Mathematica 9 is now available
Services & Resources / Wolfram Forums / MathGroup Archive
-----

MathGroup Archive 2009

[Date Index] [Thread Index] [Author Index]

Search the Archive

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

  • To: mathgroup at smc.vnet.net
  • Subject: [mg97113] Re: Conjecture: 2n+1= 2^i+p ; 6k-2 or 6k+2 = 3^i+p
  • From: Tangerine Luo <tangerine.luo at gmail.com>
  • Date: Thu, 5 Mar 2009 04:57:42 -0500 (EST)
  • References: <goj2cu$2s5$1@smc.vnet.net> <golr5o$qg0$1@smc.vnet.net>

Thanks!
Because the moderator kindly told me that this group doesn't discuss
general math questions, I posted it on
http://groups.google.com/group/sci.math/browse_thread/thread/78fbaeb4c16154=
ea#

To my surprise, this question had been mentioned long years ago.
In 1950 , a mathematician proved the conjecture is false!

On 3=D4=C24=C8=D5, =CF=C2=CE=E78=CA=B111=B7=D6, "Sjoerd C. de Vries" <sjoer=
d.c.devr... at gmail.com>
wrote:
> 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 tha=
n
> 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 =A3=ACI don't know i , p . it is sure that i>30 00=
0 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?



  • Prev by Date: Re: Problems in writing into a sequential filename
  • Next by Date: Re: Re: "Do What I Mean" - a suggestion for improving
  • Previous by thread: Re: Conjecture: 2n+1= 2^i+p ; 6k-2 or 6k+2 = 3^i+p
  • Next by thread: equation numbering