [1..n]]:n in [1..60]]; // _Vincenzo Librandi_, Jul 17 2019

%o (PARI) a(n) = sum(k=1, n, floor(1/frac(k*sqrt(6)))); \\ _Michel Marcus_, Jul 17 2019

%Y Cf. A024561. Partial sums of A024559.

%K nonn

%O 1,1

%A _Clark Kimberling_