a(n) is the number of integer tuples (x_1, x_2, ..., x_(k+1)) where 0 <= x_i <= b-1, such that |x_i - x_(i+1)| = d_i for all i, where (d_1, d_2, ..., d_k) are digits of n in base b.

Animation varies base b from 2 to 50 (grid size is b^2) and the mapping is (x,y) gets a(x xor y) mod 4 which is visualised by color.

When I woke up today, the spoonerism was already in my mind, so I made this.

https://imgur.com/a/mqVMuyN

the goal is to make a spiral spiraling around a spiral thats spiraling around a spiral thats spiraling around a spiral until my computer crashes.

sometimes a loop of loops loop around the looping loops loop looped too

To be clear, I made the payment of $1,041.38 in May and the final installment payment of $890.50 will be paid on Friday. I've called three times and each time I get a little different color smoke blow up my exit hole.

Hi

If you are remotely interested in deep diving how differently quantum computers work compared to our transistor-based and also the algebra behind in a fully interactive way that teach computer science from scratch, oh boy this is for you. I am the Dev behind Quantum Odyssey (AMA! I love taking qs) - worked on it for about 10 years (3+ during PhD, the visual method I developed ended up being my thesis, it is a complete Hilbert space visualizer), the goal was to make a super immersive space for anyone to learn quantum computing through zachlike (open-ended) logic puzzles and compete on leaderboards and lots of community made content on finding the most optimal quantum algorithms. The game has a unique set of visuals capable to represent any sort of quantum dynamics for any number of qubits and this is pretty much what makes it now possible for anybody 12yo+ to actually learn quantum logic without having to worry at all about the mathematics behind.

This is a game super different than what you'd normally expect in a programming/ logic puzzle game, so try it with an open mind.

Stuff you'll play & learn a ton about

• Boolean Logic – bits, operators (NAND, OR, XOR, AND…), and classical arithmetic (adders). Learn how these can combine to build anything classical. You will learn to port these to a quantum computer.
• Quantum Logic – qubits, the math behind them (linear algebra, SU(2), complex numbers), all Turing-complete gates (beyond Clifford set), and make tensors to evolve systems. Freely combine or create your own gates to build anything you can imagine using polar or complex numbers.
• Quantum Phenomena – storing and retrieving information in the X, Y, Z bases; superposition (pure and mixed states), interference, entanglement, the no-cloning rule, reversibility, and how the measurement basis changes what you see.
• Core Quantum Tricks – phase kickback, amplitude amplification, storing information in phase and retrieving it through interference, build custom gates and tensors, and define any entanglement scenario. (Control logic is handled separately from other gates.)
• Famous Quantum Algorithms – explore Deutsch–Jozsa, Grover’s search, quantum Fourier transforms, Bernstein–Vazirani, and more.
• Build & See Quantum Algorithms in Action – instead of just writing/ reading equations, make & watch algorithms unfold step by step so they become clear, visual, and unforgettable. Quantum Odyssey is built to grow into a full universal quantum computing learning platform. If a universal quantum computer can do it, we aim to bring it into the game, so your quantum journey never ends.

Nice to watch:

Khan academy style tutorials in qm/qc: https://www.youtube.com/@MackAttackx

Physics teacher stream with 400hs in https://www.twitch.tv/beardhero

A rare sighting of the tree function in the wild

In liquid crystal they experimentally get point-like quantized topological charges with Coulomb-like interactions ( https://www.nature.com/articles/s41598-017-16200-z ), or topological vortices and their various size knots resembling nuclei ( https://www.nature.com/articles/s41567-025-03107-0 ). Here are Fable-based tests of various models of particles, and liquid crystal-based wins.

So I wanted to propose a discussion if particles around could be of topological nature?

1st is with q 2 and 3 same equation without q Last 2 is within prize rules

https://oeis.org/A065825 https://mathworld.wolfram.com/NonaveragingSequence.html

The Swatch Color Function announcement has lots of pictures, including an interactive Voronoi diagram of named colors. You can also download all the data from the Swatch function itself.

Sorry for extremely bad handwriting.

I’m very curious what would be the opposite of 0, if real number line formed into a loop. So i thought of this diagram. What’s yall’s thoughts on this ‘Ω’?

