A260482 -id:A260482 - 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)/(152^k)) = D(A(p)/(152^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

Bill Gosper, Jul 30 2015

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.

			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

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

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

(* 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]]

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

A260750 Dragon Curve triple point upper inverses. If D:[0,1] is a Dragon curve, then if k is any integer > log_2(A(n)/15), besides n there are two smaller integers p and q with D(A(p)/(152^k)) = D(A(q)/(152^k)) = D(A(n)/(15*2^k)).

Original entry on oeis.org

23, 46, 47, 92, 83, 94, 107, 173, 184, 163, 143, 166, 188, 167, 203, 214, 329, 346, 341, 368, 333, 227, 331, 326, 293, 283, 263, 286, 287, 332, 376, 377, 323, 334, 369, 347, 383, 406, 428, 407, 658, 659, 692, 682, 736, 671, 666, 663, 661, 443, 454, 569, 662, 652, 586, 581, 573, 467, 571, 566, 533, 523, 503, 526, 527, 572, 563, 574, 587, 653, 664, 643, 752, 623, 754, 753, 646, 751, 668, 739, 761, 738, 647, 737, 683, 694, 729, 707, 743, 766, 767, 812, 856, 857, 803, 814, 849, 827, 863

Offset: 1

Bill Gosper, Jul 30 2015

nonn

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

For the triples grouped, use Dragon(A260748(n)) = Dragon(A260749(n)) = Dragon(A260750(n)). (I.e., they're "conformal".)

			For definiteness, we choose the Dragon in the complex plane with Dragon(0) = 0, Dragon(1) = 1, Dragon(1/3) = 1/5+2i/5
(I^2:=-1)  Then using A(3) = 47, for k=2,3,4, {dragun[47/60], dragun[47/120],dragun[47/240]}
-> {{2/3 + I/6}, {1/4 + (5 I)/12}, {-(1/12) + I/3}}
These have inverse images undrag/@First/@%
{{37/60, 13/20, 47/60}, {37/120, 13/40, 47/120}, {37/240, 13/80, 47/240}}
dragun[47/15/2^k] = dragun[39/15/2^k] = dragun[37/15/2^k], which empirically = (5/3 - I) (1 + I)^k 2^(-1 - k)
so every eighth point is 5/6-I/2 over a power of 16.

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

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

(* 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 &}]
DeleteDuplicates[Reap[Do[If[Length[#] > 2, Sow[15*64*#[[3]]]] &@
     undrag[dragun[k/15/64][[1]]], {k, 0, 288*3}]][[2, 1]]]
(* or 128 or 256 or ... *)

A260748 Dragon Curve triple point lower inverses. If D:[0,1] is a Dragon curve, then besides n, there are two larger integers p, q (with p < q) with D(A(n)/(152^k)) = D(A(p)/(152^k)) = D(A(q)/(15*2^k)), where k is any integer > log_2(A(q)/15).

Original entry on oeis.org

13, 26, 37, 52, 73, 74, 97, 103, 104, 111, 133, 146, 148, 157, 193, 194, 199, 206, 207, 208, 209, 217, 221, 222, 223, 231, 253, 266, 277, 292, 296, 307, 313, 314, 317, 337, 373, 386, 388, 397, 398, 409, 412, 414, 416, 417, 418, 419, 431, 433, 434, 439, 442, 444, 446, 447, 449, 457, 461, 462, 463, 471, 493, 506, 517, 532, 553, 554, 577, 583, 584, 591, 592, 613, 614, 619, 626, 627, 628, 629, 631, 634, 637, 667, 673, 674, 677, 697, 733, 746, 757, 772, 776, 787, 793, 794, 797, 817, 853

Offset: 1

Bill Gosper, Jul 30 2015

For the triples grouped, use Dragon(A260748(n)) = Dragon(A260749(n)) = Dragon(A260750(n)).  (I.e., they're "conformal".)
Unlike A260747, A260749, A260750, and A260482, the first differences show a fractal pattern of repetitions.
See dragun in the MATHEMATICA section for an exact evaluator of a continuous, spacefilling Dragon function, and undrag, its multivalued inverse.

