I've been building a puzzle game called Rule Hunter — each level gives you a

few example pairs and you have to figure out the hidden rule connecting them,

then apply it to solve the final case.

Here's one from the Normal difficulty:

3 → 9

5 → 25

8 → 64

7 → ?

What's the rule, and what's the answer?

(It's a free Android game, link in comments if anyone's curious — didn't want

to make this feel like an ad post.)

4 digits, 0-9. I was volunteering at an animal charity stall the other day, the lovely lady who put this together thought it was a great idea and she genuinely hoped someone would win it. No one did. I didn't have the heart to tell her the actual odds.

Second episode of a five part series, each about a different proof

Prime sieving via modular projection: Theta(sqrt(N)/log N) memory. Sieve operator spectrum: primes = null space (lambda=0), excited levels = GOE (discrete Hilbert-Polya). Paper + code (Python/Lean4): https://github.com/NachoPeinador/modular-projection-sieve

(I'm not a mathematician, my math is not that good, but i do my own stuff).

You all know converting decimals to base-2 or higher base (b) systems.

It is boring, so i sat down to make a formula for it(note/ i do not know if it's exit the way i did or not, i just do).

The only thing i do in math like 95% of the time is looking for a pattern, so this is the way i was thinking on how to make a formula to convert decimals into base-2 (Binary):

1- i wrote the first 17 decimal numbers(0 to 16) and their binary values, any one will see that they are all powers of 10, like 5 is 101 ( 100 + 1 or 10² + 10⁰), but i set it like this to make the math have sense, ( 10²(a) + 10¹(b) + 10⁰(c) ).

2- now we can see that it is just a sum of powers of 10 with a coefficient for each power, by setting a = 1, b = 0, c = 1, we will have the sum for 5 in binary as i showed earlier.

3- i did the same thing for other decimal numbers, i have found that the upper bound of the sum must be a function that counts how many powers of 10 must be for a number m to convert it into base-2(or how many digits in a number in base-2) , any number m can be shown as a power of 2 (2^alpha), so m = 2^alpha where 2ⁿ ≤ 2^alpha < 2^(n+1).

For example 5 = 2^alpha where alpha > than 2 slightly.

4- i took that power and did some operations on it, first we need to round it to a natural number by using the floor or ceiling function(round up or down), to deal with the power ( alpha ) itself we need to use the log_2(m) to extract that power, it also can be shown like this log_2(2^alpha).

A- If we're gonna round down ( floor(log_2(m)) ) we need to add 1 to count for the digits, like m = 5, in base_2 101 (3 digits), floor(log_2(5)) = 2 so we add 1.

B- to make math simple and avoid adding one, we can use the ceiling function instead, so f(m) = ceil(log_2(m) is gonna be the upper bound of the sum, for the lower bound i set it to k = 1, and we know from earlier that powers of 10 get smaller and smaller by 1, so it is wiser to make the power of 10 to be f(m) - k.

So the sum looks like this:

Bi(m) = sum(k=1 to f(m)) of ((10)^(f(m)-k)) * (beta_k(m)).

5- the only thing left is the coefficients of each power of 10, in m = 5 we get a = 1, b = 0, c = 1

in m = 4 we get a = 1, b = 0, c = 0

in m = 7 we get a = 1, b = 1, c = 1

Do you see the thing? If we set abcd... As digits, they are just the binary representative of that m = 5,4,7...

So we need an equation that gives us each digit of m in binary as k gets larger by 1, so we can use the floor function ( rounding down ) and modular forms for that ( i hate modular forms ) or a pure algebra for it.

After some looking and testing:

the beta_k(m) = floor(m/(2^(alpha-k)) mod 2

And we are done.

(Note/ the photo i provide have the algebraic beta_k(m) not the mod one, it has f before the () to show floor because my note app doesn't have the symbol for it)

Then i want to move on to base-3, duo to less knowledge in math, i suspect that it needs more that just replacing 2 with 3 (because of other things i worked on), but it turns out it is just replacing numbers, befor that i asked that to an AI and he explains that just replacing works, and that stupid AI generalize the whole thing, i was not hard to do that but i want it to do it.

