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Beatty sequence for (6+sqrt(26))/5.
3

%I #5 Feb 16 2025 08:33:58

%S 2,4,6,8,11,13,15,17,19,22,24,26,28,31,33,35,37,39,42,44,46,48,51,53,

%T 55,57,59,62,64,66,68,71,73,75,77,79,82,84,86,88,91,93,95,97,99,102,

%U 104,106,108,110,113,115,117,119,122,124,126,128,130,133,135

%N Beatty sequence for (6+sqrt(26))/5.

%C Let r = (4+sqrt(26))/5. Then (floor(n*r)) and (floor(n*r + 2r/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="https://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 = (6+sqrt(26))/5.

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

%t Table[Floor[r*n], {n, 1, 200}] (* A329837 *)

%t Table[Floor[s*n], {n, 1, 200}] (* A329838 *)

%Y Cf. A329825, A329837 (complement).

%K nonn,easy,changed

%O 1,1

%A _Clark Kimberling_, Dec 31 2019