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

%I #14 Jul 17 2019 04:43:28

%S 2,3,12,14,15,19,20,21,24,25,46,48,49,55,56,57,60,61,62,64,65,75,77,

%T 78,83,84,85,88,89,116,118,119,126,127,128,132,133,134,136,137,149,

%U 151,152,157,158,159,162,163,201,203,204,211,213,214,218,219,220,222,223,236,238,239

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

%H Clark Kimberling, <a href="/A024579/b024579.txt">Table of n, a(n) for n = 1..1000</a>

%t Table[Sum[Floor[1/FractionalPart[k/E]], {k, n}], {n, 100}] (* _Clark Kimberling_, Aug 17 2012 *)

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

%Y Cf. A024580. Partial sums of A024578.

%K nonn

%O 1,1

%A _Clark Kimberling_