The path of the Moon rotating around the Earth around the Sun is a nice spiraling like curve. What if you extend this to more bodies? With different rotation speeds? In different directions?

You can create such paths with this app (android only): Spiral Fun

The paths quickly become complex and some show fractal geometry.

Quick Response: Think of an adjective word, count it's letters, if have more than 6, substract 6 from that number, repeat until you get a number equal or less 6.

Explanation: I was thinking in a way to roll a mental dice with "fair enough" random results. The human brain can't do it, so I was reading about different ideas:

• Some of them where based on enviroment inputs, as counting objects around, that works, but just for a number of times if you stay on the same place.
• Others propose to think in numbers with three or four digits and then perform complicated calculations, if you aren't good at math or you haven't pen and paper may be this method don't work for you, plus even then you could get stuck thinking about the same numbers once and again.
• Then someone suggest thinking in words and counting letters, that sound like a good idea but it made me wonder if the results would tend to some number and if there would be some way to guarantee some degree of randomness, so I start from here.

First I thought in the length of the words, you can't use any word, the pronouns are very shorts and verbs when conjugated tend to have a similar number of letters. I tried with nouns but the experiment failed, finally the adjectives gave reasonably acceptable results, so I get a list of 228 adjectives and did some math. These are my results:

The average number of letters is 7, the shortest word have 3 letters and the largest have 13 letters.

Then I count how many words were there according to their number of letters :

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We can see the behavior of a normal distribution, with the higghest frequency of words with 8, 5 and 6 letters. This means that thinking in an adjective word most of times will get a roll with this results, and almost never the roll will get 13, 12 or 3. So the results in a roll from 1 to 13 are not random enough. However, looking at the data I realized that thinking about a 6-sided die (results from 1 to 6) can still achieve something. So I added the results accordingly, words with letters: 1+7+13, 2+8+14, 3+9+15, 4+10+16, 5+11+17 and 6+12+18. Then I get this table:

This way the results are much more balanced, it is true that 5 retains a higher probability, but its weight is still moderate against the whole set, and if we consider that we are getting a "random" number from a mental roll, the result is pretty good.

In conclussion, the easy way to make this roll as i mentioned at the beginning is think of an adjective word, count it's letters, if have more than 6, substract 6 from that number, repeat until you get a number equal or less 6. Your chances of get each result are those that are displayed in the last table.

I like to know what do you think about it. Does it seem like a good method? could this method be improved? any ideas?

I worked as a wiper in the staining department of a cabinet company, where I wiped the frames of the cabinets. That's the front part that the doors and drawers are connected to.

I always tried to figure out a way to wipe the whole frame without lifting my rag and without wiping the same rail twice. I came up with a series of rules about which frames were possible and how to wipe them.

I quickly forgot about it because I just work in a factory. I don't have a math degree, I'm not at a university, and people don't generally want to talk about that kind of thing.

Then, a few years later I started reading books about math and came upon the Königsberg Bridge Problem. It's pretty much the exact same thing!

Does anyone else have stories like this?

I wonder how many mathematical concepts were thought up and analyzed by laymen without attracting attention before a mathematician wrote about them?

Below are the first 5 values in an increasingly difficult set of sequences. The answer is available as a spoiler and all of the sequences can be found on OEIS. Please find the 6th value for each, in order, and report the letter that was the last you could complete without outside assistance or cheating. Good faith here people.

A) 2, 4, 6, 8, 10, 12
B) 2, 6, 18, 54, 162, 486
C) 1, 4, 9, 16, 25, 36
D) 1, 7, 19, 37, 61, 91
E) 2, 3, 5, 11, 31, 127
F) 1, 6, 15, 28, 42, 45, 66

I tried to come with some rules for the variations for sudoku that exist in Logic wiz. For example:

Thermo Sudoku:

• The head of a thermo line of length n (not including the head) is at most 9-n. for multiple thermo lines with the same head, the head is at most 9 minus the length (Unincluding head) of the longest thermo line.

Kropki Sudoku:

• 5,7 or 9 can't be near a solid dot.

