Season's Greetings from Susanna Cuyler and Neil Sloane

[Neil Sloane]

[David desJardins]

I haven't thought about this much, but I wonder if you could explore the entire space, for problems like this, by constructing a graph of possible configurations. A node corresponds to a configuration, and the set of all other configurations that can be connected to it by smooth movements that don't create or destroy any regions. An edge links two configurations if one has exactly one less region than the other, can be transformed to the other by smooth movements that only create that one region and do not create or destroy any other regions.

There are a bunch of annoying details, but it seems like you should be able to build this graph iteratively until every possible configuration has been mapped.

On Tue, Dec 23, 2025 at 6:57 PM Neil Sloane <njas...@gmail.com> wrote:

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[Neil Sloane]

There is a fact that will help with that approach. Say there are 4
lollipops, numbered 1 to 4. Let f(i,j) be the number of points where
lollipops i and j intersect. Then to maximize the number of regions,
it is necessary and sufficient to maximize Sum_{i<j} f(i,j). (This is
proved in the article I referenced.) It is much easier to count
intersections than regions!

Best regards
Neil

Neil J. A. Sloane, Chairman, OEIS Foundation.
Also Visiting Scientist, Math. Dept., Rutgers University,
Email: njas...@gmail.com

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[Neil Sloane]

Attached is an updated version. Further versions will appear on my homepage, http://neilsloane.com

[Alex Violette]

I did want to say, this is quite an interesting problem. I do think it's possible for one to code a simulation thing where one adds n lollipops to a 2d grid and it auto-calculates how many regions are on the board based on the size and placement of the lollipops. Heck, it can even be a website. Worth noting I do not have the coding capabilities at the moment for this task but I'm sure someone can pull it off! I'm also curious what happens for higher dimensions.

-Alex V.

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[Neil Sloane]

Dear SeqFans, At 1:02pm EST on Dec 25 2025, Jonas Karlsson, a member of this list, found the best solution so far to the four-lollipop problem: it has 44 regions. I am posting the details here with his permission.

The lollipops all have circle radius 1, and their positions are specified by three numbers [x, y, theta], where (x,y) is the exact center of the circle, and theta is the angle in radians of the ray, relative to the x-axis:
[ 0.95 , 0. , 2.61799388],
[-0.95 , 0. , 0.52359878],
[ 0.02 , 1.57 , 4.71238898],
[-0.875, 0. , 0.55850536]

Exact values in radians for the four angles are:
theta1 = 5*pi/6
theta2 = pi/6
theta3 = 3*pi/2
theta4 = 8*pi/45
The total number of intersections between the four lollipops is 39, so (by the formula in my paper with David Cutler, Eq. (6)) the number of regions is R = 39 + 4 + 1 = 44. In fact, 44 may be best-possible (the upper bound is 47), but this is not known.

Attached is Jonas's .png plot of the configuration, which can serve as a guide to the positions of the lollipops, although for a proof that the configuration is correct, one must carefully examine the 39 intersection points.

For this purpose, Jonas has provided a text file (also attached here), giving the 39 intersection points in high precision. A comment that "provenance = [('C1', 'C2')]" means that this point was obtained by intersecting circle 1 and circle 2, etc.
Jonas has verified that the resulting planar graph has all the properties that are needed.

The sequence (max. no. of regions obtainable with n lollipops) thus begins 1, 2, 10, 25, 44? 

 It is not yet in the OEIS!

I have to say that Jonas's idea of getting the fourth lollipop by jiggling one of the first three is brilliant!  I had not even dared to try it!

[Harry Neel]

Neil, et.al;

Did not do anything with the Four Lollipops at first, but went back and picked up on the stick being of infinite length and beginning at the origin of each 'Lollipop.' I sketched out a rough drawing of four lollipop's being tangencial to one another (no overlap) with the 'sticks' going outward from the center and away from each other. Then by rotating the sticks of opposing circles and not having the sticks pass through the center of the circle opposite to itself each of the four circles can be divided into two (2) regions and the space in the center portion will have six (6) regions.

That arrangement provides a total of 14 regions without involving any additional influence from two to the sticks and the circles have yet to be overlapped. Of course this only works if passing the sticks through another lollipop is allowed.

I tried to scan my sketch so that it could be provided as an attachment but was not able to do so.  (Was able to scan fine in the past now. Unless it gets created on the computer I am not currently able to do so.) 

Neil, if you wish to see the sketch I can try photographing it with my cell phone and emailing it to you directly from my phone. Trying to get photographs from my phone into the computer is almost as hard as pulling hen's teeth. (And yes, I actually met someone earlier this year that thought chickens had teeth!)

Repards and safe journeys.

Harry

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[Allan Wechsler]

The state space for two lollipops has eight dimensions -- two each for the centers of the two lollipop, one for the size of each lollipop, and one for the orientation. For larger numbers N of lollipops, the state space has 4N dimensions.

The state space is divided into cells, within which you can get from one configuration to another by a connected, continuous path through state space without changing the number or arrangement of intersections.

Each cell has a boundary: a boundary configuration is one for which any neighborhood contains configurations in more than one cell.

This whole system is a cell complex, and it feels to me like a program could probably be written to explore the complex exhaustively, enumerating all the cells and their incidence relations. Such a program could easily be modified to report the number of regions pertaining to configurations in each cell, and therefore could find, in principle, the maximum possible number of regions.

