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

*To*: mathgroup at smc.vnet.net*Subject*: [mg97067] Re: Conjecture: 2n+1= 2^i+p ; 6k-2 or 6k+2 = 3^i+p*From*: "Sjoerd C. de Vries" <sjoerd.c.devries at gmail.com>*Date*: Wed, 4 Mar 2009 07:13:36 -0500 (EST)*References*: <goj2cu$2s5$1@smc.vnet.net>

While I think questions like these may be great fun, I feel that they don't belong in this forum as long as they are not connected with Mathematica. I tried the brute force method for your particular example of 2293 and the formal approach for the general conjecture. By brute force using Monitor[ While[! PrimeQ[2^i - 2293], i++], i ] for 2 n +1 ==2293 I find after a long wait that i must be larger than 43 032 if the conjecture is correct. A formal way of stating the problem in Mathematica would be ForAll[n, n \[Element] Integers, Exists[{i, p}, i \[Element] Integers \[And] p \[Element] Primes, 2 n + 1 == 2^i + p \[Or] 2 n + 1 == 2^i - p]] Resolve or FullSimplify could then be used to find out whether or not there is any truth in this statement. Alas, they both return unsolved. Cheers -- Sjoerd On Mar 3, 12:56 pm, 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 ，I 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?