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a(n) = Sum_{k=1..n} [ 1/{k*Pi} ], where {x} := x - [ x ].
2

%I #15 Jul 17 2019 03:41:45

%S 7,10,12,13,14,15,16,23,26,28,29,30,31,32,40,43,45,46,47,48,49,57,60,

%T 62,63,64,65,66,75,79,81,82,83,84,85,95,99,101,102,103,104,105,116,

%U 120,122,123,124,125,126,138,142,144,145,146,147,148,162,166,168,170,171,172,173,189

%N a(n) = Sum_{k=1..n} [ 1/{k*Pi} ], where {x} := x - [ x ].

%H Clark Kimberling, <a href="/A024585/b024585.txt">Table of n, a(n) for n = 1..1000</a>

%t Table[Sum[Floor[1/FractionalPart[k*Pi]], {k, n}], {n, 100}] (* _Clark Kimberling_, Aug 18 2012 *)

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

%Y Cf. A024586. Partial sums of A024584.

%K nonn

%O 1,1

%A _Clark Kimberling_