Forum: Chess
2021-01-17, 12:50
|
|
Replies: 2
Views: 140
See:...
See: https://chess.stackexchange.com/questions/4490/maximum-possible-movement-in-a-turn
|
|
Forum: Puzzles
2021-01-04, 20:50
|
|
Replies: 13
Views: 544
We need for a given x>0 that...
We need for a given x>0 that x=a+..+(a+b)=(2*a+b)*(b+1)/2 with a,b>0
So: 2*x=(2*a+b)*(b+1)
Notice that 2*a+b and b+1 has different parity, hence if x is a power of two
then this equation has no...
|
|
Forum: Forum Feedback
2020-12-26, 16:25
|
|
Replies: 4
Views: 382
|
|
Forum: Programming
2020-12-25, 18:54
|
|
Replies: 37
Views: 8,866
|
|
Forum: Math
2020-12-15, 00:53
|
|
Replies: 16
Views: 635
|
|
Forum: Tales From the Crypt(o)
2020-12-13, 19:55
|
|
Replies: 7
Views: 570
Probably not found, all of them looks like...
Probably not found, all of them looks like gibberish, but here it is:
using a wordlist of size=9027 [basically a 10k list, but deleted all words with length<4 that was not in a 4k list]. All...
|
|
Forum: y-cruncher
2020-12-08, 12:25
|
|
Replies: 49
Views: 25,501
|
|
Forum: y-cruncher
2020-12-08, 10:59
|
|
Replies: 49
Views: 25,501
That's why your suggested methods is just not...
That's why your suggested methods is just not working (needs to repeat the whole computation), but my proposed way is using much less time.
For comparison in this record size the BBP takes less than...
|
|
Forum: y-cruncher
2020-12-07, 14:05
|
|
Replies: 49
Views: 25,501
|
|
Forum: y-cruncher
2020-12-05, 18:08
|
|
Replies: 49
Views: 25,501
There is no BBP type formula for Pi in base ten...
There is no BBP type formula for Pi in base ten [though there could be], https://en.wikipedia.org/wiki/Bailey%E2%80%93Borwein%E2%80%93Plouffe_formula .
But even giving only the last few bits would...
|
|
Forum: Programming
2020-11-29, 00:33
|
|
Replies: 37
Views: 8,866
Try x^720-1. On an older Chrome it is just...
Try x^720-1. On an older Chrome it is just crashing, on Firefox after some computation it has given: "index out of bounds", the 2nd run is still computing and it is in the: "Computing LLL in matrix...
|
|
Forum: Puzzles
2020-11-26, 20:58
|
|
Replies: 34
Views: 1,529
|
|
Forum: GpuOwl
2020-11-22, 16:04
|
|
Replies: 195
Views: 12,485
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: 681
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: 1,734
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: 46
Views: 8,878
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: 1,041
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: 1,313
|
|
Forum: Miscellaneous Math
2020-10-22, 11:58
|
|
Replies: 6
Views: 790
|
|
Forum: Software
2020-10-18, 22:55
|
|
Replies: 46
Views: 8,878
|
|
Forum: GpuOwl
2020-10-11, 15:16
|
|
Replies: 20
Views: 2,978
|
|
Forum: Miscellaneous Math
2020-10-10, 15:59
|
|
Replies: 18
Views: 1,762
|
|
Forum: Math
2020-10-09, 20:12
|
|
Replies: 8
Views: 840
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: 840
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: 324
Views: 49,905
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...
|
