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A260749 Dragon Curve triple point middle inverses. If D:[0,1] is a Dragon curve, then besides n, there are two other integers p and q (with p < n < q) with D(A(p)/(15*2^k)) = D(A(n)/(15*2^k)) = D(A(q)/(15*2^k)), where k is any integer > log_2(A(q)/15).

%I A260749 #14 Dec 06 2019 21:44:24
%S A260749 21,42,39,84,81,78,99,171,168,113,141,162,156,159,201,198,213,342,211,
%T A260749 336,319,219,327,226,291,233,261,282,279,324,312,309,321,318,367,339,
%U A260749 381,402,396,399,426,423,684,422,672,421,638,649,657,441,438,453,654,452,582,451,559,459,567,466,531,473,501,522,519,564,561,558,579,651,648,593,624,621,618,749,642,641,636,633,747,734,639,669,681,678,727,699,741,762,759,804,792,789,801,798,847,819,861
%N A260749 Dragon Curve triple point middle inverses. If D:[0,1] is a Dragon curve,  then besides n, there are two other integers p and q (with p < n < q) with D(A(p)/(15*2^k)) = D(A(n)/(15*2^k)) = D(A(q)/(15*2^k)), where k is any integer > log_2(A(q)/15).
%C A260749 See dragun in the MATHEMATICA section for an exact evaluator of a continuous, spacefilling Dragon function, and undrag, its multivalued inverse.
%C A260749 For the triples grouped, use Dragon(A260748(n)) = Dragon(A260749(n)) = Dragon(A260750(n)).  (I.e., they're "conformal".)
%H A260749 Brady Haran and Don Knuth, <a href="https://www.youtube.com/watch?v=v678Em6qyzk">Wrong turn on the Dragon</a>, Numberphile video (2014)
%H A260749 Wikipedia, <a href="http://en.wikipedia.org/wiki/Dragon_curve">Dragon curve</a>
%e A260749 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 A260749 Then using A(1) = 21, for k=1,2,3, {dragun[21/30], dragun[21/60], dragun[21/120]}
%e A260749 -> {{1/2 + I/6}, {1/6 + I/3}, {-1/12 + I/4}}
%e A260749 These have inverse images undrag/@First/@%
%e A260749 {{13/30, 7/10, 23/30}, {13/60, 7/20, 23/60}, {13/120, 7/40, 23/120}}
%e A260749 dragun[21/15/2^k] = dragun[13/15/2^k] = dragun[23/15/2^k], which empirically = (2/3 - I/3) (1/2 + I/2)^k
%t A260749 (* by Julian Ziegler Hunts *)
%t A260749 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 A260749 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 A260749 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 A260749 DeleteDuplicates[Reap[Do[If[Length[#] > 2, Sow[15*64*#[[2]]]] &@
%t A260749      undrag[dragun[k/15/64][[1]]], {k, 0, 288*3}]][[2, 1]]]
%t A260749 (* or 128 or 256 or ... *)
%Y A260749 A260747 = A260748 U A260749 U A260750 = Superset of 3*A260482.
%K A260749 nonn
%O A260749 1,1
%A A260749 _Bill Gosper_, Jul 30 2015