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Beatty sequence for (5+sqrt(17))/4.
2

%I #8 Nov 23 2019 13:42:15

%S 2,4,6,9,11,13,15,18,20,22,25,27,29,31,34,36,38,41,43,45,47,50,52,54,

%T 57,59,61,63,66,68,70,72,75,77,79,82,84,86,88,91,93,95,98,100,102,104,

%U 107,109,111,114,116,118,120,123,125,127,130,132,134,136,139

%N Beatty sequence for (5+sqrt(17))/4.

%C Let r = (3+sqrt(17))/4. Then (floor(n*r)) and (floor(n*r + r/2)) are a pair of Beatty sequences; i.e., every positive integer is in exactly one of the sequences. The sequence (a(n) mod 2)) of 0's and 1's has only two run-lengths: 3 and 4. 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 = (5+sqrt(17))/4.

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

%t Table[Floor[r*n], {n, 1, 200}] (* A329825 *)

%t Table[Floor[s*n], {n, 1, 200}] (* A329826 *)

%Y Cf. A329825 (complement).

%K nonn,easy

%O 1,1

%A _Clark Kimberling_, Nov 22 2019