       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 \$B!\$(BI 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?

```

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