A329826 Beatty sequence for (5+sqrt(17))/4.

2, 4, 6, 9, 11, 13, 15, 18, 20, 22, 25, 27, 29, 31, 34, 36, 38, 41, 43, 45, 47, 50, 52, 54, 57, 59, 61, 63, 66, 68, 70, 72, 75, 77, 79, 82, 84, 86, 88, 91, 93, 95, 98, 100, 102, 104, 107, 109, 111, 114, 116, 118, 120, 123, 125, 127, 130, 132, 134, 136, 139

Offset: 1

Author

Clark Kimberling, Nov 22 2019

Keywords

• nonn
• easy

Comments

Let r = (3+sqrt(17))/4. Then (floor(n*r)) and (floor(n*r + r/2)) are a pair of Beatty sequences; i.e., every positive integer is in exactly one of the sequences. The sequence (a(n) mod 2) of 0's and 1's has only two run-lengths: 3 and 4. See the Guide to related sequences at A329825.

Links

• Eric Weisstein's World of Mathematics, Beatty Sequence.
• Index entries for sequences related to Beatty sequences

Crossrefs

Cf. A329825 (complement).

Programs

Mathematica

t = 1/2; r = Simplify[(2 - t + Sqrt[t^2 + 4])/2]; s = Simplify[r/(r - 1)];
Table[Floor[r*n], {n, 1, 200}]   (* A329825 *)
Table[Floor[s*n], {n, 1, 200}]   (* A329826 *)

Formula

a(n) = floor(n*s), where s = (5+sqrt(17))/4.