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_{n1}+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_{n1}+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. 
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] 
mostly playing with the reduced LL test today
[CODE]2(k)(n)+1 > 2(2(k)(n)+1)^21 > 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))^21 > (a^2)(2(2(k)(n)+1)^21 >(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. 
The connection between primes and the riemann zetafunction 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=MisterBitcoin;456342]The connection between primes and the riemann zetafunction 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. 
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 Lfunction stuff exist in PARI/gp but I'm not good enough at knowing what to put in to get anything back useful to me. 
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 ?

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=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. 
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). 
Better than emirp ?
[url]https://www.ctvnews.ca/canada/bcboysinventedwordgainingtractioncelebrityendorsements1.3778283[/url]

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