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

%I A329825 #30 Mar 19 2025 05:54:43
%S A329825 1,3,5,7,8,10,12,14,16,17,19,21,23,24,26,28,30,32,33,35,37,39,40,42,
%T A329825 44,46,48,49,51,53,55,56,58,60,62,64,65,67,69,71,73,74,76,78,80,81,83,
%U A329825 85,87,89,90,92,94,96,97,99,101,103,105,106,108,110,112,113
%N A329825 Beatty sequence for (3+sqrt(17))/4.
%C A329825 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.
%C A329825 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.
%C A329825 ************
%C A329825 Guide to related sequences:
%C A329825 t = 1:   A000201 and A001950 (Wythoff sequences), r = (1+sqrt(5))/2
%C A329825 t = 1/2: A329825 and A329826, r = (3 + sqrt(17))/4
%C A329825 t = 1/3: A329827 and A329828, r = (5 + sqrt(37))/6
%C A329825 t = 2/3: A329829 and A329830, r = (2 + sqrt(10))/3
%C A329825 t = 1/4: A329831 and A329832, r = (7 + sqrt(65))/8
%C A329825 t = 3/4: A329833 and A329834, r = (5 + sqrt(73))/8
%C A329825 t = 1/5: A329835 and A329836, r = (9 + sqrt(101))/10
%C A329825 t = 2/5: A329837 and A329838, r = (4 + sqrt(26))/5
%C A329825 t = 5/2: A329839 and A329840, r = (-1 + sqrt(41))/4
%C A329825 t = 3/5: A329841 and A329842, r = (7 + sqrt(109))/10
%C A329825 t = 5/3: A329843 and A329844, r = (1 + sqrt(61))/6
%C A329825 t = 5/4: A329847 and A329848, r = (3 + sqrt(89))/8
%C A329825 t = 4/5: A329845 and A329846, r = (3 + sqrt(29))/5
%C A329825 t = 6/5: A329923 and A329924, r = (2 + sqrt(34))/5
%C A329825 t = 8/5: A329925 and A329926, r = (1 + sqrt(41))/5
%C A329825 t = 2:   A001951 and A001952, r = sqrt(2)
%C A329825 t = 3:   A001961 and A004976, r = -1 + sqrt(5)
%C A329825 t = 4:   A001961 and A001962, r = -1 + sqrt(5)
%C A329825 t = 5:   A184522 and A184523, r = (-3 + sqrt(29))/2
%C A329825 t = 6:   A187396 and A187395, r = -2 + sqrt(10).
%C A329825 Starts to deviate from A059565 at a(73). - _R. J. Mathar_, Nov 26 2019
%C A329825 Sequences for t = 5/4, 4/5 and 3 corrected by _Georg Fischer_, Aug 22 2021
%H A329825 Clark Kimberling, <a href="/A329825/b329825.txt">Table of n, a(n) for n = 1..10000</a>
%H A329825 Eric Weisstein's World of Mathematics, <a href="https://mathworld.wolfram.com/BeattySequence.html">Beatty Sequence.</a>
%H A329825 <a href="/index/Be#Beatty">Index entries for sequences related to Beatty sequences</a>
%F A329825 a(n) = floor(r*n), where r = (3+sqrt(17))/4.
%t A329825 t = 1/2; r = Simplify[(2 - t + Sqrt[t^2 + 4])/2]; s = Simplify[r/(r - 1)];
%t A329825 Table[Floor[r*n], {n, 1, 200}]  (* A329825 *)
%t A329825 Table[Floor[s*n], {n, 1, 200}]  (* A329826 *)
%o A329825 (PARI) a(n)=(sqrtint(17*n^2)+3*n)\4 \\ _Charles R Greathouse IV_, Jan 25 2022
%Y A329825 Cf. A188485, A329826 (complement), A182760.
%K A329825 nonn,easy
%O A329825 1,2
%A A329825 _Clark Kimberling_, Nov 22 2019