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

*To*: mathgroup at smc.vnet.net*Subject*: [mg97047] Re: [mg97037] Conjecture: 2n+1= 2^i+p ; 6k-2 or 6k+2 = 3^i+p*From*: DrMajorBob <btreat1 at austin.rr.com>*Date*: Wed, 4 Mar 2009 07:09:07 -0500 (EST)*References*: <200903031056.FAA02950@smc.vnet.net>*Reply-to*: drmajorbob at bigfoot.com

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

**Follow-Ups**:**Re: Re: Conjecture: 2n+1= 2^i+p ; 6k-2 or 6k+2 = 3^i+p***From:*Andrzej Kozlowski <akoz@mimuw.edu.pl>

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