- --- In primenumbers@yahoogroups.com, "gordon_as_number"

<gordon_as_number@y...> wrote:>

Finite images (about which I assume you speak) are just countable

>

>

> Here is just a comment.

>

> It is possible to compile a data image as a number,

data, so can be mapped to and from N.

> as in

Not my kind of place. Better than arxiv's GM section though; barely.

> xxx.arXiv.org/physics/0510148.

> That number, when factored

Why? Data compression _solely_ relies of redundancy.

> in primes, can be stored in the latter to data compress that

> image by a factor of an approximate 10^12.

Losslessly converting data from one format _never_ introduces

redundancy. (But in the case of the BWT, may reveal redundancies

in ways that are easier to exploit.)

> As the labeling

a) What proportion of large numbers are factorable into primes that

> of primes is about a digit or two below the number in accordance

> with numbers up to 10^17 (the known primes), with density of

> an approximate x/ln(x) 10^17.

are <10^17.

b) You've not thought about how you are going to separate your tokens.

It takes nearly lg(n)+lglg(n)+2lg(n) bits to transmit an arbitrary n

in a uniquely decodable way.

c) What did you think was actually /more redundant/ about the

expression of a number as its prime factorisation than in any other

form? Why?

> The point is,

That you've reinvented "compression by coincidence", as I like to call

it, just like hundreds before you who wanted to secrete data into the

digits of Pi, square roots of non-squares, radix-n expansions of

ratios of coprime integers, etc., etc.

> with the prime factorization of a number up to

Wrong, as indicated above.

> 10^15 or so, and each prime in accordance with integers, a

> digit is lost with each prime factor.

> And assuming that there

So wrong that it beggars belief.

> a maximum ln(N) prime factors, or rather less, most data

> compression schemes can be achieved with a factor of 10^12,

> which is actually quite amazing.

Can't you even see that it even contradicts your previous wrong

statement? You're not just wrong - you're internally consistent as well.

>(Images could be written as

Ah, so this did belong in GM after all.

> a number with approximately 10^260 digits.)

>

> So, depending on how fast you can factor a number with 10^100

> digits or so, compressions of 10^10 are achievable.

Please learn at least a little about data compression before making

statements about data compression. Start with the comp.compression

FAQ. Don't even begin to _think_ about making outlandish claims until

you have read Cover and Thomas from cover to cover. And then don't

make them here, as they will _not_ be concerned with prime numbers.

Compression (at an intellectual level, rather than an "oooh shiny"

level) was my bag a long time before prime numbers were. Just don't go

there.

Phil - On 10/20/2012 10:07 AM, primenumbers@yahoogroups.com wrote:
> 2.2. Re: Question

Thank you David.

> Posted by: "djbroadhurst"d.broadhurst@... djbroadhurst

> Date: Fri Oct 19, 2012 9:02 am ((PDT))

>

>

>

> --- Inprimenumbers@yahoogroups.com,

> Kermit Rose<kermit@...> wrote:

>

>> >Methods to solve Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0

>> >

>> >for what values of x would

>> >x^2 + 5 x + 6 be a perfect square.

> Set A = 1, B = 0, C = -1, D = 5, E = 0, F = 6.

>

> and you will get the obvious answer from Dario:

>

> x = -2

> y = 0

> and also:

> x = -3

> y = 0

> Calculation time: 0h 0m 0s

>

> David

I had been confused by consideration of the following:

if x y = z and y > x,

then if we set t = (y+x)/2 and s = (y-x)/2

then z = t^2 - s^2

and x = y - 2 t.

So z = x y = (y - 2t) y = y^2 - 2 t y

y^2 - 2 t y - z = 0

Transforming w = y - k yields

(w + k)^2 - 2 t (w + k) - z = 0

w^2 + (2 k - 2 t) w + (k^2 - 2 t k - z) = 0

which has integral solution if and only if

(2k - 2 t)^2 + (z + 2 t k - k^2) is an integral square

In all that derivation, I forgot that I still did not know what value t was.

I confused myself too easily.

Kermit

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