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A276391
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G.f. satisfies A(x) - 4*A(x^2) = x/(1+x).
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3
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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
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OFFSET
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1,2
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COMMENTS
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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).
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LINKS
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FORMULA
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G.f.: A(x) - 4*A(x^2) = x/(1+x).
(a(2*n)+1)/4 = a(n) for n>0. (End)
G.f.: Sum_{k>=0} 4^k * x^(2^k) / (1 + x^(2^k)). - Ilya Gutkovskiy, Dec 14 2020
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EXAMPLE
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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.
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MAPLE
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a:= proc(n) option remember; `if`(n=0, 0,
`if`(n::odd, 1, 4*a(n/2)-1))
end:
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MATHEMATICA
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(* 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 *)
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CROSSREFS
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KEYWORD
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nonn,easy,mult
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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