We'd have to think really hard to figure out how to represent the cells and their incidence relationships, but this doesn't feel impossible to me

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[David desJardins]

On Fri, Dec 26, 2025 at 12:19 PM Allan Wechsler <acw...@gmail.com> wrote:

The state space for two lollipops has eight dimensions -- two each for the centers of the two lollipop, one for the size of each lollipop, and one for the orientation. For larger numbers N of lollipops, the state space has 4N dimensions. 

This whole system is a cell complex, and it feels to me like a program could probably be written to explore the complex exhaustively, enumerating all the cells and their incidence relations.

I would say it only has 3N dimensions -- the first lollipop can be fixed. Also, without loss of generality, the first lollipop can have the largest radius, which makes the space compact, at least if we assume all of the other circles must intersect the first circle, which certainly seems likely for the maximum.

I agree that exploring (exhausting) the cells seems challenging but not impossible. If you want to prove that you have completely explored the boundary of a cell, that is going to be harder than empirically just probably exploring the boundary.

  -- David desJardins

[Allan Wechsler]

Probably we would want to test this technique on the analogous A250001 first. I think it will be possible to methodically explore cell complexes of this kind, and prove rigorously that the resulting census is complete, rather than hoping that a Monte Carlo exploration of a cell boundary has inspected all the weird bits. Especially for larger numbers of intersecting objects, a methodical, exhaustive technique will be necessary, since some crucial features of configurations are likely to be microscopic.

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[Joey Wheat]

Not to be a stickler for details, but wouldn't cutting a pancake with an exotic knife imply that the size of the cuts remains fixed?

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[Neil Sloane]

The lollipop sequence is now A389624. Allan (or David): how about
adding a comment there describing your ideas for using a state space
approach to actually prove something? That seems pretty important,
and should be put on record. I could try to summarize those emails,
but it would be better if you did it!

Best regards
Neil

Neil J. A. Sloane, Chairman, OEIS Foundation.
Also Visiting Scientist, Math. Dept., Rutgers University,
Email: njas...@gmail.com

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[David desJardins]

As Joey suggests, maybe you also want a sequence for fixed-radius lollipops (the known terms are the same).

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[Neil Sloane]

Joey Wheat asked:

> Not to be a stickler for details, but wouldn't cutting a pancake with an exotic knife imply that the size of the cuts remains fixed?

No, the knife or cookie cutter can be chosen from a wide range of possibilities:  take a look at the paper:
http://arxiv.org/abs/2511.15864
Title: Cutting a Pancake with an Exotic Knife
Authors: David O. H. Cutler, Neil J. A. Sloane

For a lollipop shape, the radii of the circles and the angles of the sticks can be anything you want.

Best regards
Neil

Neil J. A. Sloane, Chairman, OEIS Foundation.
Also Visiting Scientist, Math. Dept., Rutgers University,
Email: njas...@gmail.com

On Fri, Dec 26, 2025 at 4:20 PM Joey Wheat <jraym...@gmail.com> wrote:
>

Best regards

Neil 

Neil J. A. Sloane, Chairman, OEIS Foundation.

Also Visiting Scientist, Math. Dept., Rutgers University, 

Email: njas...@gmail.com

On Fri, Dec 26, 2025 at 4:20 PM Joey Wheat <jraym...@gmail.com> wrote:

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[Joey Wheat]

I had one more nagging question if it's not too late. Would the intersection of two infinite rays divide the plane into two distinct regions?

P.S. Merry Xmas to everyone :)

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[David desJardins]

On Fri, Dec 26, 2025 at 5:09 PM Joey Wheat <jraym...@gmail.com> wrote:

I had one more nagging question if it's not too late. Would the intersection of two infinite rays divide the plane into two distinct regions?

Sure, of course. Why not? For example, the ray starting at (-1,0) and pointing in the positive X direction, and the ray starting at (0,-1) and pointing in the positive Y direction, divide the plane into two distinct regions, one which comprises the upper right quadrant and the other which comprises the other three quadrants.

[Joey Wheat]

Thanks for the reply, David. I was just trying to think of some related sequences you could make by tweaking the parameters a little bit. For instance, if infinite sticks were required then the maximum number of regions for two lollipops would be eight.

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[Amarnath Krishnamurthy]

Happy New Year to all SEQ Fans

Here is a new set of sequences 

a(n) = ((f(n+1) + (f(n+2) +...+ f(n+k)) / (f(n)) is the smallest such an integer

sequences f(n) = n

2/1=1,(3+4+5)/2 = 6,(5+6+7+8+9+10+11)/4 = 14

2, 6, 3, 14, 6, 4, 9, 41, 12, 5, 15,...

sequences f(n) = T(n)= nth triangular number

3, 2, 35,10, 28, 19,...

sequences f(n) = n^2

4,70,9,1104,...

we can form another sequence set  for values of k.

Warm regards

Amarnath Murthy

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[h...@crypt.org]

David desJardins <da...@desjardins.org> wrote:
:I would say it only has 3N dimensions -- the first lollipop can be fixed.

Unless I have misunderstood what you are saying, that would be 4(N-1).

Hugo

[David desJardins]

Oops, yes of course. I was focused on N=4 so it’s the same!

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[Allan Wechsler]

It's true, I did not try to factor anything out. It wouldn't change the structure of the cell complex anyway, it would only slide the index numbers up and down. If we nail the first lollipop so that the circle is centered at the origin, say, and the stick points along the positive X axis, then we are down to 4(N-1). I don't think anything else can be factored out. I am tempted to start charting the N=2 case.