|
[Date Index]
[Thread Index]
[Author Index]
Re: Re: Conjecture: 2n+1= 2^i+p ; 6k-2 or 6k+2 = 3^i+p
Well, actually there is a funny story behind this. In 1849, Alphonse
de Polignac conjectured that every odd number is the sum of a prime
and power of 2. He claimed to have verified this up to 3,000,000.
However, much earlier Euler had already noted that 127 and 959 are
not. In fact, Erd=F6s proved that there is an infinite arithmetic
progression of odd integers that cannot be represented as sum of a
prime and 2^n. Numbers that refute Polignac's conjecture are known as
obstinate numbers.
It seems that people are still interested in wasting computing power
on finding more obstinate numbers.
http://sprott.physics.wisc.edu/pickover/obstinate.html
Why? Beats me.
On the other hand, the OP question was a bit different as he included
the possibility of "negative prime", in other words his question was:
is it true that for every obstinate number x one can find an integer n
such that 2^n-x is prime. Actually, one could ask an even stronger
question: is it true that for every odd number a there exists an
integer n such that 2^n - a is prime? It seems very likely that the
answer is yes, and probably the proof is easy but I can't spend any
more time on this...
Andrzej Kozlowski
On 4 Mar 2009, at 13:09, DrMajorBob 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[18])
>
> {{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.luo 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?
>>
>
>
>
> -- =
>
> DrMajorBob at bigfoot.com
>
Prev by Date:
Re: Conjecture: 2n+1= 2^i+p ; 6k-2 or 6k+2 = 3^i+p
Next by Date:
Re: Epilog doesn't work with loglinearplot
Previous by thread:
Re: Conjecture: 2n+1= 2^i+p ; 6k-2 or 6k+2 = 3^i+p
Next by thread:
Re: Re: Conjecture: 2n+1= 2^i+p ; 6k-2 or 6k+2 = 3^i+p
|