... Will Edgington says many consider it to be an open question rather than a conjecture, but there are no counterexamples below 4e12. He has information...
If z is an odd integer between 2 and 16, and 1 + 2 z1 + 4 z2 + 8 z3 is its base two representation, then z is prime if and only if z3 * [ z2 + (1 - z1) * (1 -...
... I'm relieved. I was trying to get up some nerve to ask you what sort of ... I just tested the other PAP-8 from the search and ... equivalent to the above...
w_sindelar@...
Oct 2, 2007 1:21 pm
19113
... I would expect your method to be much slower based on how "randomly" consecutive prime gaps appear to be distributed. ... No, and also no to the only-part....
If z = 1 + 2 z1 + 4 z2 + 8 where z1 and z2 are variables that take on only the values of 0 or 1, then 8 < z < 16 z is odd if z1 is different than z2, then z is...
... Trivial. ... Trivial. ... Also trivial. All odd composites < 16 are multiples of 3 and 5. ... Quite possibly true, but in the absence of an explanation of...
I discovered how to algebraically trace the carry in base 2 multiplication. This enables us to set up equations, which when solved, find the factors of a...
I discovered how to algebraically trace the carry in base 2 multiplication. This enables us to set up equations, which when solved, find the factors of a...
I now have found a way to simplify the calculations somewhat. Setting z = x y where z = 1 + 2 z1 + 4 z2 + . . . x = 1 + 2 x1 + 4 x2 + . . . y = 1 + 2 y1 + 4 y2...
... Fermat or Fantasy ????? Let F(x)= 2^(2^x)+1 represent a Fermat number such that `x' is a /whole number/. The first few... F(0) = 3, F(1) = 5, F(2) = 17,...
There's some interesting prime-related content in this week's /This Week's Finds in Mathematical Physics/ by world famous folk-singer and physicist Joan Baez: ...
This Wikipedia edit is 10 minutes old as I write: http://en.wikipedia.org/w/index.php?title=Prime_Pages&diff=166202401&oldid=165149465 Some sceptic reverted it...
I think "Jens Kruse Andersen" is more cool from my personal experience! ... <jens.k.a@...> wrote: This Wikipedia edit is 10 minutes old as I write: ...
Hi, Congratulations to the Seventeen-or-Bust team for their latest prime: Sturle Sunde discovers prime #11!, 33661•2^7031232+1 with 2116617 decimal digits: ...
... Special congratulations to Sturle. He's been hunting for big primes for a decade. Patience pays off - he's finally snared a big one. [Non-text portions of...
Using Robert Gerbicz program, I found new primitive solution for Euler Quartic Conjecture 1871713857^4=1593513080^4+1553556440^4+92622401^4 At first Fray using...
So fast we have 2 a new solution and numbers very close. Hmm... 1787882337^4=1662997663^4+1237796960^4+686398000^4 http://www.euler413.narod.ru Leonid Durman...
Does anyone know of some code (Fortran preferably) which solves a single linear Diophantine equation? Thanks Hugo Scolnik [Non-text portions of this message...
PQ? Cito un passo del discorso di H.W. Lenstra jr. al Congresso Internazionale di Matematica, tenutosi a Berkley (USA) nel 1986: "Supponiamo di avere...
Hi all! I am using Proth.exe version 7.1 to search for Keller primes. I am a little concerned about the ranges of numbers I'm testing though. Will Proth.exe...
Hi, Group, et al. the famous Fermat function: F(x) = 2^(2^x) +1 let x be from the set of whole numbers: Does F(x) have Q= 2*floor(sqrt((x+3)/2)) number of...
Hmmmmm... 3 | 1562533*2^6250134-1, but what for 1562533*2^6250134+1 ? At first sight, this Cullen candidate has no small divisor... Mr Karsten Haesslich, I...
... I guess this is an arithmetical error, and the number should be something like: woodall(3125066). This would fit in with PrimeGrid's current testing...
If I submit the posted candidate to LLR, I get immediatly that : 1562533*2^6250134-1 has a small factor : 3 !! So, if this number has really been tested using...
Seeing the second red line on the status page : 5345289*2^5345289-1 1609100 L498 Nov 2007 Woodall I am now almost convinced there are hoaxes... Sorry for him,...
I suppose... if GCD(n! , n + 1) = 1 if GCD(n! , n + 1 ) != 1 also (n+1)+1 so I do GCD(n!, n+2) = 1? if yes I do the sum n! + n+2 also I do another GCD with n+3...
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