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a(n) = [ sum of 1/{k*sqrt(5)} ], k = 1,2,...,n, where {x} := x - [ x ].
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%I #10 Apr 07 2019 22:04:51

%S 4,6,7,8,14,16,18,19,27,30,31,33,47,50,52,54,130,134,136,137,138,143,

%T 146,147,148,155,158,160,161,173,176,178,179,217,221,223,225,226,230,

%U 233,234,235,242,244,246,247,258,261,262,264,289,293,295,296,297,302,304,305,306

%N a(n) = [ sum of 1/{k*sqrt(5)} ], k = 1,2,...,n, where {x} := x - [ x ].

%H Clark Kimberling, <a href="/A024555/b024555.txt">Table of n, a(n) for n = 1..1000</a>

%t Table[Floor[Sum[1/FractionalPart[k*Sqrt[5]], {k, 1, n}]], {n, 1, 100}]

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

%t Floor[Accumulate[1/FractionalPart[Range[60]Sqrt[5]]]] (* _Harvey P. Dale_, Apr 07 2019 *)

%Y Cf. A024554.

%K nonn

%O 1,1

%A _Clark Kimberling_