cp's OEIS Frontend

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).

%I A260747 #24 Dec 06 2019 21:44:09
%S A260747 13,21,23,26,37,39,42,46,47,52,73,74,78,81,83,84,92,94,97,99,103,104,
%T A260747 107,111,113,133,141,143,146,148,156,157,159,162,163,166,167,168,171,
%U A260747 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
%N 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).
%C A260747 It appears that every Dragon triple point is an image of A(n)/(15*2^k) for three different n and some k.
%C A260747 For the triples grouped, use
%C A260747 Dragon(A260748(n)) = Dragon(A260749(n)) = Dragon(A260750(n)).  (I.e., they're "conformal".)
%C A260747 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.
%C A260747 See dragun in the MATHEMATICA section for an exact evaluator of a continuous, spacefilling Dragon function, and undrag, its multivalued inverse.
%C A260747 Even excluding multiples of 5, it is NOT the case that A260747 contains 7*A260747, e.g., 7*13=91 is missing.
%H A260747 Brady Haran and Don Knuth, <a href="https://www.youtube.com/watch?v=v678Em6qyzk">Wrong turn on the Dragon</a>, Numberphile video (2014)
%H A260747 Wikipedia, <a href="http://en.wikipedia.org/wiki/Dragon_curve">Dragon curve</a>
%e A260747 For definiteness, we choose the Dragon in the complex plane with Dragon(0) = 0, Dragon(1) = 1, Dragon(1/3) = 1/5+2i/5
%e A260747 Then using A(1) = 13, for k=0,1,2, {dragun[13/15], dragun[13/30], dragun[13/60]}
%e A260747 -> {{2/3 - I/3}, {1/2 + I/6}, {1/6 + I/3}} (where I^2:=-1)
%e A260747 These have inverse images undrag/@First/@%
%e A260747 {{13/15}, {13/30, 7/10, 23/30}, {13/60, 7/20, 23/60}}
%e A260747 k=0 is too small--7/5 and 23/15 are off the end of the curve!
%e A260747 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
%t A260747 (* by Julian Ziegler Hunts *)
%t A260747 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]));
%t A260747 dragun[t_] := piecewiserecursivefractal[t, Identity, Piecewise[{{{1}, 0 <= # <= 1/2}, {{2}, 1/2 <= # <= 1}}, {}] &, {2*# &, 2*(1 - #) &}, {(1 + I)*#/2 &, (I - 1)*#/2 + 1 &}]
%t A260747 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 &}]
%t A260747 Reap[Do[If[Length[undrag[dragun[k/15/32][[1]]]] > 2, Sow[k]],{k,0,288}]][[2, 1]]
%Y A260747 A260747 = A260748 U A260749 U A260750 = Superset of 3*A260482.
%K A260747 nonn,frac,obsc
%O A260747 1,1
%A A260747 _Bill Gosper_, Jul 30 2015
%E A260747 Corrected subtle bug in NAME section, plus three tweaks to EXAMPLE. Tweaked comment. - _Bill Gosper_, Jul 31 2015