## Re: A conjecture

Expand Messages
• Hi I found 17.41.233=162401=3^11-3^9+3^8-3^7+3^6-3^5+3^4-3-1. Oh well, what do I say? More holes! I let x be 4 and things worked again. Or should I go for
Message 1 of 5 , Jul 2 9:31 AM
Hi

I found 17.41.233=162401=3^11-3^9+3^8-3^7+3^6-3^5+3^4-3-1.
Oh well, what do I say? More holes!
I let x be > 4 and things worked again. Or should I go for c_i in
{0,1}?

Paul

> Hi
>
> Define
> f(x)=x^n-\sigma_{i=1}^{i=n-1}{c_i.x^i}-1
> where
> (1) not all zero c_i in {-1,0,1}
> (2) n>2
> (3) x in N and x>2
>
> conjecture: f|x^f-x => f is prime
>
> If anybody finds a counterexample please let me know. I think this
> has something to do with modularity over Z_f of the elementary
> symmetric functions of the roots at f(x)=0.I have checked the
> conjecture for positive c_i and 2<x<2^16 against Carmichael Numbers
> upto 10^14. Fo the simplest case: f(x)=x^3-x-1, I have checked
(with
> error rate 4^-15) all x less than 10^9.
>
> Paul Underwood
• ... Hello my friends, Here is a small conjecture on which I am pondering over from several days. Consider the following set S = {1,2,3,...,k} Consider the sum
Message 2 of 5 , Aug 9, 2002
--------------------------------------------------------------------------------

Hello my friends,

Here is a small conjecture on which I am pondering over from several days.

Consider the following set S = {1,2,3,...,k}

Consider the sum of all the integers in the above set.

That is 1+2+3+...+k,

If this is equal to rn i.e. 1+2+3+...+k = rn, where k,r,n are all positive integers.

If n is greater than or equal to k, then the conjecture says that there are exactly 'r' number of partitions of 'n' in the set 'S'

An illustration will make things more clear.

Consider the following set :

{1,2,3,4,5,6,7}. As you can see here, k=7.

1+2+3+4+5+6+7 = 28

28 can be written as 28 = 2 * 14 (i.e. r=2, n=14).
Therefore according to the conjecture there are exactly 2 partitions of 14 namely, {7,6,1} and {5,4,3,2}.

or

28 = 4 * 7 (i.e. r=4, n= 7), therefore according to the conjecture there are exactly 4 partitions of 7 namely, {7}, {6,1}, {5,2},{4,3}.

But the question is whether the conjecture is true for any positive integer k.

It is enough if anyone can guide me by refering to books which can train me to solve this problem.

Greatest regards,
Sudarshan

--------------------------------------------------------------------------------

[Non-text portions of this message have been removed]
• ... {4,2,1} Phil ===== -- The good Christian should beware of mathematicians, and all those who make empty prophecies. The danger already exists that the
Message 3 of 5 , Aug 10, 2002
--- "S.R.Sudarshan Iyengar" <sudarshansr@...> wrote:
> Consider the following set S = {1,2,3,...,k}
> Consider the sum of all the integers in the above set.
> That is 1+2+3+...+k,
> If this is equal to rn i.e. 1+2+3+...+k = rn, where k,r,n are all
> positive integers.
>
> If n is greater than or equal to k, then the conjecture says that
> there are exactly 'r' number of partitions of 'n' in the set 'S'
>
>
>
> An illustration will make things more clear.
>
> Consider the following set :
> {1,2,3,4,5,6,7}. As you can see here, k=7.
> 1+2+3+4+5+6+7 = 28
>
> 28 can be written as 28 = 2 * 14 (i.e. r=2, n=14).
> Therefore according to the conjecture there are exactly 2
> partitions of 14 namely, {7,6,1} and {5,4,3,2}.
>
> or
>
> 28 = 4 * 7 (i.e. r=4, n= 7), therefore according to the
> conjecture there are exactly 4 partitions of 7 namely, {7}, {6,1},
> {5,2},{4,3}.

{4,2,1}

Phil

=====
--
The good Christian should beware of mathematicians, and all those who make
empty prophecies. The danger already exists that the mathematicians have
made a covenant with the devil to darken the spirit and to confine man in
the bonds of Hell. -- Common mistranslation of St. Augustine (354-430)

__________________________________________________
Do You Yahoo!?
HotJobs - Search Thousands of New Jobs
http://www.hotjobs.com
• ... {4,2,1} ... Might work if one says that there are at least r partitions of n. Jon Perry perry@globalnet.co.uk http://www.users.globalnet.co.uk/~perry/maths
Message 4 of 5 , Aug 11, 2002
> Consider the following set :
> {1,2,3,4,5,6,7}. As you can see here, k=7.
> 1+2+3+4+5+6+7 = 28
>
> 28 can be written as 28 = 2 * 14 (i.e. r=2, n=14).
> Therefore according to the conjecture there are exactly 2
> partitions of 14 namely, {7,6,1} and {5,4,3,2}.
>
> or
>
> 28 = 4 * 7 (i.e. r=4, n= 7), therefore according to the
> conjecture there are exactly 4 partitions of 7 namely, {7}, {6,1},
> {5,2},{4,3}.

{4,2,1}

---

Might work if one says that there are at least r partitions of n.

Jon Perry
perry@...
http://www.users.globalnet.co.uk/~perry/maths
BrainBench MVP for HTML and JavaScript
http://www.brainbench.com

-----Original Message-----
From: Phil Carmody [mailto:thefatphil@...]
Sent: 10 August 2002 09:45

--- "S.R.Sudarshan Iyengar" <sudarshansr@...> wrote:
> Consider the following set S = {1,2,3,...,k}
> Consider the sum of all the integers in the above set.
> That is 1+2+3+...+k,
> If this is equal to rn i.e. 1+2+3+...+k = rn, where k,r,n are all
> positive integers.
>
> If n is greater than or equal to k, then the conjecture says that
> there are exactly 'r' number of partitions of 'n' in the set 'S'
>
>
Your message has been successfully submitted and would be delivered to recipients shortly.