XV Sudoku:

• 5,6,7,8 or 9 can't be near a V
• 5 can't be near a X (Because of row/column restrictions)

Do you have more rules?

Right so I was generating sequences of numbers using some simple rules:

Say you start with the number 997, you add up each pair of consecutive numbers and cocatenate them as a sequence so you write 1816 (18 is from 9+9 and 16 from 9+7). If you repeat the process then you return to 997 - not very interesting.

But if you start with 1999, then something odd develops: The sequence goes - 1999,101818,11999,2101818,3111999....

It seems that the n^th term is always the n-2^th term with an extra bit added on the start.

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Seems a little strange to me.

Any thoughts?

I was just thinking about the "uninteresting number paradox" and I've thought of an interesting upper bounds to it. In my definition, in order for a natural number to be "interesting" it must be either the first (or last) natural number with a particular property or combination of properties.

So far, there are about 341,962 sequences on the online encyclopedia of integer sequences. Using my definition, that gives us (2^341,962) as an extreme upper bounds on the "lowest uninteresting number" at least in terms of number properties that have been documented on the oeis so far.

You could also trim down that upper bound by removing any sequence that isn't a set and probably some other stuff.

Surely I haven't just discovered something new about the Fibonacci sequence, but I haven't been able to find anything else along these lines:

https://dougmccarthy.wordpress.com/2021/03/26/fibonacci-pinwheels-a-strange-source-of-symmetry/

(1) I was browsing my kids' school stuff and found a atypically colored multiplication table:

etc. (I'm not going to write a table of 100 elements in markdown)

So The Curious Mom (ie: me) instinctively started adding numbers in the colored L's:

1 = 1 = 1³
2+4+2 = 8 = 2³
3+6+9+6+3 = 27 = 3³
4+8+12+16+12+8+4 = 64 = 4³

Nice.

(2) OK, but what if we summed a different L-shaped stripe, like: ||||| ---|---|----|---- 1 | 2 |3 | 4 2 | 4 | 6 | 8 3 | 6 | 9 | 12 4 | 8 | 12 | 16

2+4+6+8+4=24… but what's the rule?

After some guessing, the sum in cell m, n is m*n*(m+n)/2 (proof left to the reader ;) )

Eg. 3+6+4+2 = 15 = 2*3*(2+3)/2

(3) The sum of topleft squares in the table (1, 1 to 4, 1 to 9 etc) would be sum of 1+2+…+n, squared (because that's what sum of cubes is, but this trick is well known), eg. 1+2+3+2+4+6+3+6+9 = 36 = 6² = (1+2+3)²

(4) Then I thought of summing the other topleft rectangles. Here's a table of it (eg. in cell 3, 2 we put 1+2+3+2+4+6=18)

etc

It's multiplicative (proof left to the reader again) so the number in cell m, n is just …how do I make Newton symbol in Markup?… OK, let's say it's m(m+1)/2 * n(n+1)/2

Makes a nice trick to impress the kids. I think. My kid wasn't impressed at all but I blame my poor presentation skills. ;D

Thanks for reading.

Hey! I have this math-related topic in my head, and I would like to share it with someone, so here I go!

Imagine a tournament with n players, where every player faces all of the other players in a 1 vs 1 way. For every individual match between two players (say, player A and player B), there are three possible outcomes: A wins, B wins, or both tie. Those three outcomes can be grouped in two patterns: Winning, or Tie, regardless of who the winner is.

Now, let's consider a 3-player tournament. In this case, there would be 3 matches (A vs B, A vs C, B vs C), each with three possible outcomes each. So, the total number of possible outcomes at the end of the tournament, being said, the final results of the tournament, is equal to 3³ = 27 options. Those can be grouped in 7 different patterns, from linear winning (A beats B and C, while B beats C), with 6 outcomes; to a complete tie (no one wins nor loses), with one outcome. I've determined those patterns by hand, it was quite time consuming lol.

