Generalized Fermat numbers (in our case primes)
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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 
[QUOTE=pepi37;408313]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[/QUOTE] If it is written like this, then it is written incorrectly (or you took its meaning out of context). [URL="http://primes.utm.edu/top20/page.php?id=12"]Generalized Fermat numbers[/URL] (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=pepi37;408313]Since 4*155^n+1 is Generalized Fermat sequence...[/QUOTE]  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). 
[QUOTE=pepi37;408313]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[/QUOTE] 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.) 
(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 :wink:)

Thanks once more for explanation!

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