My theory is: I will categorize mathematicians worldwide into two major types. The first type is the group of mathematicians who research breadth, and the second type is the group of mathematicians who research depth. If you don't know the true nature of these two groups, I will explain.
The group of mathematicians researching breadth are those who study and discover major branching fields of mathematics, creating incredibly important formulas like e^(iπ) + 1 = 0 or V-E+F=2.
On the other hand, the group of mathematicians researching depth are those who focus their research on just one, two, or a few major branching fields. They create many theorems and formulas, but these do not carry the same monumental weight as the work of the breadth research group. However, like the 7 algebraic identities, the formulas in this group are smaller things used frequently. This group pieces and assembles things together from different places to create difficult, unusual, tricky, and trap problems. Drilling deep down creates many profound things.
But to me, I see that in mathematics nowadays, there are not many people researching breadth anymore; instead, everyone is researching depth. Their ways of solving problems apply several different formulas and theorems together. In the past, there were many mathematicians researching breadth, but now it is entirely depth. Furthermore, I notice that great and famous mathematicians like Gauss, Newton, and Einstein followed a pattern: they together created a few research products of broad scope first, and after that, they shifted to researching and digging deeply into that specific field and its formulas”. Before formulating this theory, I had never read Professor Freeman Dyson’s work, nor did I even know who he was. I have also never studied philosophy. It was only after coming up with this concept that I researched it and discovered a similar theory already existed. I was honestly shocked to find that my thoughts aligned with such a great professor. Even though my theory might not be a 100% match with Dyson’s, I feel incredibly happy knowing that I arrived at a concept so similar to his.
I noticed that Professor Dyson named his theory 'Birds and Frogs,' which sounds much more creative than my naming of 'Breadth and Depth.' I realize Professor Dyson was immensely creative, whereas my own creativity is not as much as his because I am currently struggling with my phone addiction. I have been exposed to phones since I was about 5 years old, and counting up to now, I am really wrestling with it, though I didn't feel as addicted before turning 15.
My passion for mathematics started when I was 13, and I am currently studying with the AoPS (Art of Problem Solving) books. My dream is to become a mathematician specifically, a frog mathematician that has wings to fly.

This is for Calculus 2 applications of integration. You can toggle between the two methods and see the graph get recut

I made this for mathpath.learnhub.com
Sometimes the question asks you to use a specific method and will only allow you to see the visual for either washers/disks/shells and the integral setup only accepts one, and sometimes (like the one above) it accepts both answers.

Find two groups of at least 2 distinct natural numbers such that the sums of their squares are equal, with no number appearing on either side more than once. Find the smallest possible solution.

Whats the value of root I
(If you aren’t a nerd I = root(-1)

[EDIT: Commenters have pointed out that 53 should be 49. Somehow I missed that.]

Consider the integer sequence 0, 2, 5, 9, 11, 53, ...

(Or, if you prefer, consider any integer sequence of your choice. Who am I to judge?)

Take the first two elements: 0, 2.

If they have the same parity and are not equal, replace them with their average: 1, 1.

Append the next element from the original sequence: 1, 1, 5.

Search from right to left for a pair of adjacent non-equal elements with the same parity. Upon finding such a pair, replace them with their average: 1, 3, 3.

Repeat, searching from right to left each time, until no such adjacent pair remains: 2, 2, 3.

Append the next element from the original sequence: 2, 2, 3, 9

Continuing we get:

2, 2, 6, 6
2, 4, 4, 6
2, 4, 5, 5
3, 3, 5, 5
3, 4, 4, 5

3, 4, 4, 5, 11
3, 4, 4, 8, 8
3, 4, 6, 6, 8
3, 4, 6, 7, 7
3, 5, 5, 7, 7
3, 5, 6, 6, 7
4, 4, 6, 6, 7
4, 5, 5, 6, 7

4, 5, 5, 6, 7, 53
4, 5, 5, 6, 30, 30
4, 5, 5, 18, 18, 30
4, 5, 5, 18, 24, 24
4, 5, 5, 21, 21, 24
4, 5, 13, 13, 21, 24
4, 5, 13, 17, 17, 24
4, 5, 15, 15, 17, 24
4, 5, 15, 16, 16, 24
4, 5, 15, 16, 20, 20
4, 5, 15, 18, 18, 20
4, 5, 15, 18, 19, 19
4, 10, 10, 18, 19, 19
4, 10, 14, 14, 19, 19
4, 12, 12, 14, 19, 19
4, 12, 13, 13, 19, 19
4, 12, 13, 16, 16, 19
8, 8, 13, 16, 16, 19

The sequence I started with isn't random. It has the property that the leftmost element undergoes a replacement at some point before each new element is appended. I believe it is the minimal sequence with that property.

Play around with that. See what you find. What comes next? (Slightly different construction rules may give different answers, pick the one you find most interesting.)