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 A329825 Beatty sequence for (3+sqrt(17))/4. 64
 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 (list; graph; refs; listen; history; text; internal format)
 OFFSET 1,2 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. ************ 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: A329845 and A329846, r = (3 + sqrt(89))/8 t = 4/5: A329847 and A329848, 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:   A184480 and A001956, 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 LINKS Clark Kimberling, Table of n, a(n) for n = 1..10000 Eric Weisstein's World of Mathematics, Beatty Sequence. FORMULA a(n) = floor(r*n), where r = (3+sqrt(17))/4. 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 *) CROSSREFS Cf. A188485, A329826 (complement), A182760. Sequence in context: A284366 A037086 A327212 * A059565 A329999 A187841 Adjacent sequences:  A329822 A329823 A329824 * A329826 A329827 A329828 KEYWORD nonn,easy AUTHOR Clark Kimberling, Nov 22 2019 STATUS approved

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Last modified June 13 05:41 EDT 2021. Contains 344981 sequences. (Running on oeis4.)