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

  • To: mathgroup at smc.vnet.net
  • Subject: [mg97047] Re: [mg97037] Conjecture: 2n+1= 2^i+p ; 6k-2 or 6k+2 = 3^i+p
  • From: DrMajorBob <btreat1 at austin.rr.com>
  • Date: Wed, 4 Mar 2009 07:09:07 -0500 (EST)
  • References: <200903031056.FAA02950@smc.vnet.net>
  • Reply-to: drmajorbob at bigfoot.com

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


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