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

• To: mathgroup at smc.vnet.net
• Subject: [mg97101] Re: Conjecture: 2n+1= 2^i+p ; 6k-2 or 6k+2 = 3^i+p
• From: DrMajorBob <btreat1 at austin.rr.com>
• Date: Thu, 5 Mar 2009 04:55:32 -0500 (EST)
• References: <200903031056.FAA02950@smc.vnet.net> <golqt7\$qa1\$1@smc.vnet.net>
• Reply-to: drmajorbob at bigfoot.com

```Yeah, I get that now. Hence, it's not a Mathematica issue.

Bobby

On Wed, 04 Mar 2009 07:02:31 -0600, Sjoerd C. de Vries
<sjoerd.c.devries at gmail.com> wrote:

> Hi Bob,
>
> You don't seem to have noticed that in the conjecture negative primes
> were allowed as well. Therefore, the conjecture reads as
> 2n+1 = 2^i+p OR 2n+1 = 2^i - p, with p now defined as positive prime.
> This is much harder to prove or disprove numerically because the
> search space is infinite contrary to the case you examined.
>
> Cheers -- Sjoerd
>
> On Mar 4, 2:07 pm, DrMajorBob <btre... at austin.rr.com> wrote:
>> The conjecture is false. It fails when the number tested is prime (but
>> not
>> the second of a twin prime pair). And it fails in other cases, too.
>>
>> Proof:
>>
>> Clear[test]
>>   test[k_?OddQ] /; k >= 3 :=
>>   Module[{n = 0},
>>    Catch[While[2^n < k, PrimeQ[k - 2^n] && Throw@{2^n, k - 2^n, True};
>>      n++]; {k, False}]]
>>
>> These are the failures up to 1000:
>>
>> failures =
>>   Cases[test /@ Range[3, 1000, 2], {k_, False} :> {k, PrimeQ@k}]
>>
>> {{127, True}, {149, True}, {251, True}, {331, True}, {337,
>>    True}, {373, True}, {509, True}, {599, True}, {701, True}, {757,
>>    True}, {809, True}, {877, True}, {905, False}, {907, True}, {959,
>>    False}, {977, True}, {997, True}}
>>
>> Primes are marked with True, and non-primes with False, so the most
>> interesting of these is the first non-prime failure, 905.
>>
>> Here's an independent test for that one:
>>
>> 905 - 2^Range[0, Log[2, 905]]
>> PrimeQ /@ %
>>
>> {904, 903, 901, 897, 889, 873, 841, 777, 649, 393}
>>
>> {False, False, False, False, False, False, False, False, False, False}
>>
>> Also, there are 2^k + 1 that fail:
>>
>> test /@ (1 + 2^Range)
>>
>> {{1, 2, True}, {2, 3, True}, {2, 7, True}, {4, 13, True}, {2, 31,
>>    True}, {4, 61, True}, {2, 127, True}, {16, 241, True}, {4, 509,
>>    True}, {4, 1021, True}, {32, 2017, True}, {4, 4093, True}, {2, 8191,
>>     True}, {4, 16381, True}, {512, 32257, True}, {16, 65521, True}, {2,
>>     131071, True}, {0, 262145, False}}
>>
>> The smallest of these is 2^18+1 == 262145.
>>
>> Bobby
>>
>> On Tue, 03 Mar 2009 04:56:33 -0600, 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 =1B\$B!\$=1B(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?
>>
>> -- =
>>
>> DrMajor... at bigfoot.com
>

--
DrMajorBob at bigfoot.com

```

• Prev by Date: Re: Convolution of a Gaussian Distribution and a discontinuous function
• Next by Date: Re: Re: Conjecture: 2n+1= 2^i+p ; 6k-2 or 6k+2 = 3^i+p
• Previous by thread: Re: Conjecture: 2n+1= 2^i+p ; 6k-2 or 6k+2 = 3^i+p
• Next by thread: Re: Conjecture: 2n+1= 2^i+p ; 6k-2 or 6k+2 = 3^i+p