Some silliness for the March break from school. Let a 2-3 number be those integers having, at most, factors of 2 and 3. Such numbers are...
19903
Sebastian Martin Ruiz
s_m_ruiz
Mar 17, 2009 5:26 pm
Hello all: Sqrt[(p^2-1)/24] is integer for the primes p=5 and p=4801. Find the next Sincerely Sebastian Martin Ruiz [Non-text portions of this message have...
19904
Norman Luhn
nluhn
Mar 17, 2009 7:02 pm
If there a next prime, so have the number more than 2600 digits. Best [Non-text portions of this message have been removed]...
19905
David Broadhurst
djbroadhurst
Mar 17, 2009 7:31 pm
... A lot more than digits than that, Norman. Linear(1,5,49,485,1) trivially factors prime!: 5 Linear(1,5,49,485,4) trivially factors prime!: 4801 ...
19906
Jack Brennen
jbrennen
Mar 17, 2009 8:05 pm
I believe that you only need to test the numbers with indexes equal to a power of 2. Note that the numbers can be expressed easily in PARI using: v(n) =...
19907
Werner D. Sand
theo2357
Mar 18, 2009 10:58 am
In words: If N is a square (N=m²) and N+1 and N-7 are primes, then N-5 is a prime, too. 3) is superfluous, because it follows from 1) by simple...
19908
Maximilian Hasler
maximilian_h...
Mar 18, 2009 12:56 pm
... Stated like this, it is indeed more or less trivial. As I see it, the nontrivial part of the assertion is: There is NO m such that * m²-7 is prime *...
19909
Maximilian Hasler
maximilian_h...
Mar 18, 2009 1:12 pm
... er... I read but you didn't write: "...and there is a prime between N-7 and N+1, ..." else we have counter-examples for m = ...
19910
Ignacio Larrosa Ca...
ilarrosa
Mar 18, 2009 1:14 pm
... Sqrt[(n^2-1)/24] is integer for n = a(k) =((5 + 2sqrt(6))^k + (5 - 2sqrt(6))^k)/2 a(k) = 10a(k-1) - a(k-2) a(k) = 5, 49, 485, 4801, 47525, 470449, 4656965,...
19911
jbrennen
Mar 18, 2009 5:52 pm
... As I wrote in my earlier response to David, it's much more constrained than that... Any other prime must be of the form a(2^c). This is because, like the...
19912
Ignacio Larrosa Ca...
ilarrosa
Mar 18, 2009 7:05 pm
... You're right. I finally reached the same conclusion after giving a big roundup ... I had not read the previous messages, because I replied to a message...
19913
David Broadhurst
djbroadhurst
Mar 19, 2009 1:16 am
... Thanks, Jack. As you say, V(10,1,n)/2 is composite if n has an odd prime factor. I remark that V(10,1,2*p)/98 may be prime if p is prime. A proof of...
19914
David Broadhurst
djbroadhurst
Mar 19, 2009 3:36 am
... T(p) = V(10,1,2*p)/98 is a proven prime for p = 3, 5, 421, 6551. Challenge for the brave cyclotomist: prove that T(8369) is prime. David...
19915
Kermit Rose
kermit1941
Mar 19, 2009 3:45 am
... If p is any odd integer, not divisible by 3, then (p**2 -1) / 24 is integer (5**2 -1)/24 = 1 (7**2 -1)/24 = 2 (11**2 -1)/24 = 5 (13**2 -1)/24 = 7 (17**2...
19916
Mike Oakes
mikeoakes2
Mar 19, 2009 7:15 am
... k=14,15,16 are also composite (18 GHz hrs with pari-gp). Here are the sizes: k digits 0 1 1 2 2 4 3 8 4 16 5 32 6 64 7 128 8 255 9 510 10 1020 11 2039 12...
19917
Werner D. Sand
theo2357
Mar 20, 2009 4:19 pm
... O.k., put it in....
19918
David Broadhurst
djbroadhurst
Mar 21, 2009 1:28 pm
For positive integer n, let J(n) = x^n + 1/x^n, with x = sqrt(3) + sqrt(2). Jack observed that J(n)/2 is not prime if n is not a power of 2. Note that J(8)/2 =...
19919
Mike Oakes
mikeoakes2
Mar 22, 2009 7:08 pm
... A truly remarkable discovery, David. And a surprising claim, as you are saying that the prime values are restricted to the weird union of "(odd) prime OR...
19920
Bryan Bartlett
valareos
Mar 22, 2009 9:55 pm
Compressed from my previous mailings, designed to limit calculations to find if a number is prime using pasquals triangle. Knowledge of structure of triangle...
19921
Bryan Bartlett
valareos
Mar 22, 2009 10:02 pm
edit... last line should read {XnX^n + Xn-1X^n-1 + ... + X1X}/X is an integer for all primes n in range 1>n>=–X ... -- Bryan Bartlett...
19922
David Broadhurst
djbroadhurst
Mar 22, 2009 10:31 pm
... Let J(n) = x^n + 1/x^n, with x = sqrt(3) + sqrt(2). For positive odd integer n, A(n) = J(n)/sqrt(3) + 1 B(n) = J(n)/sqrt(3) - 1 are coprime integers. It is...
19923
Bryan Bartlett
valareos
Mar 22, 2009 11:36 pm
sighs, teaches me to store it on paper napkins in my room :P I need to go back and figure out where I put a wrong number. Will repost at a later date...
19924
Bryan Bartlett
valareos
Mar 22, 2009 11:54 pm
Knew i forgot a key part of the formula Here it is, fixed and completed Given that X1 = (-X + 1)! And Xn = Xn-1 + [(X-1)n * -(X-1)] X is a prime number if ...
19925
David Broadhurst
djbroadhurst
Mar 23, 2009 12:19 am
Here are the PRPs found so far. Let A(n) = J(n)/sqrt(3) + 1 B(n) = J(n)/sqrt(3) - 1 where J(n) = x^n + 1/x^n with x = sqrt(3) + sqrt(2). A(n) is a prime (or...
19926
Bryan Bartlett
valareos
Mar 23, 2009 12:27 am
Ive noticed (and gone back to my archives) that there is a message sent from a place called freeml that has a copy of the message i send to this group,...
19927
Maximilian Hasler
maximilian_h...
Mar 23, 2009 1:27 am
That's indeed annoying. I have already signaled these two bouncing japanese addresses, mihhou@... robotics@... to some moderators or owners of...
19928
Ali Poland
alipoland
Mar 23, 2009 6:37 am
Dear All, Please advise me if there is an already names for following two groups of prime categrories. I would like to call primes that have digit sums...
19929
Ali Poland
alipoland
Mar 23, 2009 6:37 am
Dear All, I would like to share with you 2 small utilities I developed for the Pocket PC for prime factorization. A faster one but upto MAX LONG values...
