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Beatty sequence for (7+sqrt(109))/10.
3

%I #4 Jan 02 2020 08:23:05

%S 1,3,5,6,8,10,12,13,15,17,19,20,22,24,26,27,29,31,33,34,36,38,40,41,

%T 43,45,47,48,50,52,54,55,57,59,61,62,64,66,68,69,71,73,74,76,78,80,81,

%U 83,85,87,88,90,92,94,95,97,99,101,102,104,106,108,109,111

%N Beatty sequence for (7+sqrt(109))/10.

%C Let r = (7+sqrt(109))/10. Then (floor(n*r)) and (floor(n*r + 3r/5)) 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(109))/10.

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

%t Table[Floor[r*n], {n, 1, 200}] (* A329841 *)

%t Table[Floor[s*n], {n, 1, 200}] (* A329842 *)

%Y Cf. A329825, A329842 (complement).

%K nonn,easy

%O 1,2

%A _Clark Kimberling_, Dec 31 2019