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what is the GCD algorithm

gcd(a, b) = gcd(a-b, b) = gcd(a, b-a)

i have been messing with that
but no success..

1/2, 2, 1/2, 1

okay
(p+1)/2 = p/2 + 1/2,
and p-1 = 2*(p/2 - 1/2)

|dw:1440466619213:dw|

i think showing this pattern recurring is better

we can show that all oodds will be 1 and 2 repeating

let p=2n+1

GCD((2n+1)+1)/2,2n-1)

\(p\) is given as an odd prime
so..

GCD((2n+1)+1)/2,2n-1) = 1 or 2 in fact we can evn do 1 step better by changing n=2k and 2k+1

to show every 2nd add is 2 and every 1st odd is 1

Oh! for odd numbers also it is giving 1 or 2
didn't notice before.. interesting..

right

oops i mean (2n-1)-1

|dw:1440467089588:dw|

or 1

lets see every 4th odd is 1
when odd of form 4k+1
when odd of form 4k+3 we have 2s

Nice! it works perfectly!
GCD((2n+1)+1)/2,2n-1) = GCD(n+1, 2n) = GCD(n+1, n-1) = GCD(2, n-1)

ooo okay that looks neater

when odd of form 4k+1 we have 1s as GCD
when odd of form 4k+3 we have 2s

if p = 4k+1, then
(p+1)/2 = (4k+1 + 1)/2 = 2k+1
clearly 2 doesn't divide 2k+1, so gcd = 1

yep what u solved for already shows

(2,n-1)

every 2nd n has 2, every 1st n has 1 gcd

4k+3*

http://prntscr.com/88jluw this line u did here pretty much is 100% completed work

okay now to figure out something how the order of 1s and 2s follow for primes that will be something

feels almost as complex as knowing the next prime number itself

yeah

but u dont know if its a 1 or a 2

it would be interesting if it was like always alternating perfectly

i am pretty sure its quite random, just like primes

ohk you're thinking of sorting them..

it would be interesting if u could prove something based on density of primes and stuff

|dw:1440468566168:dw|

the mysteriess

lol

oh true

oh cool

I wonder if there are more primes of the form 6n+1 or 6n+5 while we're at it.

Fascinating, the difference between them in the "race" isn't very far all the way up to 100,000 cool

Might be time to make some programs :P

we should consider log sets

Fun fact: All Mersenne primes are of the form 4n+3

like instead of considering sets of 10,20,30
how about sets of length approximately
e and e^2 , e^3

it could result in a linear growth of the different between team 1 and team 3 primes

difference*

e,pi,phi lets gooo lol

then e*pi*phi

extension of prime number theorem to arithmetic progressions is something beyong my grasp..

interesting, central limit theorem relates sampling distributions to sample distribution right ?

yeah

the more samples u take, the more steep the normal curve of the distribution of samples becomes ?

i didnt really finish it lol i just got some integral expression

http://prntscr.com/88k0av
http://prntscr.com/88k0hm
i got upto there and forgot about it xD

but i think the steps i took there is pretty logical its how someone naive would try to approach it

like if u flip 2 coins heads and tails, you will expect quite a bit of variation often

and how the variation changes can be shown with some combinatorics

and u end up with the central limit theorem i think

|dw:1440471255706:dw|

That's one way to look at it, yeah

Here's my code, check it out and play around:
https://repl.it/BD31/1

it is very interesting, how evenly they distribute themselves among 16a+b progressions!

Uh oh I think I made a tiny bug left in there

haha all i can think of is putting sqrt for that n
```
for (int i = 2; i < Math.sqrt(n); i++)
```

yes it has to do with prime number thm..

Oh right of course! XD

Ahhh also it looks like \(p_n\)~ n log n

Or any other kind of fun ideas we should try looking at?