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

*To*: mathgroup at smc.vnet.net*Subject*: [mg97105] Re: Conjecture: 2n+1= 2^i+p ; 6k-2 or 6k+2 = 3^i+p*From*: "m.g." <mg at michaelgamer.de>*Date*: Thu, 5 Mar 2009 04:56:16 -0500 (EST)*References*: <goj2cu$2s5$1@smc.vnet.net>

On 3 Mrz., 11:56, 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? The conjecture ist false. A (o.k. brute force) approach is: primes = Table[Prime[i], {i, 1, 100}]; temp = DeleteCases[ Distribute[{powers, primes}, List, List, List, Plus], _?EvenQ] // Union // Sort; Complement[DeleteCases[Range[1, 1500], _?EvenQ], temp] // Short Which gets: {1,3,127,149,251,<<259>>,1489,1493,1495,1497,1499} So, Mathematica could in an easy way help to show that it, the conjecture, ist not true. Greetings m.g.