Adam, your winning candidate was c ~ 0.553742984573881 - 0.220041001470351 i . In view of the tessellation <http://gosper.org/In&Out.png>, have you not simultaneously solved the inradius question? r = (√3-1) |c| = .4361*999*1440836026564516021182720360465 —rwg On Fri, Apr 12, 2019 at 4:48 AM Bill Gosper <billgosper@gmail.com> wrote:
On 2019-04-11 11:05, Adam P. Goucher wrote:
Bill,
The first few decimal places of the circumradius are:
0.595860164210417909024063032937823238789
Amazing!
That's enough to see that RIES can't find a simple closed form.
Also amazing! How can such a simple construction compute an entirely new constant? Do we get another one from the same base but the "triskelion" digits triskelion <http://gosper.org/tril7image2.gif> , gosper.org/tril7image3.gif, gosper.org/tril7image4.gif? At least the inradius is easy!
And therefore the circumradius! —rwg
—rwg
Here's some Wolfram Mathematica code to obtain accurate lower bounds:
omega = Simplify[(-1)^(2/3)]; base = 3 + omega; candidates = {1}; unit = base; Do[ unit = Simplify[base unit]; candidates = DeleteDuplicates@ Flatten@Table[{1, omega, omega^2, -1, -omega, -omega^2} + base e, {e, candidates}]; candidates = Map[Simplify, candidates]; maxnorm = Max[Map[Norm, candidates]]; candidates = Select[candidates, (Norm[#] + 1 >= maxnorm) &]; Print[{Length[candidates], N[maxnorm/Norm[unit], 50]}]; , {i, 1, 100}];
It does this by a tree search on sequences of digits (from the set of sixth roots of unity, because 0 can't be in the digit expansion of an extremal point). It prunes branches whose upper bound is smaller than the currently established lower bound. Fortunately, the number of candidates seems to be O(1), independent of the iteration, so this runs relatively quickly.
(Also, the first digit is wlog assumed to be a 1.)
Best wishes,
Adam P. Goucher
On Thu, Apr 11, 2019 at 7:29 AM Bill Gosper <billgosper@gmail.com> wrote:
On 2019-04-10 06:10, Joerg Arndt wrote:
Pics or it didn't happen!
Agreed: https://www.wolframcloud.com/objects/797a1110-f0cd-47a5-90f6-f0b3d1d8ce4b
Joking aside: I'd like to understand, but don't.
Best regards, jj
* Bill Gosper <billgosper@gmail.com> [Apr 08. 2019 15:50]:
On Sat, Apr 6, 2019 at 9:20 AM Bill Gosper <billgosper@gmail.com> wrote:
Mandelbrot detested the name. Then he discovered a spacefill of Koch's Snowflake. The worthy claimant to the Flowsnake name! Which he still detested. I recently realized how little I know about my own curve, which fills the FranceFlake, the fundamental region for the positional number system Base: 2 + i^(2/3) = √7 e^(i arccos(11/14)/2) and the (necessarily) seven digits 0 and the 6th roots of 1.
SPOILERS
If FranceFill(0)= 0+0i, FranceFill(1) = 1 + 0i: 1) What is the area of a FranceFlake?
Only √3/2.
2) Where is its centroid?
1/2+i/2/√3
3) Is its circumradius √(52/147) ?
No. ≥ √(6140317/3)/2401. (What the heck is this approaching?) The same as the FranceFlake, duh.
4) Ironically, it was popular partly for how completely its canonical sampling (at 0/7, 1/7,...,1) self-avoided. But like all true spacefills, it is *dense* with *triple* points. E.g., what are the preimages of (9 + i √3)/21?
{5/42, 11/42, 17/42} Are there quadruple points? How can we prove "No."?
—rwg