login
Beatty sequence for (1+sqrt(41))/5.
3

%I #4 Jan 02 2020 16:15:32

%S 1,2,4,5,7,8,10,11,13,14,16,17,19,20,22,23,25,26,28,29,31,32,34,35,37,

%T 38,39,41,42,44,45,47,48,50,51,53,54,56,57,59,60,62,63,65,66,68,69,71,

%U 72,74,75,76,78,79,81,82,84,85,87,88,90,91,93,94,96,97

%N Beatty sequence for (1+sqrt(41))/5.

%C Let r = (1+sqrt(41))/5. Then (floor(n*r)) and (floor(n*r + 8r/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*r), where r = (1+sqrt(41))/5.

%t t = 8/5; r = Simplify[(2 - t + Sqrt[t^2 + 4])/2]; s = Simplify[r/(r - 1)];

%t Table[Floor[r*n], {n, 1, 200}] (* A329925 *)

%t Table[Floor[s*n], {n, 1, 200}] (* A329926 *)

%Y Cf. A329825, A329926 (complement).

%K nonn,easy

%O 1,2

%A _Clark Kimberling_, Jan 02 2020