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Beatty sequence for (7+sqrt(65))/8.
4

%I #4 Jan 02 2020 08:20:11

%S 1,3,5,7,9,11,13,15,16,18,20,22,24,26,28,30,32,33,35,37,39,41,43,45,

%T 47,48,50,52,54,56,58,60,62,64,65,67,69,71,73,75,77,79,80,82,84,86,88,

%U 90,92,94,96,97,99,101,103,105,107,109,111,112,114,116,118

%N Beatty sequence for (7+sqrt(65))/8.

%C Let r = (7+sqrt(65))/8. Then (floor(n*r)) and (floor(n*r + r/4)) 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 = (7+sqrt(65))/8.

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

%t Table[Floor[r*n], {n, 1, 200}] (* A329831 *)

%t Table[Floor[s*n], {n, 1, 200}] (* A329832 *)

%Y Cf. A329825, A329832 (complement).

%K nonn,easy

%O 1,2

%A _Clark Kimberling_, Dec 31 2019