			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}}
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[21/15/2^k] = dragun[23/15/2^k], which empirically = (2/3 - I/3) (1/2 + I/2)^k

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

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

(* 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 &}]
DeleteDuplicates[Reap[Do[If[Length[#] > 2, Sow[15*64*#[[1]]]] &@
     undrag[dragun[k/15/64][[1]]], {k, 0, 288*3}]][[2, 1]]]
(* or 128 or 256 or ...*)

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)/(152^k)) = D(A(n)/(152^k)) = D(A(q)/(15*2^k)), where k is any integer > log_2(A(q)/15).

Original entry on oeis.org

21, 42, 39, 84, 81, 78, 99, 171, 168, 113, 141, 162, 156, 159, 201, 198, 213, 342, 211, 336, 319, 219, 327, 226, 291, 233, 261, 282, 279, 324, 312, 309, 321, 318, 367, 339, 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

Offset: 1

Bill Gosper, Jul 30 2015

nonn

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

For the triples grouped, use Dragon(A260748(n)) = Dragon(A260749(n)) = Dragon(A260750(n)). (I.e., they're "conformal".)

			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) = 21, for k=1,2,3, {dragun[21/30], dragun[21/60], dragun[21/120]}
-> {{1/2 + I/6}, {1/6 + I/3}, {-1/12 + I/4}}
These have inverse images undrag/@First/@%
{{13/30, 7/10, 23/30}, {13/60, 7/20, 23/60}, {13/120, 7/40, 23/120}}
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

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

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

(* 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 &}]
DeleteDuplicates[Reap[Do[If[Length[#] > 2, Sow[15*64*#[[2]]]] &@
     undrag[dragun[k/15/64][[1]]], {k, 0, 288*3}]][[2, 1]]]
(* or 128 or 256 or ... *)

A261120 The number of distinct triple points in the set of function values FLSN(m/6/7^n), m in 0, 1, 2... 6*7^n, where FLSN:[0,1] is the "flowsnake" plane filling curve.

Original entry on oeis.org

2, 17, 134, 989, 7082, 50057, 351854, 2467349, 17284562, 121031297, 847337174, 5931714509, 41523064442, 290664639737, 2034662044094, 14242663006469, 99698727138722, 697891348251377, 4885240212600614, 34196683812727229, 239376793662659402, 1675637576559322217

Offset: 1

Bradley Klee, Aug 08 2015

One derives recurrence equations for the numbers of tiles, internal edges, internal vertices, and triple point vertices--respectively t(n), e(n), v(n), a(n)--by creating a set of planar substitution rules and proving that two triple points occur on the interior of every supertile, and that other triple points only occur at the intersection of three supertiles.
Restricting the domain to [0,1] introduces flowsnake deceptions along the boundary: the set of function values FLSN(m/6/7^n), m in 0, 1, 2... 6*7^n contains some points that would be exactly triple points if [0,1] were extended to [-infinity,infinity]. Extending the system of linear recurrence equations constrains the deception-free count to equal a(n) + 3^n . - Bradley Klee, Aug 30 2015
This sequence counts all triple points of the Q-function, up to the boundary deceptions ( cf. Klee, "A Pit of Flowsnakes" ). - Bradley Klee, Aug 30 2015

			Define one particular snowflake, or slowfake, or flowsnake on [0,1] with values:
FLSN(m/6) = {{0, 0}, {1/2, -Sqrt[3]/6}, {4/7, 2 Sqrt[3]/7}, {1/6, Sqrt[3]/6}, {1/2, Sqrt[3]/2}, {1, Sqrt[3]/3}, {1, 0}}.
There exists a(1) = 2 triple points when the denominator is 42 = 6*7:
FLSN(5/42) = FLSN(11/42) = FLSN(17/42) = {3/7, Sqrt[3]/21},
FLSN(13/42) = FLSN(31/42) = FLSN(37/42) = {5/7, 4 Sqrt[3]/21}.

Colin Barker, Table of n, a(n) for n = 1..1000
M. Beeler, R. W. Gosper, and R. Schroeppel, HAKMEM, (1972), Item 115.
B. Klee, A Pit of Flowsnakes, Complex Systems, 24, 4 (2015).
B. Klee, Flowsnake Q-Function, Wolfram Demonstrations(2015).
Index entries for linear recurrences with constant coefficients, signature (11,-31,21).

