Re: Re: Conjecture: 2n+1= 2^i+p ; 6k-2 or 6k+2 = 3^i+p

*To*: mathgroup at smc.vnet.net*Subject*: [mg97104] Re: [mg97047] Re: [mg97037] Conjecture: 2n+1= 2^i+p ; 6k-2 or 6k+2 = 3^i+p*From*: Andrzej Kozlowski <akoz at mimuw.edu.pl>*Date*: Thu, 5 Mar 2009 04:56:05 -0500 (EST)*References*: <200903031056.FAA02950@smc.vnet.net> <200903041209.HAA27014@smc.vnet.net>

Well, actually there is a funny story behind this. In 1849, Alphonse de Polignac conjectured that every odd number is the sum of a prime and power of 2. He claimed to have verified this up to 3,000,000. However, much earlier Euler had already noted that 127 and 959 are not. In fact, Erd=F6s proved that there is an infinite arithmetic progression of odd integers that cannot be represented as sum of a prime and 2^n. Numbers that refute Polignac's conjecture are known as obstinate numbers. It seems that people are still interested in wasting computing power on finding more obstinate numbers. http://sprott.physics.wisc.edu/pickover/obstinate.html Why? Beats me. On the other hand, the OP question was a bit different as he included the possibility of "negative prime", in other words his question was: is it true that for every obstinate number x one can find an integer n such that 2^n-x is prime. Actually, one could ask an even stronger question: is it true that for every odd number a there exists an integer n such that 2^n - a is prime? It seems very likely that the answer is yes, and probably the proof is easy but I can't spend any more time on this... Andrzej Kozlowski On 4 Mar 2009, at 13:09, DrMajorBob wrote: > The conjecture is false. It fails when the number tested is prime > (but not > the second of a twin prime pair). And it fails in other cases, too. > > Proof: > > Clear[test] > test[k_?OddQ] /; k >= 3 := > Module[{n = 0}, > Catch[While[2^n < k, PrimeQ[k - 2^n] && Throw@{2^n, k - 2^n, True}; > n++]; {k, False}]] > > These are the failures up to 1000: > > failures = > Cases[test /@ Range[3, 1000, 2], {k_, False} :> {k, PrimeQ@k}] > > {{127, True}, {149, True}, {251, True}, {331, True}, {337, > True}, {373, True}, {509, True}, {599, True}, {701, True}, {757, > True}, {809, True}, {877, True}, {905, False}, {907, True}, {959, > False}, {977, True}, {997, True}} > > Primes are marked with True, and non-primes with False, so the most > interesting of these is the first non-prime failure, 905. > > Here's an independent test for that one: > > 905 - 2^Range[0, Log[2, 905]] > PrimeQ /@ % > > {904, 903, 901, 897, 889, 873, 841, 777, 649, 393} > > {False, False, False, False, False, False, False, False, False, False} > > Also, there are 2^k + 1 that fail: > > test /@ (1 + 2^Range[18]) > > {{1, 2, True}, {2, 3, True}, {2, 7, True}, {4, 13, True}, {2, 31, > True}, {4, 61, True}, {2, 127, True}, {16, 241, True}, {4, 509, > True}, {4, 1021, True}, {32, 2017, True}, {4, 4093, True}, {2, 8191, > True}, {4, 16381, True}, {512, 32257, True}, {16, 65521, True}, {2, > 131071, True}, {0, 262145, False}} > > The smallest of these is 2^18+1 == 262145. > > Bobby > > On Tue, 03 Mar 2009 04:56:33 -0600, Tangerine Luo > <tangerine.luo at gmail.com> wrote: > >> I have a conjecture: >> Any odd positive number is the sum of 2 to an i-th power and a >> (negative) prime. >> 2n+1 = 2^i+p >> >> for example: 5 = 2+3 9=4+5 15=2^3+7 905=2^12-3191 .... >> as to 2293=2^i +p =1B$B!$=1B(BI don't know i , p . it is sure = that >> i>30 000 = > >> if >> the conjecture is correct. >> >> More, >> n = 3^i+p, (if n=6k-2 or n=6k+2) >> for example:8 = 3+5 16=3^2+7 100=3+97, 562 = 3^6 -167 >> >> I can't proof this. Do you have any idea? >> > > > > -- = > > DrMajorBob at bigfoot.com >

**References**:**Conjecture: 2n+1= 2^i+p ; 6k-2 or 6k+2 = 3^i+p***From:*Tangerine Luo <tangerine.luo@gmail.com>

**Re: Conjecture: 2n+1= 2^i+p ; 6k-2 or 6k+2 = 3^i+p***From:*DrMajorBob <btreat1@austin.rr.com>