cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Showing 1-5 of 5 results.

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.

Original entry on oeis.org

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
Offset: 1

Views

Author

Bill Gosper, Jul 26 2015

Keywords

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

Examples

			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

Links

Crossrefs

Cf. A260747..A260750, A261120.

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 &}]
Do[If[Length[undrag[dragun[k/80][[1]]]] > 2, Print[k]], {k, 0, 68}]
(* same as, e.g. *)
Do[If[Length[undrag[dragun[k/20480][[1]]]] > 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. *)

Extensions

Name simplified by Bradley Klee, Aug 18 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)/(15*2^k)) = D(A(q)/(15*2^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

Views

Author

Bill Gosper, Jul 30 2015

Keywords

Comments

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".)

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
(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.

Links

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 &}]
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)/(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).

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

Views

Author

Bill Gosper, Jul 30 2015

Keywords

Comments

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.

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}}
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

Links

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 &}]
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)/(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).

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

Views

Author

Bill Gosper, Jul 30 2015

Keywords

Comments

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".)

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) = 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

Links

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 &}]
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

Views

Author

Bradley Klee, Aug 08 2015

Keywords

Comments

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

Examples

			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}.

Links

Crossrefs

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

Programs

  • Magma
    [1/14*(7-7*3^n+6*7^n): n in [1..25]]; // Vincenzo Librandi, Aug 10 2015
  • Maple
    A261120:=n->(7-7*3^n+6*7^n)/14: seq(A261120(n), n=1..30); # Wesley Ivan Hurt, Aug 27 2015
  • Mathematica
    1/14 (7 - 7*3^# + 6*7^#) & /@ Range[1, 20]
LinearRecurrence[{11, -31, 21}, {2, 17, 134}, 20]
  • PARI
    Vec(-x*(9*x^2-5*x+2)/((x-1)*(3*x-1)*(7*x-1)) + O(x^30)) \\ Colin Barker, Aug 17 2015

Formula

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)
Showing 1-5 of 5 results.

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