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Beatty sequence for (1+sqrt(61))/6.
3

%I #4 Jan 02 2020 16:14:27

%S 1,2,4,5,7,8,10,11,13,14,16,17,19,20,22,23,24,26,27,29,30,32,33,35,36,

%T 38,39,41,42,44,45,46,48,49,51,52,54,55,57,58,60,61,63,64,66,67,69,70,

%U 71,73,74,76,77,79,80,82,83,85,86,88,89,91,92,93,95,96

%N Beatty sequence for (1+sqrt(61))/6.

%C Let r = (1+sqrt(61))/6. Then (floor(n*r)) and (floor(n*r + 5r/3)) are a pair of Beatty sequences; i.e., every positive integer is in exactly one of the sequences. See the Guide to related sequences at A329825.

%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/BeattySequence.html">Beatty Sequence.</a>

%H <a href="/index/Be#Beatty">Index entries for sequences related to Beatty sequences</a>

%F a(n) = floor(n*r), where r = (1+sqrt(61))/6.

%t t = 5/3; r = Simplify[(2 - t + Sqrt[t^2 + 4])/2]; s = Simplify[r/(r - 1)];

%t Table[Floor[r*n], {n, 1, 200}] (* A329843 *)

%t Table[Floor[s*n], {n, 1, 200}] (* A329844 *)

%Y Cf. A329825, A329844 (complement).

%K nonn,easy

%O 1,2

%A _Clark Kimberling_, Jan 02 2020