Prime abc conjecture b == (a1)/(2^c)
Prime numbers generated by the prime abc conjecture when c=4: suppose a is positive, odd and not a multiple of 3 and b is the cycle length of a as defined below. Then if b == (a1)/(2^c) for some positive integer c then a is prime.
The cycle length of 2n1 is OEIS A179382(n). Example: 11 = 5*2^1+1 11 (1,3, 7, 9, 5) Prime numbers generated by the prime abc conjecture when c=4,see OEIS A225759. 
Conjecture on cycle length and primes prime abc conjecture final version: Suppose a is positive odd, and b=A179382((a+1)/2), if b=(a1)/(2^c) for some c>0, as a approaches infinity, the possibility of a is prime approaches 1.
Counter seq: 92673,143713,3579553,4110529,28688897,127017857,141127681,157648097,212999489,663414881 
:popcorn:

[QUOTE=miket;341060]Conjecture on cycle length and primes prime abc conjecture final version: Suppose a is positive odd, and b=A179382((a+1)/2), if b=(a1)/(2^c) for some c>0, as a approaches infinity, the possibility of a is prime approaches 1.
Counter seq: 92673,143713,3579553,4110529,28688897,127017857,141127681,157648097,212999489,663414881[/QUOTE] You said: 1  if a is a positive odd 2  and b = A179382, c>0 3  then the possibility of a is prime approaches 1 as a approaches infinity. Did you mean that, as a grows, the possibility that a is prime approaches 1? In that case, what is the use of A179382? Luigi 
[QUOTE=miket;341060]Conjecture on cycle length and primes prime abc conjecture final version: Suppose a is positive odd, and b=A179382((a+1)/2), if b=(a1)/(2^c) for some c>0, as a approaches infinity, the possibility of a is prime approaches 1.
Counter seq: 92673,143713,3579553,4110529,28688897,127017857,141127681,157648097,212999489,663414881[/QUOTE] Gibberish 
[QUOTE=R.D. Silverman;341087]Gibberish[/QUOTE]Seconded.

Why make it so complicate? Let x be a 2prp, the probability of x to be prime approaches 1 as x goes to infinity :razz:
So what? 
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