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a(n) = [Sum_{k=1..n} (k*r)], where [ ]=floor, ( )=fractional part, and r=sqrt(6).
3

%I #12 Nov 12 2017 03:03:08

%S 0,1,1,2,2,3,3,4,4,4,5,6,6,7,7,8,8,8,9,10,10,11,12,12,13,13,13,14,14,

%T 15,15,16,17,17,18,18,18,19,19,20,21,21,22,22,23,23,24,24,24,25,26,26,

%U 27,27,28,28,29,29,29,30,30,31,32,32,33,33,33,34,34,34,35,36,37

%N a(n) = [Sum_{k=1..n} (k*r)], where [ ]=floor, ( )=fractional part, and r=sqrt(6).

%H G. C. Greubel, <a href="/A194173/b194173.txt">Table of n, a(n) for n = 1..5000</a>

%t r = Sqrt[6];

%t a[n_] := Floor[Sum[FractionalPart[k*r], {k, 1, n}]]

%t Table[a[n], {n, 1, 90}] (* A194173 *)

%t s[n_] := Sum[a[k], {k, 1, n}]

%t Table[s[n], {n, 1, 100}] (* A194174 *)

%o (PARI) a(n) = floor(sum(k=1, n, frac(sqrt(6)*k))); \\ _Michel Marcus_, Nov 12 2017

%Y Cf. A194173.

%K nonn

%O 1,4

%A _Clark Kimberling_, Aug 18 2011