Cf. A260482, A260747, A260748, A260749, A260750, A261180, A261185.

[1/14*(7-7*3^n+6*7^n): n in [1..25]]; // Vincenzo Librandi, Aug 10 2015

A261120:=n->(7-7*3^n+6*7^n)/14: seq(A261120(n), n=1..30); # Wesley Ivan Hurt, Aug 27 2015

1/14 (7 - 7*3^# + 6*7^#) & /@ Range[1, 20]
LinearRecurrence[{11, -31, 21}, {2, 17, 134}, 20]

Vec(-x*(9*x^2-5*x+2)/((x-1)*(3*x-1)*(7*x-1)) + O(x^30)) \\ Colin Barker, Aug 17 2015

t(0)=1, e(n)=v(n)=a(n)=0,
t(n)= 7 t(n-1),
e(n)= 12 t(n-1)+ 3 e(n-1),
v(n)= 6 t(n-1) + 2 e(n-1) + v(n-1),
a(n)= 2 t(n-1) + 1/2 v(n-1).
G.f.: 1/14 (7/(1 - x) - 7/(1 - 3 x) + 6/(1 - 7 x)).
From Colin Barker, Aug 17 2015: (Start)
a(n) = (7-7*3^n+6*7^n)/14.
a(n) = 11*a(n-1)-31*a(n-2)+21*a(n-3) for n>3.
G.f.: -x*(9*x^2-5*x+2) / ((x-1)*(3*x-1)*(7*x-1)).
(End)

A276391 G.f. satisfies A(x) - 4*A(x^2) = x/(1+x).

Original entry on oeis.org

1, 3, 1, 11, 1, 3, 1, 43, 1, 3, 1, 11, 1, 3, 1, 171, 1, 3, 1, 11, 1, 3, 1, 43, 1, 3, 1, 11, 1, 3, 1, 683, 1, 3, 1, 11, 1, 3, 1, 43, 1, 3, 1, 11, 1, 3, 1, 171, 1, 3, 1, 11, 1, 3, 1, 43, 1, 3, 1, 11, 1, 3, 1, 2731, 1, 3, 1, 11, 1, 3, 1, 43, 1, 3, 1, 11, 1, 3, 1, 171, 1, 3, 1, 11, 1, 3, 1, 43, 1, 3, 1, 11, 1, 3, 1, 683, 1, 3, 1, 11, 1, 3

Offset: 1

Bill Gosper, Sep 07 2016

Describes one of the two patterns of spacings of preimages of quadruple points of the Hilbert curve, H(t), 0 <= t <= 1. If H fills the complex unit square [0,1] X [0,i], H(0)=0, H(1)=1, then 1/2 + i/4 is a quadruple point with preimages t in {5/48, 7/48, 41/48, 43/48}. If we can characterize the rest of the quadruple points along the vertical bisector 1/2 + iy, all the rest are generated recursively by the to-quadrant maps (H/i + i)/2, (H + i)/2, (H + i + 1)/2, and (i H + 2)/2. Julian Ziegler Hunts has privately observed that H = 1/2 + ir is a quadruple point for all dyadic rational r in (0,1/2). E.g., the 31 r with denominator 64, i.e., 1/64, 3/64, ..., 31/64 generate preimage 4-tuples
   {{1025, 1027, 11261, 11263}, {1037, 1039, 11249, 11251},
   {1073, 1075, 11213, 11215}, {1085, 1087, 11201, 11203},
   {1217, 1219, 11069, 11071}, {1229, 1231, 11057, 11059},
   {1265, 1267, 11021, 11023}, {1277, 1279, 11009, 11011},
   {1793, 1795, 10493, 10495}, {1805, 1807, 10481, 10483},
   {1841, 1843, 10445, 10447}, {1853, 1855, 10433, 10435},
   {1985, 1987, 10301, 10303}, {1997, 1999, 10289, 10291},
   {2033, 2035, 10253, 10255}, {2045, 2047, 10241, 10243}}/12288
  with differences
   {{1, 1, -1, -1}, {3, 3, -3, -3}, {1, 1, -1, -1}, {11, 11, -11, -11},
   {1, 1, -1, -1}, {3, 3, -3, -3}, {1, 1, -1, -1}, {43, 43, -43, -43},
   {1, 1, -1, -1}, {3, 3, -3, -3}, {1, 1, -1, -1}, {11, 11, -11, -11},
   {1, 1, -1, -1}, {3, 3, -3, -3}, {1, 1, -1, -1}}/1024
But the r in (1/2,1) are 1/6th as dense. The relevant quadruple points with denominator 2^n are 1/2 + i (6k - mod(5^n, 12))/2^n, 1 <= k < 2^n/6. E.g., if n = 6, then r is in {37/64, 43/64, 49/64, 55/64, 61/64} and the preimage 4-tuples of 1/2 + ir have differences {{-11, -11, 11, 11}, {-1, -1, 1, 1}, {-3, -3, 3, 3}, {-1, -1, 1, 1}}5/1024 (the reverse of) probably just -5*(this sequence).