In this diagram i wanted none of the infinitesimals to equal to 0, 0 is absolutely 0. (It should’ve been dotted line around 0)

I saw this posted on Reddit recently as a ripped clip with zero credit to either the mathematician or the YouTube channel it came from, so wanted to put up a proper version.

What it is: Take any 4-digit number that isn't all the same digit (leading zeros allowed, so 0026 counts). Arrange its digits into the biggest number possible and the smallest number possible, then subtract the smaller from the bigger. Repeat with the result. However you start, you'll land on 6174 within 7 steps — and once you're there, it just keeps producing 6174 forever (7641 − 1467 = 6174). It's named after D. R. Kaprekar, the Indian mathematician who discovered it in 1949.

Credits:

• Discovered by D. R. Kaprekar (1955, Scripta Mathematica)
• Explained by Roger Bowley (University of Nottingham) for Numberphile, created by Brady Haran: https://youtu.be/d8TRcZklX_Q
• Numberphile's channel: https://www.youtube.com/@numberphile
• Background/further reading: https://en.wikipedia.org/wiki/6174

I got curious enough to build a spreadsheet running every possible 4-digit number through it — turns out the fastest converge in 1 step (e.g. 0026) and the slowest take the full 7 (e.g. 9985). I've written all of this up on my own site with full credit, some background on the maths, and a live calculator where you can type in any 4-digit number and watch it walk to 6174 step by step, plus a browsable table of all 9,990 results: [https://eyepeasea.org/KaprekarConstant.aspx]

Thumbnail picture is curtesy of Wikipedia.

From

———————————————————————

Wikimedia Commons — Möbius Ladder

https://upload.wikimedia.org/wikipedia/commons/7/71/Moebius-ladder-16-animated.svg

———————————————————————

✦

Surely there's a correspondence between this equivalence & the projective plane commonality of a Möbius strip & the sphere with antipodal points identified.

From

———————————————————————

Small cycle covers of 3-connected cubic graphs

by

Fan Yang & Xiangwen Li

https://www.sciencedirect.com/science/article/pii/S0012365X10004000?ref=cra_js_challenge&fr=RR-1

———————————————————————

❝

Theorem 1.1. For n ≥ 8 , a 3-connected simple cubic graph G with n vertices has a cycle cover of size at most ⌈ n/6 ⌉ if and only if G ∉ F .

{My interposition: F being the set of five graphs shown here.}

Theorem 1.1 is sharp in the sense that there are 3-connected simple cubic graphs on n vertices having no cycle cover of size less than the upper bound ⌈ n/6 ⌉ . As examples, let n = 2m and let Cₘ × K₂ denote the Cartesian product of an m-cycle and K₂ . When m ∈ {4, 6} , it can be verified that Cₘ × K₂ has no cycle cover with fewer than ⌈ n/6 ⌉ cycles, and so the upper bound ⌈ n/6 ⌉ cannot be decreased. However, we do not know any infinite families of graphs for which the bound of Theorem 1.1 cannot be improved.

❞

From

———————————————————————

COMPUTING SMALL UNIT-DISTANCE GRAPHS WITH CHROMATIC NUMBER 5

by

MARIJN JH HEULE

https://arxiv.org/abs/1805.12181

———————————————————————

✦

The Hadwiger–Nelson problem queries the number of colours reauired for a colouring of the plane such that there shall be no two points unit distance apart & @ the same colour, which is often referenced as the chromatic number of the plane . This problem has transpired to be incredibly difficult to solve ᐝ ... & for a long time the best result was that it's @least 4 & @most 7 .

ᐝ ... & is, ImO, an outstanding of one of those problems that're colossally disproportionately difficult to solve relative to how difficult one's intuition might lead one to imagine they would be to solve.

But a few years ago the goodly Aubry de Grey found a unit-distance graph on 1581 vertices that has a chromatic № of 5 . This actually shows that the chromatic number of the plane is @least 5 , because if the plane could be coloured with 4 colours such that there shall be no two points unit distance apart & @ the same colour, then such a graph could not exist.

But it's then natural to ask whether there are unit-distance graphs on fewer vertices & yet still having a chromatic № of 5 . And I didn't look for quite a while ... but I find, on looking again now, that the goodly Marijn Heule has found a couple: there's the one shown here; & there's also one on 610 vertices, also shown in the paper that's the source of this one.

