A187396 -id:A187396 - cp's OEIS frontend

A187395 a(n) = floor(r*n), where r = 4 + sqrt(10); complement of A187396.

7, 14, 21, 28, 35, 42, 50, 57, 64, 71, 78, 85, 93, 100, 107, 114, 121, 128, 136, 143, 150, 157, 164, 171, 179, 186, 193, 200, 207, 214, 222, 229, 236, 243, 250, 257, 265, 272, 279, 286, 293, 300, 307, 315, 322, 329, 336, 343, 350, 358, 365, 372, 379, 386, 393, 401, 408, 415, 422, 429, 436, 444, 451, 458, 465, 472, 479, 487, 494, 501

Offset: 1

Clark Kimberling, Mar 09 2011

nonn

A187395 and A187396 are the Beatty sequences based on r = 4 + sqrt(10) and s = -2 + sqrt(10); 1/r + 1/s = 1.

Cf. A187396.

r=4+10^(1/2); s=-2+10^(1/2);
Table[Floor[r*n],{n,1,80}]  (* A187395 *)
Table[Floor[s*n],{n,1,80}]  (* A187396 *)

from sympy import integer_nthroot
def A187395(n): return 4*n+integer_nthroot(10*n**2,2)[0] # Chai Wah Wu, Mar 16 2021

a(n) = floor(r*n), where r = 4 + sqrt(10).

Edited by Clark Kimberling, Mar 17 2021

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

Original entry on oeis.org

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

Clark Kimberling, Nov 22 2019

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.
************
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

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

Cf. A188485, A329826 (complement), A182760.

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 *)

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

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

cp's OEIS Frontend

A187395 a(n) = floor(r*n), where r = 4 + sqrt(10); complement of A187396.

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