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Forum: Miscellaneous Math 2020-10-22, 11:58
Replies: 6
Views: 130
Posted By R. Gerbicz
Combined Theorem 1 is enough from...

Combined Theorem 1 is enough from https://primes.utm.edu/prove/prove3_3.html
with F1=1, F2=10^8668.
Forum: Software 2020-10-18, 22:55
Replies: 22
Views: 2,790
Posted By R. Gerbicz
Basically in the algorithm there is "almost" only...

Basically in the algorithm there is "almost" only squarings and very few multplications (at the stuff for error computation).

For N=(k*b^n+c)/d using a^d as "base" with Fermat's little theorem if...
Forum: GpuOwl 2020-10-11, 15:16
Replies: 20
Views: 1,424
Posted By R. Gerbicz
What processor bug you would detect with my error...

What processor bug you would detect with my error check, actually it would not give any "correct or incorrect" answer because in a non-random FFT implementation it would not pass the error check with...
Forum: Miscellaneous Math 2020-10-10, 15:59
Replies: 18
Views: 728
Posted By R. Gerbicz
On page 39: "The Euler Phi Function is useful...

On page 39:
"The Euler Phi Function is useful since for all a
a^(1+φ(n)) ≡ a (mod n)."

This is very false, a small counter-example is a=2, n=4.
Forum: Math 2020-10-09, 20:12
Replies: 8
Views: 291
Posted By R. Gerbicz
Really not read, in general I know the shortest...

Really not read, in general I know the shortest proofs. Another way:
If 2^(2^m)==-1 mod d then for n>m taking this to the 2^(n-m)-th power:
2^(2^n)==1 mod d from this F(n)==2 mod d and the rest is...
Forum: Math 2020-10-09, 08:52
Replies: 8
Views: 291
Posted By R. Gerbicz
I've learnt in this nice way:...

I've learnt in this nice way: F(n)=2+F(0)*F(1)*F(2)*...*F(n-1), so if we have a common divisor d of F(n) and F(m), where m<n then d|2, hence d=1 because all F(k) is odd.
Forum: FermatSearch 2020-10-06, 22:39
Replies: 318
Views: 40,620
Posted By R. Gerbicz
As I can remember pfgw is doing this (or...

As I can remember pfgw is doing this (or something like that, the constant=50 is not that interesting,
we'd catch all known factors if n is not small: n>50): let N=k*2^n+1 and
r(u)=u^(2^(n-50)) mod...
Forum: Programming 2020-10-02, 16:23
Replies: 26
Views: 3,241
Posted By R. Gerbicz
OK, more inputs: (3*x^2)/(2*x+1) Your...

OK, more inputs:

(3*x^2)/(2*x+1)
Your polynomial
3x2

Clearly wrong. Following your way it would mean that 3*x^2=(2*x+1)*3*x^2+R(x) and in this case the remainder would be 3rd degree?...
Forum: Programming 2020-10-01, 20:46
Replies: 26
Views: 3,241
Posted By R. Gerbicz
Thanks. More examples: polynom: x^5 - 1615*x^4...

Thanks. More examples:
polynom: x^5 - 1615*x^4 + 861790*x^3 - 174419170*x^2 + 14998420705*x - 465948627343

Irreducible polynomial factors
The 4 factors are:
x − 653
(x − 103)2
x2 − 756⁢x +...
Forum: Programming 2020-10-01, 15:24
Replies: 26
Views: 3,241
Posted By R. Gerbicz
There is a bug even in the polynom evaluation...

There is a bug even in the polynom evaluation part, try for: x*(x+1)-(x+1)
it shows:
Your polynomial
−1
Clearly wrong.
Forum: Abstract Algebra & Algebraic Number Theory 2020-10-01, 14:36
Replies: 8
Views: 583
Posted By R. Gerbicz
Found 47668 positive integers x for that x^2+1 is...

Found 47668 positive integers x for that x^2+1 is 757-smooth, downloadable at:
https://drive.google.com/file/d/1etrDpTXK1gF6Z7N9-9PIrUijmvLIXRW-/view?usp=sharing
An old link giving all 200-smooth...
Forum: enzocreti 2020-09-30, 14:17
Replies: 2
Views: 193
Posted By R. Gerbicz
For n>3 the a(n)=10^n+8==8 mod 16 hence it...