			A(4) = 11. Thus
Table[unbert[1/2 + (2*4+1) I/2^n] - unbert[1/2 + (2*4-1) I/2^n], {n, 5, 9}]
{{11/256, 11/256, -11/256, -11/256},
{11/1024, 11/1024, -11/1024, -11/1024},
{11/4096, 11/4096, -11/4096, -11/4096},
{11/16384, 11/16384, -11/16384, -11/16384},
{11/65536, 11/65536, -11/65536, -11/65536}}
where unbert(H(t)) = {t}, the multivalued inverse Hilbert function (with I = sqrt(-1). See the definition of unbert[] in the MATHEMATICA section.
Note that this table must have n > 4, lest (2*4+1)/2^n > 1/2.

Alois P. Heinz, Table of n, a(n) for n = 1..16383
Bill Gosper, Connect-the-dots exact samplings of Hilbert's spacefiller
Nicholas J. Rose, Hilbert-Type Space-Filling Curves. [Wayback Machine link]

Cf. A000007, A000695, A001511, A007583, A115716, A163361, A260482.

a:= proc(n) option remember; `if`(n=0, 0,
      `if`(n::odd, 1, 4*a(n/2)-1))
    end:
seq(a(n), n=1..100); # Alois P. Heinz, Sep 07 2016

(* Cf. the numerators of Out[339], below*)
hilbert[t_] :=
piecewiserecursivefractal[t, Identity, {Min[4, 1 + Floor[4*#]]} &,
    {1 - 4*# &, 4*# - 1 &, 4*# - 2 &, 4 - 4*# &},
    {I*(1 - #)/2 &, (I + #)/2 &, (I + 1 + #)/2 &, 1 + #*I/2 &}]
(* E.g., hilbert[1/2] {1/2 + I/2} *)
unbert[z_] :=
piecewiserecursivefractal[z, Identity,
     If[0 <= Re[#] <= 1 && 0 <= Im[#] <= 1,
   Range[4], {}] &,
    {1 - 2*#/I &, 2*# - I &, 2*# - I - 1 &, (# - 1)*2/I &},
    {(1 - #)/4 &, (# + 1)/4 &, (# + 2)/4 &, 1 - #/4 &}]
(* unbert[1/2 + I/2] {1/6, 1/2, 5/6} a triple point: hilbert/@% {{1/2 + I/2}, {1/2 + I/2}, {1/2 + I/2}} *)
ClearAll[piecewiserecursivefractal];
piecewiserecursivefractal[x_, f_, which_, iters_, fns_] :=
CheckAbort[
  Check[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])),
   Abort[], {$RecursionLimit::reclim, $RecursionLimit::reclim2}],
  piecewiserecursivefractal[x, g_, which, iters, fns] =.; Abort[]]
(* For a simpler but less bulletproof version, see the MATHEMATICA section of A260482 *)
In[338]:= unbert /@ (1/2 + I Range[1/32, 15/32, 1/16])
Out[338]= {{257/3072, 259/3072, 2813/3072, 2815/3072},
             {269/3072, 271/3072, 2801/3072, 2803/3072},
             {305/3072, 307/3072, 2765/3072, 2767/3072},
             {317/3072, 319/3072, 2753/3072, 2755/3072},
             {449/3072, 451/3072, 2621/3072, 2623/3072},
             {461/3072, 463/3072, 2609/3072, 2611/3072},
             {497/3072, 499/3072, 2573/3072, 2575/3072},
             {509/3072, 511/3072, 2561/3072, 2563/3072}}
In[339]:= Differences@%
Out[339]= {{1/256, 1/256, -1/256, -1/256},
             {3/256, 3/256, -3/256, -3/256},
             {1/256, 1/256, -1/256, -1/256},
             {11/256, 11/256, -11/256, -11/256},
             {1/256, 1/256, -1/256, -1/256},
             {3/256, 3/256, -3/256, -3/256},
             {1/256, 1/256, -1/256, -1/256}}
(* Check that %338[[1]] is a quadruple point *)
In[340]:= hilbert /@ %%[[1]]
Out[340]= {{1/2 + I/32}, {1/2 + I/32}, {1/2 + I/32}, {1/2 + I/32}}
In[341]:= Select[Range[0, 1, 1/512], Length[unbert[# + I/2] > 3] &]
Out[341]= {}
(* I.e., there aren't any quadruple points on the horizontal bisector of the unit square! Other such horizontal and vertical lines of dyadic rationals intersect a dense set of quadruple points. *)
a[n_] := (2^(2*IntegerExponent[n, 2]+1) + 1)/3; Array[a, 100] (* Amiram Eldar, Dec 18 2023 *)

