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Beatty sequence for (11+sqrt(73))/8.
3

%I #4 Jan 02 2020 08:20:30

%S 2,4,7,9,12,14,17,19,21,24,26,29,31,34,36,39,41,43,46,48,51,53,56,58,

%T 61,63,65,68,70,73,75,78,80,83,85,87,90,92,95,97,100,102,105,107,109,

%U 112,114,117,119,122,124,127,129,131,134,136,139,141,144,146

%N Beatty sequence for (11+sqrt(73))/8.

%C Let r = (5+sqrt(73))/8. Then (floor(n*r)) and (floor(n*r + 3r/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*s), where s = (11+sqrt(73))/8.

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

%t Table[Floor[r*n], {n, 1, 200}] (* A329833 *)

%t Table[Floor[s*n], {n, 1, 200}] (* A329834 *)

%Y Cf. A329825, A329833 (complement).

%K nonn,easy

%O 1,1

%A _Clark Kimberling_, Dec 31 2019