For n>3 the
a(n)=10^n+8==8 mod 16 hence it won't be divisible by even 16=2^4 so not by 6^4.
And you can check the n<=3 cases easily since 6^4=1296>1008.
Forum: Abstract Algebra & Algebraic Number Theory 2020-09-29, 17:20
Replies: 8
Views: 583
Posted By R. Gerbicz
Don't know why not use this colossal 57 terms(!)...

Don't know why not use this colossal 57 terms(!) arctan formula what I've found, its efficiency isn't competing with Chudnovsky, but using a bunch of computers we could get the result quicker in Wall...
Forum: GpuOwl 2020-09-28, 10:06
Replies: 108
Views: 3,668
Posted By R. Gerbicz
If you want a simple formula, counting everything...

If you want a simple formula, counting everything in squaremod:
total cost=Time(p,B1,B2)+(1-Prob(p,B1,B2))*(p-B1) and you should minimize this, where Time(p,B1,B2) is the cost of doing this new...
Forum: Analysis & Analytic Number Theory 2020-09-21, 15:41
Replies: 8
Views: 564
Posted By R. Gerbicz
It is a triviality, once you know the sum of a...

It is a triviality, once you know the sum of a geometric serie, could be known for 2400 years. To make things easier split the proof for odd/even n.
Forum: Abstract Algebra & Algebraic Number Theory 2020-09-20, 20:33
Replies: 8
Views: 583
Posted By R. Gerbicz
See also for lots of big arctan formulas:...

See also for lots of big arctan formulas: http://www.machination.eclipse.co.uk/FSChecking.html
(last update was at 2013)
Forum: Abstract Algebra & Algebraic Number Theory 2020-09-20, 20:17
Replies: 8
Views: 583
Posted By R. Gerbicz
See: http://oeis.org/A185389 (somewhere there...

See: http://oeis.org/A185389 (somewhere there could be a longer computed list also).
This problem is solvable by https://en.wikipedia.org/wiki/St%C3%B8rmer%27s_theorem using a Pell type equation,...
Forum: Analysis & Analytic Number Theory 2020-09-20, 16:58
Replies: 8
Views: 564
Posted By R. Gerbicz
It says that there are floor(2^n+10)/6 quadratic...

It says that there are floor(2^n+10)/6 quadratic residues mod 2^n.


You need 4 squares for x iff x=4^e*(8*k+7) it is pretty old fact.
If you want the count for this for x<2^n then you can set...
Forum: Miscellaneous Math 2020-09-15, 14:50
Replies: 12
Views: 446
Posted By R. Gerbicz
I've thought that I understand you, but not....

I've thought that I understand you, but not. Learn more Maths, and more English.
Forum: Miscellaneous Math 2020-09-15, 13:58
Replies: 12
Views: 446
Posted By R. Gerbicz
Yes, that is true, since if N denotes your number...

Yes, that is true, since if N denotes your number then N+2 is divisible by 3 and N-2 is divisible by 107.
Forum: GpuOwl 2020-09-08, 23:24
Replies: 63
Views: 2,596
Posted By R. Gerbicz
Hm, but if you use my prp cf method then you can...

Hm, but if you use my prp cf method then you can save that annoying large division. So compute just
(3^mp mod mp) mod 2^2048 (or 3^(1+mp)... ) just as if we'd know only d=1, and then apply my prp...
Forum: GpuOwl 2020-09-08, 22:59
Replies: 63
Views: 2,596
Posted By R. Gerbicz
Exactly, what I don't understand is that why...

Exactly, what I don't understand is that why anyone tries to run something that is clearly not supported, and you can check this easily at https://github.com/preda/gpuowl . It means nothing that it...
Forum: Puzzles 2020-08-30, 15:17
Replies: 20
Views: 1,297
Posted By R. Gerbicz
Congrats, and you could easily generalize it:...

Congrats, and you could easily generalize it: this time using only 6 and 9 in the last 50 places

? Mod(2,10^50)^5342684466535587793434484498741672
%26 =...
Forum: Puzzles 2020-08-30, 12:49
Replies: 20
Views: 1,297
Posted By R. Gerbicz
Like these challenges, and how would you kill...

Like these challenges, and how would you kill this example:
power of two ending last 50 digits using only 1-2:

? Mod(2,10^50)^18530118576724016889332506805003089
%36 =...
Forum: Factoring 2020-08-26, 19:50
Replies: 6
Views: 795
Posted By R. Gerbicz
And then msieve is faster than light? Save out...

And then msieve is faster than light?
Save out logs.length to a parameter. Don't know Java, but similar code in c++ is also slower when I'm using .size() in a loop.
Showing results 1 to 25 of 1000

 
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