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Old 2017-01-09, 19:05   #1
science_man_88
 
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Default what should I post ?

most of what I can think of I probably already have posted somewhere like the trivial results that 2^m{M_n}^2 \equiv M_{n-1}+2^m \pmod {M_{n+m}}

edit: yep so trivial I messed up it's statement. basically I was just using the fact that:
1(2x+1)^2 = 4x^2+4x+1 = x(4x+3)+(x+1)
2(2x+1)^2 = 8x^2+8x+2 = x(8x+7)+(x+2)

which technically is:

2^m{M_n}^2 \equiv M_{n-1}+2^m \pmod {M_{n+m+1}} because m=0 for the 1 case. so this is just a specific case of a more general algebraic fact.

Last fiddled with by science_man_88 on 2018-01-20 at 23:56 Reason: Fixed basic algebra mistakes
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Old 2017-01-13, 01:23   #2
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Default closest thing I could think of to post

is how some of the things on these forums are connected like:
  1. mersenne prime exponents linking to cunningham chains of the first kind (as well as other primes if you don't accept length 1 chains if you accept that the following hold) a mersenne prime exponent>3 can only be the last member of a cunningham chain of the first kind or the first term in a chain that starts with a prime p=4n+1, or a 1 chain length prime of form 4n+3.
  2. the Lucas Lehmer test in reduced form is similar to TF
  3. okay not sure of what else to put here can't think of any more right now even though I could say maybe some computer science things.

Last fiddled with by science_man_88 on 2017-01-13 at 02:40 Reason: typo
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Old 2017-04-03, 23:42   #3
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Default mostly playing with the reduced LL test today

Code:
2(k)(n)+1 -> 2(2(k)(n)+1)^2-1 -> 8(k^2)(n^2)+8(k)(n)+1 -> 8(k)(n)(n+1)+8(k^2)+8(k)+1
a(2(k)(n)+1)->2(a(2(k)(n)+1))^2-1 -> (a^2)(2(2(k)(n)+1)^2-1 ->(a^2)(8(k^2)(n^2)+8(k)(n)+1) -> (a^2)(8(k)(n)(n+1)+8(k^2)+8(k)+1)
k are related to fermat quotients

a are the values that are s_n divided by the mersenne primes in that form of the test.
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Old 2017-04-07, 11:59   #4
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The connection between primes and the riemann zeta-function is very exiting. Maybe somethink for you. ;)

Check it out: https://en.wikipedia.org/wiki/Riemann_zeta_function
And one more link: https://en.wikipedia.org/wiki/Proof_..._zeta_function

Last fiddled with by MisterBitcoin on 2017-04-07 at 12:01
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Old 2017-04-07, 12:20   #5
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Quote:
Originally Posted by MisterBitcoin View Post
The connection between primes and the riemann zeta-function is very exiting. Maybe somethink for you. ;)

Check it out: https://en.wikipedia.org/wiki/Riemann_zeta_function
And one more link: https://en.wikipedia.org/wiki/Proof_..._zeta_function
I'm not quite that advanced at last check. I would need to read up on complex exponentiation again I think.
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Old 2017-04-14, 20:57   #6
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closest to that I have are numberphile videos mostly right now:




it relates to too much for me to deal with and L-function stuff exist in PARI/gp but I'm not good enough at knowing what to put in to get anything back useful to me.
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Old 2017-12-16, 21:44   #7
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Default twitter update

At least on my account, there's now an option to post 25 tweets in a tweet thread. Each one can contain a poll of up to 4 options. What are some potentially useful topics to poll about, other than worst intersection in a municipality ?
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Old 2017-12-16, 22:57   #8
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I assume they make their money by selling information about users.
So the question is what you want that information about you to say, perhaps?
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Old 2017-12-16, 23:26   #9
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Quote:
Originally Posted by Nick View Post
I assume they make their money by selling information about users.
So the question is what you want that information about you to say, perhaps?
I was thinking at one point of making a poll of all 101 products my job placement place makes and seeing what people like. of course I have so few followers, that it probably wouldn't hit any of the bakery's customers.

Last fiddled with by science_man_88 on 2017-12-16 at 23:27
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Old 2018-01-20, 23:10   #10
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Default What's the significance of the fact...

That the residues mod some mersenne in the LL test, are quadratic residues mod the next mersenne.
Reason I think this to be true:

Sqr(A*p+b)= A^2(p^2)+ b^2(1^2)-b mod 2p+1 and two parts of that simplify to quadratic residues. Failure would only happen if the sums/ differences of quadratic residues wasn't a quadratic residue ( guess I may be wrong).

Last fiddled with by science_man_88 on 2018-01-21 at 00:16
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Old 2018-01-27, 19:27   #11
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Default Better than emirp ?

https://www.ctvnews.ca/canada/b-c-bo...ents-1.3778283
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