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

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  • 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.


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