David thanks, Sorry I left off the "Harold" in Harold N. Shapiro. John W. Nicholson...
21582
djbroadhurst
Jul 14, 2010 6:42 pm
... A proof, for integer x > 2, was given by Harold Shapiro in 1953 and may be found here: http://tinyurl.com/3yycvms The inequality is true by inspection for...
21581
Peter Kosinar
pkosinar
Jul 14, 2010 5:53 pm
... Yes, I do. While 10 is greater than 6 and 8 is greater than 3, (10-8) is not greater than (6-3). Peter...
21580
John W. Nicholson
reddwarf2956
Jul 14, 2010 5:00 pm
pi(x) - pi(x/7) > sqrt(x) proved by N. Shapiro pi(x) > sqrt(x) + pi(x/7) let x = p_(n +1) and p_n so that pi( p_(n +1)) - pi( p_n) > sqrt( p_(n +1)) + pi( p_(n...
21579
primenumbers@yahoogro...
Jul 14, 2010 10:59 am
Hello, This email message is a notification to let you know that a file has been uploaded to the Files area of the primenumbers group. File :...
21578
djbroadhurst
Jul 13, 2010 4:40 pm
... Thanks, Jens! web.archive.org/web/20040125140056/http://www.cbau.freeserve.co.uk/ ... which seems to permit personal use. David...
21577
Jens Kruse Andersen
jkand71
Jul 13, 2010 4:02 pm
... I don't know what David Baugh uses but there is something better. At http://mersenneforum.org/showthread.php?t=13210 I wrote: "Two years ago I searched for...
21576
djbroadhurst
Jul 13, 2010 2:01 pm
... How did you compute pi(x) with x > 10^15, please, David? Andrew Booker's programme hosted at http://primes.utm.edu/nthprime/ is restricted to pi(x) with x...
21575
djbroadhurst
Jul 12, 2010 4:56 pm
... Dorogoi chitatel' / Dear reader It's worse than that: you have no "result" for m > 40291. For real m and positive integer n let Q(m,n) = m*prod(k=1,n,1 -...
21574
Ситников ...
chitatel2000
Jul 12, 2010 1:59 pm
David. You are skeptical about my result. But look: If the value (m) equals the sum of primes. We have: The error calculation of the number of primes in the...
21573
djbroadhurst
Jul 11, 2010 1:01 am
... The largest has 208601 digits and was found by PrimeMogul himself: http://primes.utm.edu/primes/page.php?id=83711 David...
21572
djbroadhurst
Jul 10, 2010 11:46 pm
... Yes. These [n, p] pairs make p = n^2 + n + 1 a Carmichael number: [2304, 5310721] [47735, 2278677961] [97944, 9593125081] [172799, 29859667201] [683808255,...
21571
bhelmes_1
Jul 10, 2010 9:34 pm
A beautifull day Do you know whether there exists a fermat pseudoprime of the form p:=n^2+n+1 By the way i started a small collection of 1000 digit primes of...
21570
Ситников ...
chitatel2000
Jul 10, 2010 6:18 am
P_n^2 <m <P _ (n+1) ^2 For the left restriction (P_n^2) value (m) can pass. It is a question of an error of calculation. For the right restriction it is...
David, while I'll deal with your comment, look at the additional condition Additional condition (m) - Prime number m_p P_n^2<m_p<P_(n+1)^2 P_n - Primes n –...
21567
Journals IJI / IJAI /...
eic_iji
Jul 8, 2010 5:02 am
International Journal of Mathematics and Computation  ISSN 0974-570X (Online), ISSN 0974-5718 (Print), www.ceser.res.in/ijmc.html ...
21566
Alberto Zelaya
albrtzlya
Jul 8, 2010 12:58 am
I'm Alberto Zelaya, lawyer and retired diplomat of the Argentine Foreign Service. I served in Algeria, Colombia and New Zealand. I write some fiction short...
21565
djbroadhurst
Jul 7, 2010 11:15 pm
... It is interesting to see how Sergey fooled himself into making his false conjecture. His mistake was to believe that we may rely on the sieve of...
21564
djbroadhurst
Jul 7, 2010 6:36 pm
... It is rather easy to prove that the number of solutions to [1] is finite, in direct contradiction to a conjecture by Sergey. Consider the ratio R(m) =...
21563
djbroadhurst
Jul 7, 2010 5:42 pm
In primenumbers@yahoogroups.com, ... http://chitatel2000.blogspot.com suggests that this condition might be something like m*prod(k=1,pi(sqrt(m)),1 -...
21562
Ситников ...
chitatel2000
Jul 7, 2010 3:10 pm
Given the value (m) p_n^2<m<p_n+1^2 p_n - primes n – prime number m*p_n+1/p_(n+1)-1=m_1 The interval (m, m_1) - one prime?...
21561
djbroadhurst
Jul 7, 2010 1:36 pm
... Let n = 5. Then p_5 = 11 and p_6 = 13. I "give the value" m = 13^2 - 1, which lies between p_5^2 and p_6^2. Then m1 = m*13/(13 - 1) = 13^2 + 13, by the...
21560
Ситников ...
chitatel2000
Jul 7, 2010 6:51 am
Given the value (m) p_n^2 < m < p_{n + 1}^2 p_n - primes n – prime number E – error in calculating the number of primes in the interval (p_n,m) By formula...
21559
kraDen
kradenken
Jul 7, 2010 1:25 am
... I thought I'd issued a retraction for both these yesterday but it appears that I replied to the poster (i.e. myself) and not to the list. Apologies to all...
21558
Jens Kruse Andersen
jkand71
Jul 6, 2010 11:20 pm
... Yes, the announced CPAP-5 and CPAP-4 were both mistakes. They are AP-5 and AP-4 but my check found 14 and 7 other primes inside the AP's. It was caused by...
21557
djbroadhurst
Jul 6, 2010 10:49 pm
... Was this a mistake? It did not yet appear chez Jens: http://users.cybercity.dk/~dsl522332/math/cpap.htm#k5 David...
21556
kraDen
kradenken
Jul 5, 2010 7:22 am
Again using cpapsieve Find_CPapn pfgw Primo Cheers Ken...
21555
kraDen
kradenken
Jul 5, 2010 6:56 am
Hi All, I noticed some "low hanging fruit" on Jens' "The Largest Known CPAP's" page. http://users.cybercity.dk/~dsl522332/math/cpap.htm - !!CPAP-5!! - (Gap...
21554
William
fatbobforman
Jul 3, 2010 6:55 am
The munchkin seive is actually 2 analogous seives, one for 6n+1 primes (npp) and the other for 6n-1 primes (npm). The 6n + 1 seive Let a be a positive integer....