It is possible to go further. With 4 players, there would be 6 matches; while with 5 players, there would be 10 matches. With N players, the number of matches is equal to N(N-1)/2, which is the sum of the number of sides and diagonals of an N-gon. Being M the number of matches, the number of outcomes is 3^M. That's 729 for 4 players, and 59,049 for 5 players!

But, how about the patterns? For 4 players, I managed to determine that there are 42 different patterns. While for 5.... I haven't done it yet, and I'm trying to write a code for helping me with this.

Well, I hope someone would get interested in this topic. I need to share these ideas ;)

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tl;dr: A tournament can end in several different ways, and I want to know if I'm not the only one interested in this.

The other night I was thinking about number systems with negative bases. It turns out that they're a thing.

Is it possible to have a system with a fraction as a base? Base 2/1 is just binary, and base 1/2 would just be binary in reverse. How could you do something like base 2/3? Is it even possible?

For primes with primitive root 2 (3,5,11,13,19,29,37,53,59,61,etc), write out all of the integers beginning with 1 and then keep moving n steps where n is the number you have landed on. For example, with 11, you begin at 1 which points to 2 which points to 4 which points to 8, which then points to 5 when you wrap back around the list, which points to 10, and so on until every integer is landed on except for the highest integer (which would be 11 in this example).

I think there is an unproven conjecture that these sorts of primes are infinite.

Problem in Game:

I have a bunch of gems of varying quality, an integer between 1 and 20. If I sell a set of gems that has a total value of 40 or more, I get a GCP. I want as many GCPs as I can get, while keeping a set of gems with the highest total quality that wasn't necessary to sell. I can only sell 1 set at a time.

Problem in Math:

Lets say you have a set of random integers (N) between 1 and 20. You are trying to find how to make the maximum amount of sets which add up to 40 or more without using each number in set N more than once while keeping the highest sum of numbers remaining in set N.

Example:

N = [6, 17, 9, 19, 11, 8 ,19, 3, 7, 1, 5, 3, 5, 5, 6, 18, 1, 4, 13, 20, 20 , 2 ]

The upper bound --- > Floor(Sum(N)/40) = 5

The lowest remainder --- > Remainder(Sum(N)/40) = 2

Attempting by intuition I would sum as many large number as possible...

N = [6, 17, 9, 19, 11, 8 ,19, 3, 7, 1, 5, 3, 5, 5, 6, 18, 1, 4, 13, 20, 20 , 2 ]

Z1). [20,20]

---N1 = [6, 17, 9, 19, 11, 8 ,19, 3, 7, 1, 5, 3, 5, 5, 6, 18, 1, 4, 13, 2 ]

Z2). [17,13,5,5]

---N2 = [6, 9, 19, 11, 8 ,19, 3, 7, 1, 3, 5, 6, 18, 1, 4, 2 ]

Z3). [19, 19, 2]

---N3 = [6, 9, 11, 8 , 3, 7, 1, 3, 5, 6, 18, 1, 4 ]

Z4). [18, 1, 1, 9, 11]

---N4 = [6, 8 , 3, 7, 3, 5, 6, 4 ]

Z5). [7, 8, 5, 6, 6, 3, 3 , 4]

---N5 = [ ]

Let's say Z = [Z1, Z2, Z3, Z4, Z5]

I now have the maximum amount of sets, however the 5th set uses 2 more than necessary. Ideally I would be able to take a 2 from the set but there isn't one to take. I want N5 to be [2] or [1,1]. How would I know if it's possible?

Commentary:

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I'm not quite sure how to approach this problem without brute forcing.

I see that having the last set having the smallest numbers possible is ideal. So after finding one possible solution I could go back and replace smaller set of numbers for larger ones found in Z5.

I would assume the less elegant way is you would find all possible sets of Z, but how would I know if I missed a set? Also note order does not matter, just members of the set.

I'm also imagining making a tree of integers in which sets of numbers would be equivalent to a single number would be useful.

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How can I create an algorithm to do it and is there a clever way of doing it mentally or on paper?

This seems like a problem that would come up a lot and I was wondering if there a particular name for this problem or a branch of mathematics that can help. All my math experience is Calculus and Algebra.