Hi, Giovanni; Thanks for the question. Yes, there is an Internet site but it's protected by username/password. Our international micro-team (Nejc Škoberne,...
Hi; Is there any known analogue of Lucas-Lehmer Test for repunits (or generalized repunits)? I would really appreciate if someone posts a link or a short ...
The other option is to pose it as a challenge for people to disprove. That would probably give more mileage than getting published in the journals. ... -- ...
Dear David-- How would your solution or approach to solution vary if the number of digits keep increasing--what if X is to the power of 100.. Secondly-what if...
The type of problems --lead to finally solving 1) ax^n+bx^n-1........ to be divisible by 10X+1 2) ax^n -bx^n-1+cx^n-2-......to be divisible by 10X+1 where...
Hi, with parallel ECPP proofs quite time consuming at ~20k digits, I find it hard to believe Tze chieh Chou has *proved* his very recent Wagstaff submissions...
... It took some searching for and now I am crestfallen: n=19051957; Q=138691; gcd(Q-1,n)=1; gcd(Q+1,n)=1; n=19051957; Q=1531540; gcd(Q-1,n)=1; gcd(Q+1,n)=1; ...
look the integer sequence a(i) from 0 to 2*n and the "symetric" sequence b(i) from 2*n to 0 , a(i)+b(i)=2*n For n great , from the prime numbers theorem the...
In primeform@yahoogroups.com, "Pierre CAMI" <pierrecami@...> ... The heuristics for Goldbach partitions are discussed in http://www.ieeta.pt/~tos/goldbach.html...
let a(1) = least prime p such that 18517# + p is = q(1) prime , then a(n+1) = least number such that q(n)*(q(n)+a(n+1))-1 is prime a(1)=39317 ( see record for...
... It can be shortened, for example by using q*(q+a)-1 = (q+a/2)^2-a^2/4-1. a is odd and Chris truncates division just like PFGW so instead I used q*(q+a)-1 =...
... That is a truly brilliant working of the happenstance that ... Chris has already entered a long "blob", but it would be amusing if he might now ... ...
... Thanks. To clarify my truncation comment, I wanted an expression which at the same time was short, mathematically exact, and accepted by Chris/PFGW. My...
... 4)/2+730843)^2-730843^2-4)/2)+1328317)^2-1328317^2)/4-1 ... ************************************************* This is a very nice job and will try if...
... fractions ... ************************************************** I work on your formula starting from q(1) and I found that the shorter expression for q(5)...
Hello, Perhaps you may be interested to check out the conjecture, which I have made lately (and info on possible directions in proving it - see below) ? NO "n"...
Hi I am Giovanni Di Maria (http://primes.utm.edu/bios/page.php?id=852) You you gave to me the following answer. Many many thanks. If I want to participate to...
I suggest that http://primes.utm.edu/primes/page.php?id=85568 should be written as 10^130048+(9*10^37077-2)/11*10^46486+1 which is an integer-valued expression...
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