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Beatty sequence for (7+sqrt(29))/5.
3

%I #4 Jan 02 2020 16:14:48

%S 2,4,7,9,12,14,17,19,22,24,27,29,32,34,37,39,42,44,47,49,52,54,56,59,

%T 61,64,66,69,71,74,76,79,81,84,86,89,91,94,96,99,101,104,106,108,111,

%U 113,116,118,121,123,126,128,131,133,136,138,141,143,146,148

%N Beatty sequence for (7+sqrt(29))/5.

%C Let r = (3+sqrt(29))/5. Then (floor(n*r)) and (floor(n*r + 4r/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*s), where s = (7+sqrt(29))/5.

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

%t Table[Floor[r*n], {n, 1, 200}] (* A329845 *)

%t Table[Floor[s*n], {n, 1, 200}] (* A329846 *)

%Y Cf. A329825, A329845 (complement).

%K nonn,easy

%O 1,1

%A _Clark Kimberling_, Jan 02 2020