mersenneforum.org Search Results
 Register FAQ Search Today's Posts Mark Forums Read

 Showing results 1 to 25 of 1000 Search took 0.34 seconds. Search: Posts Made By: R. Gerbicz
 Forum: Software 2020-08-02, 20:38 Replies: 563 Views: 72,526 Posted By R. Gerbicz With the new version we could lower the trial... With the new version we could lower the trial factoring limit for all p by roughly one bit, as you'd do only one prp test with a quick proof instead of two tests. Any thoughts?
 Forum: PARI/GP 2020-08-01, 10:03 Replies: 12 Views: 331 Posted By R. Gerbicz Ok, but for f(x)=x^3 it is weaker than the binary... Ok, but for f(x)=x^3 it is weaker than the binary search. Just try this: cnt=0;solve(x=-1,2,cnt+=1;print(cnt" "x);x^3) So it is doing at most 259 iterations.
 Forum: PARI/GP 2020-07-30, 14:45 Replies: 12 Views: 331 Posted By R. Gerbicz Good question, it is not binary search, it should... Good question, it is not binary search, it should be another root finding algorithm. Btw in some really trivial cases the solve breaks: ? solve(x=-1,2,x^3) *** at top-level: solve(x=-1,2,x^3)...
 Forum: PARI/GP 2020-07-30, 12:04 Replies: 12 Views: 331 Posted By R. Gerbicz What is missing here is that you need a... What is missing here is that you need a continuous function. Otherwise the result is crap: ? solve(x=0,3,if(x<2,-1,1)) %3 = 1.9999999999999999999999999999999999999 ?
 Forum: Math 2020-07-28, 17:32 Replies: 20 Views: 556 Posted By R. Gerbicz You have a really good chance to halve the number... You have a really good chance to halve the number of ones from the expected L/2 to L/4, assuming log2(n)=L: binomial(10*L,L/4)>39^(L/4)=n^(log(39)/log(2)/4)=n^1.321 >> n
 2020-07-18, 20:28 Replies: 4 Views: 213 Posted By R. Gerbicz See... See https://en.wikipedia.org/wiki/Zsigmondy%27s_theorem and you don't need that p,q are primes, check me, but with few exception this should be working, since there is a prime r for that r | a^n-b^n...
 2020-07-05, 17:56 Replies: 10 Views: 629 Posted By R. Gerbicz Yes, that is a divisor: ?... Yes, that is a divisor: ? d=886407410000361345663448535540258622490179142922169401; ? Mod(2,d)^(2^127-1)+1 %2 = Mod(0, 886407410000361345663448535540258622490179142922169401) ? ## *** last...
 Forum: Math 2020-06-23, 09:12 Replies: 225 Views: 9,541 Posted By R. Gerbicz With those recursive calls it will use power... With those recursive calls it will use power residues in stack memory. Is it intended? You could do this also in disk (using the same size).
 Forum: Math 2020-06-23, 01:43 Replies: 225 Views: 9,541 Posted By R. Gerbicz You could use sliding window technique in void... You could use sliding window technique in void exponentiate (gwhandle *gwdata, gwnum x, uint64_t power) .
 Forum: Puzzles 2020-06-22, 09:50 Replies: 7 Views: 277 Posted By R. Gerbicz I've also that in my mind. Generated 1e8 random... I've also that in my mind. Generated 1e8 random boards, and counted 8499941 configurations that has no 5 or more same piece in a line. So roughly we are expecting 0.085*binomial(80,40) for the whole...
 Forum: Puzzles 2020-06-22, 06:33 Replies: 7 Views: 277 Posted By R. Gerbicz So there is no at least 5 in a line? So there is no at least 5 in a line?
 Forum: Puzzles 2020-06-21, 22:56 Replies: 7 Views: 274 Posted By R. Gerbicz A faster approach: if the discriminant is not a... A faster approach: if the discriminant is not a square then you can eliminate that triplet. With that you don't even need to calculate the coefficients, just check for ~50 smallest primes if the...
 Forum: Math 2020-06-20, 10:11 Replies: 225 Views: 9,541 Posted By R. Gerbicz You can easily avoid this just set hash=2*hash+1,... You can easily avoid this just set hash=2*hash+1, even if you'd this for every computed hash then no hash value will be zero among h_i.
