The MERSENNE PRIMES formula is part of the following formula 1. Let a, b, n be natural numbers with a>b - {[a^(a-b)]-[b^(a-b)]}/[(a-b)^2] is always a natural number. If {[a^(a-b)]-[b^(a-b)]}/[(a-b)^2] is prime then a-b is prime. If a-b is composite, then {[a^(a-b)]-[b^(a-b)]}/[(a-b)^2] is composite. - If [(a^n)-(b^n)]/(a-b) is prime, then n is prime. If n is composite then [(a^n)-(b^n)]/(a-b) is composite. 2. Let a, b, n be natural numbers where n is odd - With a+b being odd, {[a^(a+b)]+[b^(a+b)]}/[(a+b)^2] is always a natural number. If {[a^(a+b)]+[b^(a+b)]}/[(a+b)^2] is prime then a+b is prime. If a+b is composite then {[a^(a+b)]+[b^(a+b)]}/[(a+b)^2] is composite. - If [(a^n)+(b^n)]/(a+b) is prime, then n is prime. If n is composite, then [(a^n)+(b^n)]/(a+b) is composite.
Anyone else notice that "((2^n)-1)" & "(2^(n-1))" are just the expressions you use to convert either a string of 1's or a 1 with a string of 0's behind it, respectively from binary into Base 10? No? Just me? Cool... Worked it out on my own, even! [While investigating some Base10 vs. Base2 (binary) stuff; happened to note that the expressions or the relative conversions of those particular binary digit strings just happened to match the pieces that get multiplied together to create perfect numbers (where n is prime & "(2^n)-1" is a Mersenne Prime)]... ;) Interesting!
Haven't actually watched the video yet, so no idea if it's covered or not... If so, cool. If not, well... Interesting factoid! For whatever it's worth. Surely I'm not the first to figure this out? [Extremely doubtful...] :P ;)
Dude! We both do similar type of content in my case i use processing language.Found u from r/mathvisualized. Anyways my only criticism is that u should make the boxes bolder the white lines were too faint to see and the "7" inside each was too small, but overall really solid video keep up the good work man!
I'm really enjoying these animated proof videos, please keep making them!
Nice channel, I am so confused because this channel is not well known.
The MERSENNE PRIMES formula is part of the following formula
1. Let a, b, n be natural numbers with a>b
- {[a^(a-b)]-[b^(a-b)]}/[(a-b)^2] is always a natural number. If {[a^(a-b)]-[b^(a-b)]}/[(a-b)^2] is prime then a-b is prime. If a-b is composite, then {[a^(a-b)]-[b^(a-b)]}/[(a-b)^2] is composite.
- If [(a^n)-(b^n)]/(a-b) is prime, then n is prime. If n is composite then [(a^n)-(b^n)]/(a-b) is composite.
2. Let a, b, n be natural numbers where n is odd
- With a+b being odd, {[a^(a+b)]+[b^(a+b)]}/[(a+b)^2] is always a natural number. If {[a^(a+b)]+[b^(a+b)]}/[(a+b)^2] is prime then a+b is prime. If a+b is composite then {[a^(a+b)]+[b^(a+b)]}/[(a+b)^2] is composite.
- If [(a^n)+(b^n)]/(a+b) is prime, then n is prime. If n is composite, then [(a^n)+(b^n)]/(a+b) is composite.
Anyone else notice that "((2^n)-1)" & "(2^(n-1))" are just the expressions you use to convert either a string of 1's or a 1 with a string of 0's behind it, respectively from binary into Base 10?
No? Just me? Cool...
Worked it out on my own, even! [While investigating some Base10 vs. Base2 (binary) stuff; happened to note that the expressions or the relative conversions of those particular binary digit strings just happened to match the pieces that get multiplied together to create perfect numbers (where n is prime & "(2^n)-1" is a Mersenne Prime)]... ;)
Interesting!
Haven't actually watched the video yet, so no idea if it's covered or not... If so, cool. If not, well... Interesting factoid! For whatever it's worth. Surely I'm not the first to figure this out? [Extremely doubtful...] :P ;)
1:20 - 1:38 u referencing polya ?
the proportion between looks and content is too large
Dude! We both do similar type of content in my case i use processing language.Found u from r/mathvisualized. Anyways my only criticism is that u should make the boxes bolder the white lines were too faint to see and the "7" inside each was too small, but overall really solid video keep up the good work man!
very nice