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

*To*: mathgroup at smc.vnet.net*Subject*: [mg97101] Re: Conjecture: 2n+1= 2^i+p ; 6k-2 or 6k+2 = 3^i+p*From*: DrMajorBob <btreat1 at austin.rr.com>*Date*: Thu, 5 Mar 2009 04:55:32 -0500 (EST)*References*: <200903031056.FAA02950@smc.vnet.net> <golqt7$qa1$1@smc.vnet.net>*Reply-to*: drmajorbob at bigfoot.com

Yeah, I get that now. Hence, it's not a Mathematica issue. Bobby On Wed, 04 Mar 2009 07:02:31 -0600, Sjoerd C. de Vries <sjoerd.c.devries at gmail.com> wrote: > Hi Bob, > > You don't seem to have noticed that in the conjecture negative primes > were allowed as well. Therefore, the conjecture reads as > 2n+1 = 2^i+p OR 2n+1 = 2^i - p, with p now defined as positive prime. > This is much harder to prove or disprove numerically because the > search space is infinite contrary to the case you examined. > > Cheers -- Sjoerd > > On Mar 4, 2:07 pm, DrMajorBob <btre... at austin.rr.com> 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.... 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? >> >> -- = >> >> DrMajor... at bigfoot.com > -- 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>