(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.
