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I made the following conjectureby almahed
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#73
Nov1108, 01:00 AM

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#74
Nov1108, 01:28 AM

P: 258

can you make an algorithm to search the counterexamples for both forms? I mean, waht is important is to find first the prime number that cannot be writen as p  2^k and then search below until the prime 3 i.e., supose that you "find" that p + 2^n is composite for all values of n, what means there is no value of n such that p = q  2^n for q = prime... now you will test the values for g + 2^k (=p) such that g = prime number (until g = 3) and there is natural k > 0 then should be interesting, if you find such prime number, to see if there is any property for the counterexamples, like: they are twin, palindromic, of the xb + h form, etc, or if it is a totaly random event as you have the list of some counterexamples for p  2^k, I ask you if you can plot some of then here, at least the small ones (10 examples, if possible) 


#75
Nov1108, 08:16 AM

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Conjecture 2: All odd prime numbers are of the form q  2^k. Conjecture 3: All odd prime numbers are of the form q + 2^k or q  2^k. The first counterexample to conjecture 1 is 127. The first *proven* counterexample I know of for the second is 3367034409844073483 (but the first true counterexample may be as small as 2131). I gave a heuristic argument supporting a counterexample for conjecture 3 below a billion trillion. Or rather, I'm searching for noncounterexamples  showing that the conjecture holds up to certain bounds. I could instead start with known counterexamples to A094076 (all of which are large) and search for ones not of the form q + 2^k; this could find a true (proven) counterexample, but the results would be very large. A proven counterexample is unlikely to be a twin prime or a palindrome, since those forms are rare. They are likely to be of 'no special form'. But some quick examples (just from one family) are 3367034409844073483 150940986999520486403 261621451441777796093 556769356621130621933 778130285505645241313 1147065166979836273613 1958721906223056544673 2106295858812732957593 2180082835107571164053 2401443763992085783433 3618928872856916190023 4799520493574327493383 5426709792080452248293 6533514436503025345193 7382064663893664719483 9079165118674943468063 9263632559412038984213 9817034881623325532663 10333543715687192977883 12768513933416853791063 13063661838596206616903 13580170672660074062123 14428720900050713436413 14613188340787808952563 15314164615588771913933 (All members of this family end in 3 or 8, and the ones ending in 8 are never prime. Other families have other residues mod 5.) 


#76
Nov1108, 03:13 PM

P: 258




#77
Nov1108, 03:31 PM

P: 258

lets call conjecture this one: All odd prime numbers are of the form q [tex]\pm[/tex] 2^k, where q is a prime number
it is more easy for your program, can you use a previous prime numbers list, and a previous 2^n values list to speed up your calculation? we are interested in all primes not of the p = q  2^k, then we should search for the forms q + 2^k, it is more faster, of course, because it is "limited below" when q = 3 ok, so backing to the both lists, each of the members of both lists will be "marked" with the total of bits they have, so you only have to compare the primes and the values of 2^n with total of bits added near to the total of bits of your prime number in question of course you need a list avaiable of very large primes, is there such list in the sloane research papers? is this helpfull somehow? or it is a dummy suggestion? 


#78
Nov1108, 04:01 PM

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5101 + 2^5760 is 1.6e1730 times 5101, but is within 1.000...0001 times the value of 2^5760. So the difficulty here is proving many large numbers composite (and one large number prime). 


#79
Nov1108, 04:20 PM

P: 258

but seriously, I was wondering... we dont know yet what will be the use of this conjecture (if its true) to number theory, specially in twin primes conjecture; you said that will be one of the biggest steps towards a proof in one of your first posts, but how? do you have an heuristcs about it? any ideas of an aproach to work on? 


#80
Nov1108, 06:21 PM

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Heuristics about what? 


#81
Nov1108, 06:27 PM

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OK, what I wrote was this:
If we call T the set of primes of the form q  2^k for q prime, then your second conjecture is that T = P. This is also false, since 3367034409844073483 is not in T. Your third conjecture is that [itex]S\cup T=\mathcal{P}.[/itex] I haven't found a counterexample to this yet, but I don't think it's true. So since those seem to not hold, I suggested (way back on page 2) a weakening of conjecture 1. Call it conjecture 1A: S is infinite. Even this hasn't been proved (AFAIK), though it would follow from the twin prime conjecture. (There's no reason to doubt this conjecture, but I imagine a proof of this would be very difficult!) 


#82
Nov1108, 06:50 PM

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OK, I finally have a counterexample, 84478029997463945033 = n.
n + 2^k is always divisible by one of {2, 3, 5, 17, 257, 641, 65537, 6700417}. As for n  2^k, there are only 66 cases to check (since n < 2^67):



#83
Nov1108, 06:55 PM

P: 258

to find a counterexample for [itex]S\cup T=\mathcal{P}.[/itex] is really pretty much more interesting, but this is beyond my computational skills... it is possible to work with only that PARI program to deal with VERY large prime numbers? like 30 bits numbers? 


#84
Nov1108, 07:00 PM

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#85
Nov1108, 07:04 PM

P: 258

there also a kind of pathern to 11 when the values of k goes from +10, +20, +10, +20
and 7 goes from +6, +6, +6, ... 


#86
Nov1108, 07:41 PM

P: 258

see http://www.research.att.com/~njas/sequences/A067760
a(n) = least positive k such that (2n+1)+2^k is prime. COMMENT Does a(1065) exist? (Is 2131+2^k composite for all positive k?) 


#87
Nov1108, 07:45 PM

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#88
Nov1108, 07:47 PM

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#89
Nov1108, 08:06 PM

P: 258

could you state, in a later moment, the exactly densitiy of counterexamples? you said n/(11511227035838054400 log n) but from where you put 11511227035838054400? 


#90
Nov1108, 08:10 PM

P: 258

still not very clear to me 


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