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Old 2015-08-19, 20:20   #1
pepi37
 
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Default Generalized Fermat numbers (in our case primes)

On page where is S reservation is written this:
Generalized Fermat numbers (GFn's), i.e. q^m*b^n+1 where b is the base, m>=0, and q is a root of the base
Since 4*155^n+1 is Generalized Fermat sequence: I am interested how it can be written to form q^m*b^n+1
Thanks for reply
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Old 2015-08-19, 21:19   #2
Batalov
 
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Quote:
Originally Posted by pepi37 View Post
On page where is S reservation is written this:
Generalized Fermat numbers (GFn's), i.e. q^m*b^n+1 where b is the base, m>=0, and q is a root of the base
If it is written like this, then it is written incorrectly (or you took its meaning out of context).

Generalized Fermat numbers (GFN's) are x^2^t+1 with t > 0, that's all. (x^2 + 1 are included. For example 37 is a GFN and a prime.)
In this case (k*155^n+1 with k=4), it is (2*155^(n/2))^2^1+1 for even n.
Furthermore, for doubly even n (n=4q), there is an algebraic factorization, so these are never prime.
What is left for prime GFNs is n = 2 (mod 4).

Quote:
Originally Posted by pepi37 View Post
Since 4*155^n+1 is Generalized Fermat sequence...
-- it is not.
Not every member of this sequence is a GFN.
A Generalized Fermat "sequence" would be one where every term is a GFN; then one could make a simple argument that for such sequences only finite number of terms can be prime (that includes 0 prime terms).
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Old 2015-10-07, 21:44   #3
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Quote:
Originally Posted by pepi37 View Post
On page where is S reservation is written this:
Generalized Fermat numbers (GFn's), i.e. q^m*b^n+1 where b is the base, m>=0, and q is a root of the base
Since 4*155^n+1 is Generalized Fermat sequence: I am interested how it can be written to form q^m*b^n+1
Thanks for reply
I thought I would clarify further on this:

In this statement:
"q^m*b^n+1 where b is the base, m>=0, and q is a root of the base"

I am saying that where all of these conditions are true, then the form is excluded. What you are missing here for 4*155^n+1 is that b is equal to 155 and q can be equal to either 2 or 4 (i.e. 4^1 or 2^2). But neither 2 nor 4 is a root of 155. Remember the last statement where I say q must be a root of b. In this case it is not. Here is an example where it is:

10*1000^n+1

In this case, 10 is a cube root of 1000 so the form is always GFN and is therefore excluded from the conjecture because there is likely a finite number of primes for it, which, as Serge stated, could be zero primes.

Of course most of the time when these GFN forms are excluded, k=b or k=b^2 or k=b^3 etc. such as 2*2^n+1, 4*2^n+1, 8*2^n+1, etc. What is not so obvious is when k^2=b or k^3=b etc. as is the case with 10*1000^n+1.

Finally: This excludes all k=1 for Sierp even bases because where k=1, then b^0=1 (i.e. m=0), which fits all criteria that was stated above for the exclusion. (Note that k=1 is not eliminated for Sierp even bases with trivial factors like it is for Riesel bases. It's the GFN's that eliminate it.)

Last fiddled with by gd_barnes on 2015-10-07 at 21:45
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Old 2015-10-08, 03:00   #4
LaurV
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(The example with 10 is not very fortunate, as it is still obvious. For a "less obvious" one, think to something like 49*343^n+1 or 27*243^n+1 )
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Old 2015-10-09, 14:49   #5
pepi37
 
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Thanks once more for explanation!
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