       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)

{{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: How to "vectorize" code
• Next by Date: Re: Head logic
• Previous by thread: 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