The OEIS Foundation is supported by donations from users of the OEIS and by a grant from the Simons Foundation. Hints (Greetings from The On-Line Encyclopedia of Integer Sequences!)
 A260482 Dragon curve triple point numerators: When a(n) in 0, 1, 2, ..., (5*2^k), Dragon(a(n)/(5*2^k)) has exactly three distinct, rational preimages. 7
 7, 13, 14, 26, 27, 28, 33, 37, 47, 52, 53, 54, 56, 57, 66, 67, 69, 71, 73, 74, 77, 87, 93, 94, 97, 103, 104, 106, 107, 108, 109, 111, 112, 113, 114, 123, 127, 132, 133, 134, 138, 139, 141, 142, 146, 147, 148, 149, 151, 153, 154, 157, 167, 173, 174, 177, 186, 187, 188, 189, 191, 193, 194, 197, 206, 207, 208, 209, 211, 212, 213, 214, 216, 217, 218, 219, 221, 222, 223, 224, 226, 227, 228 (list; graph; refs; listen; history; text; internal format)
 OFFSET 1,1 COMMENTS It appears that Dragon(a(n)/(5*2^k)) = Dragon(b/(15*2^k)) = Dragon(c/(15*2^k)) for some b and c. See dragun in the MATHEMATICA section for an exact evaluator of the continuous, spacefilling Dragon function which maps [0,1] into C, and undrag, its multivalued inverse. The first differences of this sequence appear to comprise only 1,2,3,4,5,6,9, and 10. It appears that every Dragon triple point is an image of a(n)/(5*2^k) for some n and k. The set of values DRAG(m/(14*2^k)) with m in {0, 1, 2, ..., 14*2^k} also contains points at least triple whenever k > 0. See Examples. - Bradley Klee, Aug 14 2015 Using quaternary expansions of planar coordinates and a substitution tiling, one can prove the following: If a point along the Dragon curve has rational planar coordinates, it is visited one, two, or three times. The corollary is: All rational points at least triple are exactly triple. - Bradley Klee, Aug 18 2015 LINKS Brady Haran and Don Knuth, Wrong turn on the Dragon, Numberphile video (2014) Wikipedia, Dragon curve Bill Gosper, Illustration Bill Gosper, Illustration Bill Gosper, Illustration EXAMPLE a(8) = 47, so if Dragon(0) = 0, Dragon(1) = 1, Dragon(1/3) = 1/5+2i/5, then Dragon(133/240) = Dragon(47/80) = Dragon(143/240) = 2/3+5i/12 and Dragon(133/480) = Dragon(47/160) = Dragon(143/480) = 1/8+13i/24 and ... Dragon(133/3840) = Dragon(47/1280) = Dragon(143/3840) = -1/6-5i/48 and ... DRAG(13/28) = DRAG(17/28)= DRAG(19/28) = 3/5 + 3/10 i. - Bradley Klee, Aug 11 2015 MATHEMATICA (* by Julian Ziegler Hunts *) piecewiserecursivefractal[x_, f_, which_, iters_, fns_] := piecewiserecursivefractal[x, g_, which, iters, fns] = ((piecewiserecursivefractal[x, h_, which, iters, fns] := Block[{y}, y /. Solve[f[y] == h[y], y]]); Union @@ ((fns[[#]] /@ piecewiserecursivefractal[iters[[#]][x], Composition[f, fns[[#]]], which, iters, fns]) & /@ which[x])); dragun[t_] := piecewiserecursivefractal[t, Identity, Piecewise[{{{1}, 0 <= # <= 1/2}, {{2}, 1/2 <= # <= 1}}, {}] &, {2*# &, 2*(1 - #) &}, {(1 + I)*#/2 &, (I - 1)*#/2 + 1 &}] undrag[z_] := piecewiserecursivefractal[z, Identity, If[-(1/3) <= Re[#] <= 7/6 && -(1/3) <= Im[#] <= 2/3, {1, 2}, {}] &, {#*(1 - I) &, (1 - #)*(1 + I) &}, {#/2 &, 1 - #/2 &}] Do[If[Length[undrag[dragun[k/80][]]] > 2, Print[k]], {k, 0, 68}] (* same as, e.g. *) Do[If[Length[undrag[dragun[k/20480][]]] > 2, Print[k]], {k, 0, 68}] (* Not {k, 0, 69} because undrag@@dragun[69/20480] = {69/20480, 211/61440, 341/61440} but undrag@@dragun[69/80] = {69/80, 211/240}, since 341/240 > 1, outside the Dragon's preimage = [0, 1]. Corrected by Bill Gosper, Feb 18 2018. *) CROSSREFS Cf. A260747..A260750, A261120. Sequence in context: A308525 A104217 A269163 * A328254 A191976 A135054 Adjacent sequences:  A260479 A260480 A260481 * A260483 A260484 A260485 KEYWORD nonn,frac AUTHOR Bill Gosper, Jul 26 2015 EXTENSIONS Name simplified by Bradley Klee, Aug 18 2015 STATUS approved

Lookup | Welcome | Wiki | Register | Music | Plot 2 | Demos | Index | Browse | More | WebCam
Contribute new seq. or comment | Format | Style Sheet | Transforms | Superseeker | Recent
The OEIS Community | Maintained by The OEIS Foundation Inc.

Last modified July 8 18:44 EDT 2020. Contains 335524 sequences. (Running on oeis4.)