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Beatty sequence for (9+sqrt(41))/5.
3

%I #8 Jul 09 2021 00:18:06

%S 3,6,9,12,15,18,21,24,27,30,33,36,40,43,46,49,52,55,58,61,64,67,70,73,

%T 77,80,83,86,89,92,95,98,101,104,107,110,113,117,120,123,126,129,132,

%U 135,138,141,144,147,150,154,157,160,163,166,169,172,175,178,181

%N Beatty sequence for (9+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*s), where s = (9+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, A329925 (complement).

%K nonn,easy

%O 1,1

%A _Clark Kimberling_, Jan 02 2020

%E Definition corrected by _Georg Fischer_, Jul 08 2021