A260747 Consolidated Dragon Curve triple points. If D:[0,1] is a Dragon curve, then besides n, there are two other integers p and q with D(A(n)/(15*2^k)) = D(A(p)/(15*2^k)) = D(A(q)/(15*2^k)), where k is any integer > log_2(max(A(n),A(p),A(q))/15).

13, 21, 23, 26, 37, 39, 42, 46, 47, 52, 73, 74, 78, 81, 83, 84, 92, 94, 97, 99, 103, 104, 107, 111, 113, 133, 141, 143, 146, 148, 156, 157, 159, 162, 163, 166, 167, 168, 171, 173, 184, 188, 193, 194, 198, 199, 201, 203, 206, 207, 208, 209, 211, 213, 214, 217, 219, 221, 222, 223, 226, 227, 231, 233, 253, 261, 263, 266, 277, 279, 282, 283, 286, 287

Offset: 1

Author

Bill Gosper, Jul 30 2015

Keywords

• nonn
• frac
• obsc

Comments

It appears that every Dragon triple point is an image of A(n)/(15*2^k) for three different n and some k.

For the triples grouped, use

Dragon(A260748(n)) = Dragon(A260749(n)) = Dragon(A260750(n)). (I.e., they're "conformal".)

The first differences of this sequence appear to comprise only 1, 2, 3, 4, 5, 8, 11, 20, and 21. 21 occurs only twice for A(n) < 30720.

See dragun in the MATHEMATICA section for an exact evaluator of a continuous, spacefilling Dragon function, and undrag, its multivalued inverse.

Even excluding multiples of 5, it is NOT the case that A260747 contains 7*A260747, e.g., 7*13=91 is missing.

Examples

			For definiteness, we choose the Dragon in the complex plane with Dragon(0) = 0, Dragon(1) = 1, Dragon(1/3) = 1/5+2i/5
Then using A(1) = 13, for k=0,1,2, {dragun[13/15], dragun[13/30], dragun[13/60]}
-> {{2/3 - I/3}, {1/2 + I/6}, {1/6 + I/3}} (where I^2:=-1)
These have inverse images undrag/@First/@%
{{13/15}, {13/30, 7/10, 23/30}, {13/60, 7/20, 23/60}}
k=0 is too small--7/5 and 23/15 are off the end of the curve!
dragun[13/15/2^k] = dragun[7/5/2^k] = dragun[23/15/2^k], which empirically = (2/3 - I/3) (1/2 + I/2)^k
		

Links

• Brady Haran and Don Knuth, Wrong turn on the Dragon, Numberphile video (2014)
• Wikipedia, Dragon curve

Crossrefs

A260747 = A260748 U A260749 U A260750 = Superset of 3*A260482.

Programs

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 &}]
Reap[Do[If[Length[undrag[dragun[k/15/32][[1]]]] > 2, Sow[k]],{k,0,288}]][[2, 1]]

Extensions

Corrected subtle bug in NAME section, plus three tweaks to EXAMPLE. Tweaked comment. - Bill Gosper, Jul 31 2015