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Beatty sequence for (11+sqrt(61))/6.
3

%I #4 Jan 02 2020 16:14:35

%S 3,6,9,12,15,18,21,25,28,31,34,37,40,43,47,50,53,56,59,62,65,68,72,75,

%T 78,81,84,87,90,94,97,100,103,106,109,112,115,119,122,125,128,131,134,

%U 137,141,144,147,150,153,156,159,163,166,169,172,175,178,181

%N Beatty sequence for (11+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*s), where s = (11+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, A329843 (complement).

%K nonn,easy

%O 1,1

%A _Clark Kimberling_, Jan 02 2020