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

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?

0
3/3/2009 10:56:30 AM
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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....@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?


0
3/4/2009 12:11:36 PM
On 3 Mrz., 11:56, Tangerine Luo <tangerine....@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.

0
mg7419 (21)
3/5/2009 9:56:02 AM
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...@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....@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?


0
3/5/2009 9:57:27 AM
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