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Search: Posts Made By: R. Gerbicz
Forum: Puzzles 2020-11-26, 20:58
Replies: 31
Views: 663
Posted By R. Gerbicz
For Q2 the optimal is 9 terms (using 4 pieces of...

For Q2 the optimal is 9 terms (using 4 pieces of 13 and 5 pieces 666):

13-(666-(666+666)*(666+13*(13+13)))/666

and for Q1 the optimal is the above 8 terms.
Assuming that we would use "only"...
Forum: GpuOwl 2020-11-22, 16:04
Replies: 185
Views: 7,876
Posted By R. Gerbicz
In that P1(5.5M) the 5.5M is the first stage...

In that P1(5.5M) the 5.5M is the first stage limit in the P-1 method.
The product of prime powers up to that limit has roughly 5.5*10^6/ln(2)=7934822 bits.
And there was no P-1 method inside the...
Forum: And now for something completely different 2020-11-13, 18:44
Replies: 9
Views: 255
Posted By R. Gerbicz
Exactly. There is a not hard pattern for these...

Exactly. There is a not hard pattern for these factors: if additionally you remove all lower exponents factors then for the remaining p prime factors what divides polcyclo(n,x) we have p%n=1. You...
Forum: FermatSearch 2020-11-07, 19:53
Replies: 32
Views: 1,031
Posted By R. Gerbicz
Actually for large n it is better to do a Proth...

Actually for large n it is better to do a Proth test, which takes the same amount of time, but you could discover million+ digit prime (that is not divide any Fermat number).
And if it is prime then...
Forum: Software 2020-11-02, 21:32
Replies: 35
Views: 4,729
Posted By R. Gerbicz
You shouldn't see such slowdown due to the...

You shouldn't see such slowdown due to the checking. See what gpuowl is using L=400 or L=1000 with that the slowdown is "only" 2/L part of the total running time. p95 is using a not fixed L, changing...
Forum: GPU Computing 2020-11-01, 19:33
Replies: 0
Views: 297
Posted By R. Gerbicz
Better error correction

With a probably restarted computation the task is to compute base^(2^n) mod N, so here we can have that base is "big". As usual the standard setup, let:

x[t]=base^(2^(t*L)) mod N...
Forum: Information & Answers 2020-10-26, 00:11
Replies: 12
Views: 624
Posted By R. Gerbicz
Not checked, but do you reward also the p-1 work...

Not checked, but do you reward also the p-1 work when it is hidden in the prp test during the (new) gpuowl 7 run?
Forum: Miscellaneous Math 2020-10-22, 11:58
Replies: 6
Views: 362
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: 35
Views: 4,729
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: 1,991
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: 998
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: 419
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: 419
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: 43,658
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: 4,651
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: 4,651
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: 4,651
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: 1,036
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: 410
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: 1,036
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: 185
Views: 7,876
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: 1,414
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: 1,036
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: 1,036
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: 1,414
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...
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