Bill Gosper - cp's OEIS frontend

User: Bill Gosper

Bill Gosper has authored 45 sequences. Here are the ten most recent ones:

A309206 a(n) = (2*n)!/A309205(n).

1, 1, 1, 1, 5, 1, 1, 1, 7, 1, 1, 1, 13, 19, 5, 1, 221, 1, 1, 1, 1, 1, 13, 17, 5, 1, 47, 4913, 29, 7, 11, 53, 1, 47, 325, 13, 1147, 41, 1, 1, 41, 1081, 11, 1, 5, 1, 1, 83, 1, 1, 133, 1, 2491, 97, 5, 103, 61, 1, 1, 19, 226493, 1, 1, 1, 5, 31, 1, 1, 1, 1271, 289

Offset: 0

N. J. A. Sloane, Jul 28 2019, following a suggestion from Bill Gosper

Bill Gosper points out that this is a better fingerprint for the series than A309205.

Cf. A309204, A309205.

F[n_] := Module[{p}, p = 1 + O[x]; For[k=2, k <= n, k++, p = Cos[x p]]; p];
seq[n_] := Module[{v}, v = CoefficientList[F[n], x]; Table[(2(k - 1))!/ Denominator[v[[2k - 1]]], {k, 1, n}]];
seq[71] (* Jean-François Alcover, Aug 27 2019, from PARI *)

\\ here F(n) gives n terms of power series.
F(n)={my(p=1+O(x)); for(k=2, n, p=cos(x*p)); p}
seq(n)={my(v=Vec(F(n))); vector(n, k, (2*(k-1))!/denominator(v[2*k-1]))} \\ Andrew Howroyd, Aug 17 2019

Terms a(31) and beyond from Andrew Howroyd, Aug 17 2019

A286854 Numbers k such that k == 1 or -1 (mod 6) but k does not divide the numerator of Bernoulli(2*k).

Original entry on oeis.org

55, 253, 275, 385, 605, 715, 935, 1045, 1081, 1265, 1375, 1595, 1705, 1711, 1771, 1925, 2035, 2255, 2365, 2485, 2585, 2695, 2783, 2915, 3025, 3245, 3289, 3355, 3403, 3575, 3685, 3905, 4015, 4235, 4301, 4345, 4565, 4675, 4807, 4895, 5005, 5225, 5335, 5405, 5555

Offset: 1

Bill Gosper, Aug 01 2017

nonn

Cf. A000367, A286853 (1st differences).

isa := n -> abs(mods(n, 6)) = 1 and modp(numer(bernoulli(2*n)), n) <> 0:
select(isa, [$1..2255]); # Peter Luschny, Aug 02 2017

Select[Range@9999,0 != Mod[Numerator@BernoulliB[2 #], #] && MemberQ[{1, 5}, Mod[#, 6]] &]

isok(n) = (((n % 6) == 1) || ((n % 6) == 5)) && (numerator(bernfrac(2*n)) % n); \\ Michel Marcus, Aug 02 2017

A286853 First differences of A286854.

Original entry on oeis.org

198, 22, 110, 220, 110, 220, 110, 36, 184, 110, 220, 110, 6, 60, 154, 110, 220, 110, 120, 100, 110, 88, 132, 110, 220, 44, 66, 48, 172, 110, 220, 110, 220, 66, 44, 220, 110, 132, 88, 110, 220, 110, 70, 150, 110, 6, 148, 66, 110, 220, 110, 220, 110, 220, 110, 220, 110, 22, 198, 32, 78, 198, 22, 110, 220, 110, 210, 10, 30, 80, 220, 110, 220, 110, 66, 154, 110, 220, 110

Offset: 1

Bill Gosper, Aug 01 2017

nonn

Empirical observations:
The values seem to be a subset of {2k: k=3..110}.
Segments of consecutive values sum up to 330, like: | 198,22,110 | 220,110 | 220,110 | 36,184,110 | 220,110 | 6,60,154,110 | ... - Peter Luschny, Aug 06 2017

Peter Luschny, Table of n, a(n) for n = 1..500

Cf. A286854.

Differences@Select[Range@9999,0!=Mod[Numerator@BernoulliB[2#],#]&&MemberQ[{1,5},Mod[#,6]]&]

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

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 ... *)

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 ... *)

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 ...*)

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).

Original entry on oeis.org

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

A260482 Dragon curve triple point numerators: When a(n) in 0, 1, 2, ..., (52^k), Dragon(a(n)/(52^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

Bill Gosper, Jul 26 2015

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

			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

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

Cf. A260747..A260750, A261120.

(* 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. *)

Name simplified by Bradley Klee, Aug 18 2015

A145682 The value of the sum shown in the display appears to 2, 8, 32 - sqrt(2), 113, 382, 833, 1822, 3713, 7582, ... for n = 1, ..., 9.

Original entry on oeis.org

2, 8, 32, 113, 382, 833, 1822, 3713, 7582

Offset: 1

Bill Gosper, Apr 12 2005

nonn

This is an unusual sequence mentioned on the Math Fun Mailing list. It does not quite fit the format of regular OEIS entries, but is too interesting to be forgotten. - N. J. A. Sloane, Mar 29 2009
The definition arises from the Fourier series for the Snowflake curve.
..................................................................
.............................. inf ...............................
.............................. ==== ............ m ...............
............................ k \ ....... %pi (k 2..- 5) ..... n ..
....................... (- 1).. > .. tan(--------------) (- 1) ...
............. inf ............ / ............ n + m ..............
............. ==== ........... ==== ........ 2 ...................
..... 5 m - 1 \ .............. n = 1 .............................
.. 3 2 ....... > ...... --------------------------------------- ..
............. / .................................... m ...........
............. ==== .......... m ... 3 ... 10 %pi (k 2 .- 5) ......
............. k = - inf . (k 2..- 5)..csc(-----------------) .....
................................................ 2 m .............
............................................... 2 ................
.. ------------------------------------------------------------ ..
........................... 3 ... 5 %pi ..........................
........................ %pi..csc(-----) .........................
................................... m ............................
.................................. 2 .............................

cp's OEIS Frontend

User: Bill Gosper

A309206 a(n) = (2*n)!/A309205(n).

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A286854 Numbers k such that k == 1 or -1 (mod 6) but k does not divide the numerator of Bernoulli(2*k).

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Maple

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A286853 First differences of A286854.

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

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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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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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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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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)/(152^k)) = D(A(q)/(152^k)) = D(A(n)/(15*2^k)).

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).

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).

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).

A260482 Dragon curve triple point numerators: When a(n) in 0, 1, 2, ..., (52^k), Dragon(a(n)/(52^k)) has exactly three distinct, rational preimages.