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-   -   what should I post ? (https://www.mersenneforum.org/showthread.php?t=21914)

science_man_88 2017-01-09 19:05

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

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:

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

science_man_88 2017-01-13 01:23

closest thing I could think of to post
 
is how some of the things on these forums are connected like:
[LIST=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. [*]the Lucas Lehmer test in reduced form is similar to TF[*] 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. [/LIST]

science_man_88 2017-04-03 23:42

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)[/CODE]

k are related to [URL="https://oeis.org/A007663"]fermat quotients [/URL]

a are the values that are s_n divided by the mersenne primes in that form of the test.

MisterBitcoin 2017-04-07 11:59

The connection between primes and the riemann zeta-function is very exiting. Maybe somethink for you. ;)

Check it out: [URL]https://en.wikipedia.org/wiki/Riemann_zeta_function[/URL]
And one more link: [URL]https://en.wikipedia.org/wiki/Proof_of_the_Euler_product_formula_for_the_Riemann_zeta_function[/URL]

science_man_88 2017-04-07 12:20

[QUOTE=MisterBitcoin;456342]The connection between primes and the riemann zeta-function is very exiting. Maybe somethink for you. ;)

Check it out: [URL]https://en.wikipedia.org/wiki/Riemann_zeta_function[/URL]
And one more link: [URL]https://en.wikipedia.org/wiki/Proof_of_the_Euler_product_formula_for_the_Riemann_zeta_function[/URL][/QUOTE]

I'm not quite that advanced at last check. I would need to read up on complex exponentiation again I think.

science_man_88 2017-04-14 20:57

closest to that I have are numberphile videos mostly right now:

[YOUTUBE]d6c6uIyieoo[/YOUTUBE]
[YOUTUBE]VTveQ1ndH1c[/YOUTUBE]

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.

science_man_88 2017-12-16 21:44

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 ?

Nick 2017-12-16 22:57

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?

science_man_88 2017-12-16 23:26

[QUOTE=Nick;474185]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?[/QUOTE]

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.

science_man_88 2018-01-20 23:10

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).

science_man_88 2018-01-27 19:27

Better than emirp ?
 
[url]https://www.ctvnews.ca/canada/b-c-boy-s-invented-word-gaining-traction-celebrity-endorsements-1.3778283[/url]


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