... IMHO, the "sprint start" will provide the most efficient route to providing the top candidates. I am not certain what the maximum # of primes by n=10,000...
23550
mikeoakes2
Oct 30, 2011 11:44 am
... Two years ago I did Sierpinski and Riesel "drag racing", for both normal and dual forms. I investigated ALL k < 10^9, and used the sprint-start performance...
23551
djbroadhurst
Oct 30, 2011 3:09 pm
... I have updated http://physics.open.ac.uk/~dbroadhu/cert/mu.txt which now covers all 33 of the p-smooth cases up to p = prime(33) = 137. Since numbpart(33)...
23552
Mark
marku606
Oct 30, 2011 3:15 pm
... 137 smooth! Frankly I find this astonishing, that sorting through this level of complexity is doable. You must have a formula of sorts or methodology to...
23553
djbroadhurst
Oct 30, 2011 3:36 pm
... Perhaps it's the absence of comment statements and pretty-printing that makes it run faster :-) To solve the 137-smooth problem, in exact integer...
23554
Phil Carmody
thefatphil
Oct 30, 2011 6:02 pm
From: djbroadhurst <d.broadhurst@...> ... modified to return the "actual" and expected number: ...
23555
ajo
sopadeajo2001
Oct 30, 2011 6:09 pm
Puzzle: Prove (but not experimentally) that there is a value of t for which (2/3)*sqrt(6*t+1)*cos(arc cos(27*t^2+18*t-25)/(2*(6*t+1)^(3/2)))/3)+1/3 is a prime...
23556
robert44444uk
Oct 30, 2011 6:38 pm
... Hi Mike, I would concur duals are slightly more dense. I did not do much on this back in the day, but the following were the best I found on the Sierpinski...
23557
robert44444uk
Oct 30, 2011 6:41 pm
... Ugh, keyboard issues. I checked about 1 trillion (!) k values of payam proths, mainly on the Riesel side, but only a few million for the duals....
23558
djbroadhurst
Oct 30, 2011 6:43 pm
... Indeed, as I had previously suggested, in http://tech.groups.yahoo.com/group/primenumbers/message/23544 By choosing a base coprime to 6 and by re-checking ...
23559
Phil Carmody
thefatphil
Oct 30, 2011 7:34 pm
... I guess if there's only a few million terms to the sum, then you only need to add half a dozen extra digits of precision to make sure that none of the...
23560
djbroadhurst
Oct 30, 2011 8:32 pm
... (2/3)*sqrt(6*t+1)*cos(acos((27*t^2+18*t-25)/(2*(6*t+1)^(3/2)))/3)+1/3 ... Proof: Let x denote the ugly obfuscation above. Then by expressing cos(3*theta)...
23561
Robin Garcia
sopadeajo2001
Oct 30, 2011 10:33 pm
Nothing can be hiddden to you, David. [Non-text portions of this message have been removed]...
23562
djbroadhurst
Oct 31, 2011 4:12 am
Since Robin seems to enjoy trisections, here is an easy one with the square on the hypotenuse equal to a prime power. Exercise: Consider a triangle with AB =...
23563
Robin Garcia
sopadeajo2001
Oct 31, 2011 7:25 am
Experimentally on my slow computer: ...
23564
mikeoakes2
Oct 31, 2011 8:54 am
... Construct lines of the following lengths:- L1=1388-803=585 L2=803-585=218 L3=585-2*218=149 L4=218-149=69 L5=149-2*69=11 L6=149+13*11=292 Construct the...
23565
paulunderwooduk
Oct 31, 2011 2:09 pm
Hi, congratulation are in order for René Dohmen for discovering the biggest yet, 712355 digit factorial prime 150209!+1: ...
23566
djbroadhurst
Oct 31, 2011 2:51 pm
... Well done Mike. The cube and square of a prime p = 1 mod 4, in this case 137^3 = 1388^2 + 803^2 and 137^2 = 4^2 + 11^2, allow the trisection of an angle,...
23567
djbroadhurst
Oct 31, 2011 2:51 pm
... Well done Mike. The cube and square of a prime p = 1 mod 4, in this case 137^3 = 1388^2 + 803^2 and 137^2 = 4^2 + 11^2, allow the trisection of an angle,...
23568
djbroadhurst
Oct 31, 2011 4:15 pm
... but should have written 137 = 4^2 + 11^2. David...
23569
Robin Garcia
sopadeajo2001
Oct 31, 2011 7:24 pm
On Paris gp; What is the best way to determine in Pari gp if a real number is an integer, up to teh choosen precision ? I did this and got adequate results for...
23570
Maximilian Hasler
maximilian_h...
Oct 31, 2011 7:41 pm
Of course I know that Newton converges quadratically, (I teach this to students...) but I found it somehow remarkable that the simple substitution process for...
23571
mikeoakes2
Oct 31, 2011 8:07 pm
... Actually, primality doesn't enter into it. Take any coprime positive integers p and q. If a right-angled triangle has sides of length AB = q*abs(q^2-3*p^2)...
23572
Maximilian Hasler
maximilian_h...
Oct 31, 2011 8:15 pm
A real number is never an integer, they are equivalence classes modulo completely different relations in completely different sets ... ;-) In PARI/gp you might...
23574
paulunderwooduk
Oct 31, 2011 9:30 pm
... PrimeGrid's http://primes.utm.edu/primes/page.php?id=100445 (110059!+1) took 2.16 days to be verfied by PP. Any guesses for the ETA of René's amazing...
23575
paulunderwooduk
Oct 31, 2011 9:34 pm
... and a further 12 hours for trial division! Paul...
23576
djbroadhurst
Oct 31, 2011 10:26 pm
... Primality of p = 1 mod 4 ensures the /uniqueness/ of the construction for a given cube p^3 equal to the square on the hypotenuse CA of a right-angled...
23577
djbroadhurst
Nov 1, 2011 3:37 am
... PS: If I had replaced N by N-188, this rider would be far easier. Its present severity is my response to Mike's contention: "Actually, primality doesn't...
23578
mikeoakes2
Nov 1, 2011 6:15 am
... Here is Exercise 2: Consider a triangle with AB = 90050, BC = 49001 and ABC a right angle. Explain how to quinquisect the angle CAB using only ruler and...
23579
mikeoakes2
Nov 1, 2011 6:44 am
... N is divisible by every prime p = 1 (mod 4) and p<=137. The quotient is composite, but too hard for pari-GP to factorise (as yet). If accomplished, it...
