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#1 |
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May 2013
32 Posts |
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 == (a-1)/(2^c) for some positive integer c then a is prime.
The cycle length of 2n-1 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. Last fiddled with by ewmayer on 2013-05-21 at 19:32 Reason: remove annoying xtra-large font |
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#2 |
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May 2013
32 Posts |
Conjecture on cycle length and primes prime abc conjecture final version: Suppose a is positive odd, and b=A179382((a+1)/2), if b=(a-1)/(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 |
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#3 |
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6809 > 6502
"""""""""""""""""""
Aug 2003
101Γ103 Posts
230708 Posts |
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#4 | |
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Banned
"Luigi"
Aug 2002
Team Italia
32·5·107 Posts |
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 |
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#5 | |
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Nov 2003
22·5·373 Posts |
Quote:
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#6 |
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Bamboozled!
"πΊππ·π·π"
May 2003
Down not across
29·3·7 Posts |
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#7 |
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Romulan Interpreter
Jun 2011
Thailand
258B16 Posts |
Why make it so complicate? Let x be a 2-prp, the probability of x to be prime approaches 1 as x goes to infinity
![]() So what? Last fiddled with by LaurV on 2013-05-22 at 05:26 |
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