Chris Caldwell <caldwell@...> wrote: > It is highly probable that this is a well known fact, but I ... Look up prime number race instead. Indeed it is...
I have been having a mathematical long weekend, first the prime race 1mod3 and 2 mod3 which went nowhere new, and now, 2modp, which will probably also go...
... than p)" ... Phil, as always, to the rescue. Good man! Think I'll stick to prime searching rather than mathematics. I am not yet at class 101 stage, and it...
... Easily. Show that for some positive constant C, we have: 1/n - 1/p(n) >= C/n Then your sum exceeds the infinite sum of C/n, which diverges for any...
Note for below: negatives included as primes. Given positive a,b, a<b, we want b+a and b-a both prime. First solution: (a,b): (1,4) Given positive a,b,c,...
... all ... b+a, d- ... c+b+a ... (3,5,8,13) ... to be ... evaluate to the ... ways to ... You're right Phil! As it turns out my method of counting works only ...
Well surprise of surprises: The six numbers (5,30,33,42,60,63) yield primes for all 32 additive combinations of the six numbers.(!) I thought it would have...
prove that any 4 consecutive primes can't be in the form 3k+1 or 3k-1 ex: 47,53,59 is 3k-1 , but 61 is 3k+1 ... Yahoo! Messenger with Voice. PC-to-Phone calls...
A small entertainment: "Three brothers are reunited celebrating the birthday of one of them. They realize that their ages in years are three different prime...
... number ... equal ... Perhaps. Anyways, the numbers are relatively factor rich. A prime guru on very quick order came up with 7 more examples where all 32...
... time? ... third ... I think I have an answer -- scroll down to see. The only solution I could find was when we consider 1 to be a prime (which it not...
Monday, May 08, 2006 8:47 PM [GMT+1=CET], ... But 1 isn't prime ... ... \/ ... \/ ... \/ The three brothers hasn't the same birthday ... Best regards, Ignacio...
... You don't need to take 1 to be a prime to find a solution. The brother who is 1 year old reaches his 2nd birthday while the older two still have ages which...
A colleague of mine claimed the other day that 5, followed by one billion 9's, and 6, followed by 999,999,999 zeroes, with a further last digit being 1, are...
... Use a math package which supports modular arithmetic directly; PARI/GP is one freely available package: Then show that the first number of that pair is...
... With Pari/GP in a fraction of a second: ? test(p)=centerlift(6*Mod(10,p)^1000000000)^2 ? forprime(p=2,100000,if(test(p)==1,print(p))) 31 293 2861 12547 ...
I'm looking at the function lf(n) = largest prime factor of (n^2+1). Specifically I'm looking at loops in the iteration of this function. Loops of length 1,...
... with ... Impressive timings! This is the only response that actually seemed to answer the original question, *how does one go about it* rather than just ...
... Woo-woo! Brownie-points for Phil! I was thinking of answering "by evaluating the expressions for the two numbers modulo 31", which could have been a...
Thanks to everyone for the great response - now for the joys of PARI/GP Regards Bob ... Woo-woo! Brownie-points for Phil! I was thinking of answering "by...
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