a(n)= fromdigits(binary(n), 4)-fromdigits(binary(n-1), 4) \\ Bill McEachen, Dec 20 2024

a(n) = (2 + 4^A001511(n))/6.
G.f.: A(x) - 4*A(x^2) = x/(1+x).
From Alois P. Heinz, Sep 07 2016: (Start)
a(2^n) = A007583(n).
a(2^n+n) = a(n) + A000007(n).
(a(2*n)+1)/4 = a(n) for n>0. (End)
a(n) = A000695(n) - A000695(n-1). - Bill McEachen, Oct 30 2020
G.f.: Sum_{k>=0} 4^k * x^(2^k) / (1 + x^(2^k)). - Ilya Gutkovskiy, Dec 14 2020

Keyword:mult added by Andrew Howroyd, Aug 06 2018

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).

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A260750 Dragon Curve triple point upper inverses. If D:[0,1] is a Dragon curve, then if k is any integer > log_2(A(n)/15), besides n there are two smaller integers p and q with D(A(p)/(15*2^k)) = D(A(q)/(15*2^k)) = D(A(n)/(15*2^k)).

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A260748 Dragon Curve triple point lower inverses. If D:[0,1] is a Dragon curve, then besides n, there are two larger integers p, q (with p < 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(A(q)/15).

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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).

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A261120 The number of distinct triple points in the set of function values FLSN(m/6/7^n), m in 0, 1, 2... 6*7^n, where FLSN:[0,1] is the "flowsnake" plane filling curve.

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Magma

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A276391 G.f. satisfies A(x) - 4*A(x^2) = x/(1+x).

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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)/(152^k)) = D(A(p)/(152^k)) = D(A(q)/(15*2^k)), where k is any integer > log_2(max(A(n),A(p),A(q))/15).

A260750 Dragon Curve triple point upper inverses. If D:[0,1] is a Dragon curve, then if k is any integer > log_2(A(n)/15), besides n there are two smaller integers p and q with D(A(p)/(152^k)) = D(A(q)/(152^k)) = D(A(n)/(15*2^k)).

A260748 Dragon Curve triple point lower inverses. If D:[0,1] is a Dragon curve, then besides n, there are two larger integers p, q (with p < q) with D(A(n)/(152^k)) = D(A(p)/(152^k)) = D(A(q)/(15*2^k)), where k is any integer > log_2(A(q)/15).

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)/(152^k)) = D(A(n)/(152^k)) = D(A(q)/(15*2^k)), where k is any integer > log_2(A(q)/15).