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 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]

 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]

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:

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

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 ?

 CRGreathouse 2018-01-27 22:04

[QUOTE=science_man_88;478006]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).[/QUOTE]

1 is a quadratic residue but 2 is a quadratic nonresidue mod 3.

 science_man_88 2018-01-27 22:14

[QUOTE=CRGreathouse;478570]1 is a quadratic residue but 2 is a quadratic nonresidue mod 3.[/QUOTE]

Yeah, I messed up the sqr to begin with. It was mostly to find a tranform between residues mod different mersennes.

 science_man_88 2018-01-31 23:32

Superpermutations

Still can't believe the lowest number of instruction needed to have all permutations of n instructions present in a code isn't well known.

 CRGreathouse 2018-02-01 03:29

[QUOTE=science_man_88;478952]Still can't believe the lowest number of instruction needed to have all permutations of n instructions present in a code isn't well known.[/QUOTE]

True, but the bounds are pretty tight: n! + (n-1)! + (n-2)! + n-3 <= [url=https://oeis.org/A180632]A180632[/url](n) <= n! + (n-1)! + ... + 2! + 1!.

 science_man_88 2018-02-01 22:02

[QUOTE=CRGreathouse;478967]True, but the bounds are pretty tight: n! + (n-1)! + (n-2)! + n-3 <= [url=https://oeis.org/A180632]A180632[/url](n) <= n! + (n-1)! + ... + 2! + 1!.[/QUOTE]

I see why it's so hard, just annoying. It's hard to brute force when options for endpoint pairs grow by a factor of n^2.

 CRGreathouse 2018-02-01 22:26

[QUOTE=science_man_88;479026]I see why it's so hard, just annoying. It's hard to brute force when options for endpoint pairs grow by a factor of n^2.[/QUOTE]

Sure. When you have
[$$]n^{n! + (n-1)! + (n-2)! + n-3}[/$$]
possibilities to check in the best case, it's hard to reduce the options to something reasonable. Googol to the sixth power and all that.

 science_man_88 2018-02-01 23:28

[QUOTE=CRGreathouse;479030]Sure. When you have
[$$]n^{n! + (n-1)! + (n-2)! + n-3}[/$$]
possibilities to check in the best case, it's hard to reduce the options to something reasonable. Googol to the sixth power and all that.[/QUOTE]

I do see ways to cut down the number to search. Because, you can use symmetries picking (a,b) and (b',a') where apostrophies mean reversed permutations, will not change much at the same length, it can in theory reverse the superpermutation.

 CRGreathouse 2018-02-02 02:38

[QUOTE=science_man_88;479034]I do see ways to cut down the number to search. Because, you can use symmetries picking (a,b) and (b',a') where apostrophies mean reversed permutations, will not change much at the same length, it can in theory reverse the superpermutation.[/QUOTE]

That cuts it down to a little more than
[$$]n^{n! + (n-1)! + (n-2)! + n-3}/2.[/$$]

You can do better: reassign the variables so the first one you use is 1, the first one you use other than that is 2, and so on. This cuts it to about
[$$]n^{n! + (n-1)! + (n-2)! + n-3}/n! \approx n^{n! + (n-1)! + (n-2)! - 3}.[/$$]

But these numbers are huge, we need to do much, much better.

 science_man_88 2018-02-02 03:25

[QUOTE=CRGreathouse;479049]That cuts it down to a little more than
[$$]n^{n! + (n-1)! + (n-2)! + n-3}/2.[/$$]

You can do better: reassign the variables so the first one you use is 1, the first one you use other than that is 2, and so on. This cuts it to about
[$$]n^{n! + (n-1)! + (n-2)! + n-3}/n! \approx n^{n! + (n-1)! + (n-2)! - 3}.[/$$]

But these numbers are huge, we need to do much, much better.[/QUOTE]

Only patterns I see is that when the current lengths known n<6 are divided by 3 we get 1 mod 4, 3 mod 4, 1 mod 4, 3 mod 4. Wonder if that holds.

 science_man_88 2018-02-03 18:49

At least we can use derangement math to show as n gets larger about 37% of orderings will have at least 1 permutaion placed correctly.

 science_man_88 2018-02-04 03:23

[YOUTUBE]5SfXqTENV_Q[/YOUTUBE]. Not sure it wouldn't be solved already, I found a solution for the equation for n=14 just take the coeffcients on 1001 to add to 391 Mod 432 and the coefficient on 432 is 906 mod 1001.

 science_man_88 2018-02-24 17:30

https://oeis.org/draft/A158709

[QUOTE]Apart from the first term, primes such that a Goldbach partition with a pair of them, will lead to a Goldbach partition for the next even number in their sums Collatz chain.[/QUOTE] I was told to blog this in PM. Examples:

6=3+3 ->10=5+5; 10=3+7->16=5+11 did have a rough 21 digit example, but may have to find it again.

[QUOTE] Example: 100000000000000008947+ 100000000000000008899 =200000000000000017846
->150000000000000013349+150000000000000013421 =300000000000000026770[/QUOTE]

 science_man_88 2018-06-14 15:44

[CODE]my(a=0,b=[]);for(x=1,n,if((a-x)>0&&setminus(b,[a-x])==[],b=concat(b,[a-x]);a-=x;print(a),if(setminus(b,[a+x])==[],b=concat(b,[a+x]);a+=x;print(a),print("fails at x="x);break())))[/CODE]

Fails at x=26 apparently.

 science_man_88 2018-10-19 23:00

Yes I'm weird

[CODE]my(s=self());return(select(r->r=="e",Vec(Str(s))))[/CODE]

A code that searches itself.

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