When you go up the natural numbers ladder, there are (on average) bigger and bigger gaps between consecutive prime numbers.

If you look at primes as islands, there are  lakes of composites between them.

The lake between 19 and 23 is 3 composite numbers long, for example. Or between 89 and 97 there are 7 composites.

Those lakes are greater and greater on average, as it is well known. Albeit there are always small ponds of just 5 composite numbers long, no matter how far you go. (There are likely infinity many lakes with just 1 composite, too. Which is another form of the so called Twin Primes Conjecture. A few years ago it has been proved that there are always less than 70000000 large lakes as you go up the naturals. Some time later, that 70 million has been improved to just 5, but not yet to 1.)

Say, that you have just sailed across some recordly wide composite lake and you are on a prime island again. What can you expect, how much wider will the next record lake be? Usually 2 more composites wider than the previous largest, right?

No. Every new record is usually quite greater from the previous one.

1,2,2,2,6,4,2,2,12,2,8,8,20,14,10,16,2,4,14,16,6,26,30,10,2,12,14,2,32 ...

These are the increments of every new record composite lake.

P.S. Haven’t been able to Google this story.


Then, have been able to Google some about this. The so called maximal prime gap goes like this, says Wikipedia:

Not as smoothly and gradually as those three coloured functions predict, but quite abruptly in reality.













Set Theory Problem

There are at most aleph-zero disjunct 3D spheres in 3D space. And there are at least aleph-one disjunct 2D circles in every finite volume part of the 3D space?

The number of points in N dimensional space is always aleph-one. And you can also divide this space into aleph-one disjunct N-1 dimensional spheres.

Is there a way to divide 3D space into aleph-one cubes with no common volume? They may touch each other, but may not share some common volume.

If you can find one such space division, you brought down the ZF Set Theory.


Discussion there:

logic, mathematics, physics

Another Intermezzo Intermezzo Problem

The first one was solved by Oscar Cunningham and the point of the solution is that every probable black hole lasts for at least a million years longer, if you add just one hydrogen atom into it. Of course, a supermassive black hole acquires even much longer additional time with every hydrogen atom acquisition, or shall we say — a merger with a single hydrogen atom.

It is reasonable to assume, that our domestic SMBH’s lifespan is prolonged by 10^60+ years per second, due the regular infall. Assuming also, that there will be no Big Rip or something equally unpleasant for black holes in the future.

While we still don’t know the definite answer to the Intermezzo problem, here is another inter-intermezzo.

Are there non synthesizable molecules, which would be stable if they somehow managed to pop up into existence by magic or by some weird quantum fluctuation? (Non-synthesizable-repeatedly from chemistry/physics/mathematics/logic reasons.)








artificial intelligence, mathematics

Intermezzo Problem Solution(s)

Below there is a problem I invented a month or so ago. I published a link on Lesswrong and the solving began.

The first solution was 4 hexagons by LW user 9eB1. He later improved it by, quote:

 right triangle that’s half the area of the equilateral triangle. You can fit 4 in the square and two in the triangle, and the score is point six

Oscar Cunningham came with this, quote:

why not get a really good score by taking a completely gigantic shape that doesn’t cover anything

This is a clever but also somehow trivial solution, we have agreed.

So he came with thi:


The thin red line is the uncovered area of the square, while the triangle can be tiled perfectly. The score is 0.57756.

Then I made I little promise, which has been broken already, that I will publish my solution on Monday, which is about two weeks ago by now.

The whole time, I was digitally evolving solutions on a computer. Just as I was not very far from Oscar’s solution he stroke again, with a much better one. More than twice as good as the previous one. This time he perfectly tiled the square and left some triangle uncovered. The score is 0.249..

This morning I almost decided to quit, when I saw something unusual on the screen. At first, I thought something wasn’t right. Then I realized what the damn computer is telling me.

The computer was evolving the following picture (visually rephrased by me for clearness).


The perfect score 0, albeit a bit trivial solution perhaps. Of course it is not necessary that you join the triangle and the square together. You can split the covering shape. Which computer did.


The problem is with the Oscar’s last solution. When I put his algorithm into the machine, it showed an error in the form of a negative result.

Well, the algorithm is okay, but 1/7 and 1/4 can’t be. But 1/7 and 1/12 can be. But the result isn’t that good.


Intermezzo Problem

We have the unit (edge length equal 1)  square and the unit edge equilateral  triangle.

Choose a shape, such that many instances of that shape (maybe flipped or rotated) tile both polygons. There must be no overlaps and no part of the covering shape may lie outside both polygons.

A shape that leaves a smaller percentage of its own area uncovered is a better solution.

Which shape is the best covering solution?