 Forum: News 2020-06-20, 05:53 Replies: 243 Views: 8,967 Posted By R. Gerbicz With power=5 would you need more than 3%... With power=5 would you need more than 3% verification time of the original prp test.
 Forum: News 2020-06-19, 19:58 Replies: 243 Views: 8,967 Posted By R. Gerbicz If you want a flexible check, where the... If you want a flexible check, where the L=interval used for check is not fixed then you need to save the base, where you restarted the check with a new L. You can restart at every error checked...
 Forum: Miscellaneous Math 2020-06-18, 18:47 Replies: 20 Views: 463 Posted By R. Gerbicz Let s(n)=sigma(n)-n the aliquot function, the OP... Let s(n)=sigma(n)-n the aliquot function, the OP mentioned that s(3*p^2) is a square if p!=3 prime. And the connection: if n is an even perfect(!) number then s(2*n) is a perfect(!) square.
 Forum: Math 2020-06-17, 20:19 Replies: 225 Views: 9,541 Posted By R. Gerbicz Nice colorful picture, though there is very few... Nice colorful picture, though there is very few formula (zero) in that pic. And I've already shown that it does not prove that all residues are correct.
 Forum: Math 2020-06-17, 19:19 Replies: 225 Views: 9,541 Posted By R. Gerbicz It doesn't imply for any h value: to pass the... It doesn't imply for any h value: to pass the test (for E=1) you need (A^h*M)^(2^(top/2)) == M^h*B mod N but if you fix M then you can choose B in a unique way, so it has N solutions, but that is...
 Forum: News 2020-06-17, 19:06 Replies: 243 Views: 8,967 Posted By R. Gerbicz You're right, one big advantage of setting r_i=1... You're right, one big advantage of setting r_i=1 is that it has a nice "ladder" scheme, enabling a fast product calculation and (multiple) error checking. With the proof basically you are doing one...
 Forum: Math 2020-06-14, 07:38 Replies: 225 Views: 9,541 Posted By R. Gerbicz That doesn't needs uploading 2^power full... That doesn't needs uploading 2^power full residues per test?
 Forum: Math 2020-06-14, 05:58 Replies: 225 Views: 9,541 Posted By R. Gerbicz You forget the topK/2^e squarings. And there is... You forget the topK/2^e squarings. And there is no other cost in the proof. Maybe the original paper talked about 64 bit values, I'd say 16 bits would be still OK.
 Forum: Math 2020-06-13, 22:27 Replies: 225 Views: 9,541 Posted By R. Gerbicz You can do those costly residue^product(h)... You can do those costly residue^product(h) computations in the proof much faster than the described trivial way. And basically the same trick worked what I've already written in a post. After writing...
 Forum: Factoring 2020-06-09, 11:40 Replies: 13 Views: 672 Posted By R. Gerbicz My Pari gives: ? (-1064)%109 %12 = 26 ? ... My Pari gives: ? (-1064)%109 %12 = 26 ? It is a quite antique installation, but I believe in its result.
 Forum: Math 2020-06-08, 18:17 Replies: 225 Views: 9,541 Posted By R. Gerbicz I thought that too (so for prime N there is... I thought that too (so for prime N there is nothing interesting), but it is false for the prime N=997, 2nd follow-up let's consider the 921,151 pair (again the true residue pair was 249,804). ?...
 Forum: Math 2020-06-08, 17:37 Replies: 225 Views: 9,541 Posted By R. Gerbicz The true pair was (249,804), in the follow-up... The true pair was (249,804), in the follow-up consider another pair (748,193): ? check_proof(748,193,997,500) Test#1: h=780 failed the test Test#2: h=613 passed the test Test#3: h=978 failed...
 Showing results 1 to 25 of 1000

All times are UTC. The time now is 14:56.

Tue Aug 4 14:56:43 UTC 2020 up 18 days, 10:43, 0 users, load averages: 1.89, 1.69, 1.56