Checking-out the paper itself is strongly recomment, as the images in the PDF are @ far greater resolution than can be shown as a single image here ... & they can be zoompten-into @will. There are also other graphs shown that have a bearing on the methods by which these two mentioned smaller unit-distance chromatic-№-5 graphs were found by Dr Heule.

From Get Snarky: The Cycle Double Cover Conjecture -- an AI Proof?

Playlist link : Perfect loops 🔁

I've got something that i stumbled upon and found really interesting that I'd like to share.

Let a(b,n) be the number of integer tuples (x1, x2, ..., x{k+1}) where 0 <= x{i}<= b-1, such that |x{i}- x{i+1}| = d{i} for all i, where (d1, d2, ..., d{k}) are digits of n in base b.

Now consider the iterative definition a(b{m},n) = b{m+1}, with starting value (b{0},n). For any given starting value the sequence of terms a(b{0},n),a(b{1},n),a(b{2},n),... will either enter into a loop or shoot off to infinity.

This can be visualised on a 2d grid by taking the initial values (b{0},n) as the coordinate of the cells which we'd colour black if the sequence explodes and white if the sequence falls in a loop.

Surprisingly it has the pattern as shown in image1.

changing the definition of a(b,n) to ,say a(b,n) = (b xor n) + abs(b-n) gives image2.

Image 3,4,and 5 are result of other formulas that are comparatively complex(result of algorithm made to search the state space of all possible formulas for intersting patterns)

Three images, left to right A B I, representing matrices where 0=black, 1=green, 2=blue.

When a matrix multiplied by itself in modular arithmetic generates an alternating sequence of two distinct matrices, this phenomenon is generally referred to as an involutory matrix (if the two matrices are the original matrix and the identity matrix) or a matrix with a finite cyclic period of 2.

Because modular arithmetic limits the values inside the matrix to a finite set (e.g., modulo (n)), the sequence of powers is guaranteed to become periodic by the Pigeonhole Principle.

When the sequence alternates exclusively between two matrices, A and B, it means

A x A ≡ B mod (n)

B x B ≡ A mod (n)

A x B ≡ I mod (n) (where I is the identity matrix)

This behaviour is essentially a cyclic group of order 2 acting under standard matrix multiplication restricted by a modular arithmetic system.

In any triangle the center of weight lies on the straight line joining any angle to the middle point of the opposite side.

Givens: Triangle ABC with base BC, midpoint D on BC, and centerline AD.

I say that the center of weight is somewhere on centerline AD.

The proof is a reductio ad absurdum. Suppose a point H is the center of weight. Draw HI parallel to CB meeting AD at point I. If we bisect DC, then bisect the halves, and continue the process, we eventually arrive at a length DE that is hypothetically less than HI. Then divide BD and DC into lengths each equal to DE. Through the points of division draw lines parallel to DA and meeting sides BA and AC at points K, L, M, and N, P, Q respectively. Now join points M to N, L to P, and K to Q. The lines will be parallel to BC. This gives us a series of parallelograms: FQ, TP, and SN. AD bisects opposite sides in each of them so that the center of weight- of each individually as well as of the sum of them all- is on AD. [I.9]

Suppose O is the center of weight that sum. Join points O and H. Draw CV parallel to DA and produce OH so it meets CV at V.

Now, if n stands for the number of parts the side AC was divided into, then we get these ratios:

triangle ADC:(triangle ARN+the triangle on NP+the triangle on PQ+the triangle on QC)

=AC²:(AN²+NP²+PQ²+QC²)

=n²:n

=n:1

=AC:AN.

Similarly,

triangle ABD:(triangle AMR+triangle MLS+triangle LKT+triangle KBF)

=AB:AM.

And AC:AN=AB:AM.

Therefore

the whole triangle ABC:(the sum of all the little triangles)

=CA:AN

>VO:OH. [Through parallelism.]

Now produce OV to point X so that

triangle ABC:(the sum of little triangles)

=XO:OH

which, separando, makes

(the sum of parallelograms):(the sum of little triangles)

=XH:HO.

Because the center of weight of the whole triangle ABC is supposedly at H, while the center of weight of the part of triangle ABC made up of parallelograms is at O, it follows that the center of gravity of the remaining part which is made up of little triangles is at X. [I.8]

But that's absurd since the part made up of the little triangles is now on one side of the line that passes through X parallel to AD.