A329825 Beatty sequence for (3+sqrt(17))/4.

1, 3, 5, 7, 8, 10, 12, 14, 16, 17, 19, 21, 23, 24, 26, 28, 30, 32, 33, 35, 37, 39, 40, 42, 44, 46, 48, 49, 51, 53, 55, 56, 58, 60, 62, 64, 65, 67, 69, 71, 73, 74, 76, 78, 80, 81, 83, 85, 87, 89, 90, 92, 94, 96, 97, 99, 101, 103, 105, 106, 108, 110, 112, 113

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: 4 and 5.

More generally, suppose that t > 0. There exists an irrational number r such that (floor(n*r)) and (floor(n*(r+t))) are a pair of Beatty sequences. Specifically, r = (2 - t + sqrt(t^2 + 4))/2, as in the Mathematica code below. See Comments at A182760.

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Guide to related sequences:

t = 1: A000201 and A001950 (Wythoff sequences), r = (1+sqrt(5))/2

t = 1/2: A329825 and A329826, r = (3 + sqrt(17))/4

t = 1/3: A329827 and A329828, r = (5 + sqrt(37))/6

t = 2/3: A329829 and A329830, r = (2 + sqrt(10))/3

t = 1/4: A329831 and A329832, r = (7 + sqrt(65))/8

t = 3/4: A329833 and A329834, r = (5 + sqrt(73))/8

t = 1/5: A329835 and A329836, r = (9 + sqrt(101))/10

t = 2/5: A329837 and A329838, r = (4 + sqrt(26))/5

t = 5/2: A329839 and A329840, r = (-1 + sqrt(41))/4

t = 3/5: A329841 and A329842, r = (7 + sqrt(109))/10

t = 5/3: A329843 and A329844, r = (1 + sqrt(61))/6

t = 5/4: A329847 and A329848, r = (3 + sqrt(89))/8

t = 4/5: A329845 and A329846, r = (3 + sqrt(29))/5

t = 6/5: A329923 and A329924, r = (2 + sqrt(34))/5

t = 8/5: A329925 and A329926, r = (1 + sqrt(41))/5

t = 2: A001951 and A001952, r = sqrt(2)

t = 3: A001961 and A004976, r = -1 + sqrt(5)

t = 4: A001961 and A001962, r = -1 + sqrt(5)

t = 5: A184522 and A184523, r = (-3 + sqrt(29))/2

t = 6: A187396 and A187395, r = -2 + sqrt(10).

Starts to deviate from A059565 at a(73). - R. J. Mathar, Nov 26 2019

Sequences for t = 5/4, 4/5 and 3 corrected by Georg Fischer, Aug 22 2021

Links

• Clark Kimberling, Table of n, a(n) for n = 1..10000
• Eric Weisstein's World of Mathematics, Beatty Sequence.
• Index entries for sequences related to Beatty sequences

Crossrefs

Cf. A188485, A329826 (complement), A182760.

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 *)

PARI

a(n)=(sqrtint(17*n^2)+3*n)\4 \\ Charles R Greathouse IV, Jan 25 2022

Formula

a(n) = floor(r*n), where r = (3+sqrt(17))/4.