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Beatty sequence for (9+sqrt(65))/8.
4

%I #4 Jan 02 2020 08:20:18

%S 2,4,6,8,10,12,14,17,19,21,23,25,27,29,31,34,36,38,40,42,44,46,49,51,

%T 53,55,57,59,61,63,66,68,70,72,74,76,78,81,83,85,87,89,91,93,95,98,

%U 100,102,104,106,108,110,113,115,117,119,121,123,125,127,130,132

%N Beatty sequence for (9+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*s), where s = (9+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, A329831 (complement).

%K nonn,easy

%O 1,1

%A _Clark Kimberling_, Dec 31 2019