... David [in loco parentis]: May I suggest an extension of your excellent challenge? A "k-H-B" (say) prime is defined to be a k-character Hulme-Beaman prime...
I might be missing something but isn't the solution for k=4 just 9973? i.e. another trivial solution with no symbols involved. Gary ... David [in loco...
... I think you're right, but only because the following happen to be composite:- R(5)..R(9) and because the following are all smaller (and composite):- F(9) ...
Well, I must admit I didn't check them but did check them but some # and ! are larger than 4 digits but are trivially composite. Gary mikeoakes2@... wrote:...
By inspection of Chris Caldwell's and Henri Lifchitz's databases, and David's recent correspondence, here is a first stab at a table:- k k-H-B digits...
Nice idea Mike, but I implore you to strike out ... The whole object of HB work is to encourage ingenuity in primality _proving_. Here is my hint for a...
I was amused by this tidy little proof, at 7583 digits: Calling N+1 BLS with factored part 54.38% and helper 0.16% (163.31% proof) F(36285)-3 is prime!...
... Consider them struck. However, I feel the first could be retained in the mind as a "marker", as it will before too long be provable by (what my maths...
Mike: The NSW abbreviations are allowed by PFGW but not by Chris: U(2,-1,13339) 5106 x21 2001 Generalized Lucas number, Cyclotomy, NSW prime V(2,-1,9679)/2...
... David, The 12-HB candidate U(9!,9,9999) has only 55587 digits, which is not enough for the top-5000 (min required is currently 59191). That seemed to be...
It seems that Larry and Dennis are too cunning, since neither of my good friends Bouk and Mike could decode my several hints. Here is the math. If n is an odd...
... But I had already tried relaxing the other one, by putting p = 1, and had got at best:- U(1,9,99999) 47712 digits Help, someone ... please! Mike...
\\ If n is an odd prime then \\ u(q)=U(1,-q,n) \\ v(q)=V(1,-q,n) \\ are both irreducible monic polynomials in q \\ of degree (n-1)/2 congruent to 1 mod q. ...
Mike: I was wrong about "monic" in the case V, where the leading coefficient is the odd prime n, but that does not affect the algorithm. Note that U is indeed...
... There is no way to separate their mental effort, any more than Bouk and I can be separated in x25-mode. Larry and Dennis feed off each other's ingenuity, ...
Mike: A definition is a definition: http://groups.yahoo.com/group/primeform/message/4271 Why does the Lucas quadratic satisfy you less than this cubic: ...
... It's not clear, IMHO. There could be a top-5000 prime of form k*2^(n+1)+3, for example; "all"(:) that's needed for proof is that k*2^n+1 is prime. Ok,...
Mike: Your Sophie example ending in 3 is simply 2*p+1, a linear form ending in 1. Yves always checks for that, so you can be sure that no listed g-code Proth...
PS: I forgot to mention Ralph's amusing Lucas quadratics: http://primes.utm.edu/primes/page.php?id=74384 http://primes.utm.edu/primes/page.php?id=74398 ...
While investigating U(p,q,5) I noted that U(p,q,5)-1, considered as a polynomial in p, is reducible in Z iff q is of the form fibonacci(2*k). [Note that k and...
Hi, I have been getting my feet wet with pfgw. Thanks for the xp work around of exponentiation at the dos command prompt. Ie., b^^k. I was using ** or ~ and...
... David, The "method of proving" depends how far back you want to go :) Here's an intermediate level demonstration. Write z=x*y*sqrt(5), an element of the...
Erratum (mea culpa) - I should have written this: and all solutions of s^2-5*q^2 = 4 are therefore s + q*sqrt(5) = 2*e[2*k] ...(1) with a corresponding...
... Happy to oblige. 1. 4^10 = 1 mod 11 by Fermat's Little Theorem[FLT] So 4^(10*m) = 1 mod 11 So 4^(10*m+1) = 4 mod 11 = -7 mod 11 QED 2. Proved in identical...
... Ah, thanks for that "all", Mike, which was the bit of "exhaustive" theory that I had lost. In exchange for that favour, here is a related paper that I ...
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