## Factorization

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• Oops, i did it. I found the entirely new method for factorization.Its running time so good.For small example, if C is a composite Number with least prime
Message 1 of 6 , Jun 27, 2006
Oops, i did it.
I found the entirely new method for factorization.Its running time so good.For small example, if C is a composite Number with least prime factor(N) in the order of some 100 crores i can factorize it in 6 crore calculations.(perfectly all this kind)
Can u say guys mine is best ever.Without any trial division and gcd i can factorize a given composible.Is it possible ever so.Reply soon..

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• Oops, i did it. I found the entirely new method for factorization.Its running time so good.For small example, if C is a composite Number with least prime
Message 2 of 6 , Jun 27, 2006
Oops, i did it.
I found the entirely new method for factorization.Its running time so good.For small example, if C is a composite Number with least prime factor(N) in the range of some 100 crores i can factorize(C) it in 6 crore calculations.(perfectly every kind of composite)
Can u say guys mine is best ever.Without any trial division and gcd i can factorize a given composite.Is it possible ever so.Reply soon....
Send instant messages to your online friends http://uk.messenger.yahoo.com

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• ... Can we give you some examples to check?
Message 3 of 6 , Jun 27, 2006
At 08:15 PM 6/27/2006, kadhirvel wrote:

>Oops, i did it.
>I found the entirely new method for factorization.Its running time so
>good.For small example, if C is a composite Number with least prime
>factor(N) in the range of some 100 crores i can factorize(C) it in 6 crore
>calculations.(perfectly every kind of composite)
>Can u say guys mine is best ever.Without any trial division and gcd i can
>factorize a given composite.Is it possible ever so.Reply soon....

Can we give you some examples to check?
• ... I had never heard of crore but Google knew it: An Indian crore is 10 million. So for a composite of unspecified size, you can find a prime factor around
Message 4 of 6 , Jun 27, 2006
> I found the entirely new method for factorization.Its running
> time so good.For small example, if C is a composite Number
> with least prime factor(N) in the order of some 100 crores
> i can factorize it in 6 crore calculations.

An Indian crore is 10 million.
So for a composite of unspecified size, you can find a prime factor
around 10^9 with 6*10^7 calculations of unspecified complexity.
A bit vague. There are 50,847,534 primes below 10^9 so the number
of trial divisions would be around your number of calculations.
If you post again, I suggest posting your method, or at least something
specific about composite size and running time. Otherwise people are
likely to assume you have nothing new of interest (they probably already do).
Finding 9-digit factors is considered trivial.

--
Jens Kruse Andersen
• Hi Jens, Last time i told u the small example.U didn t notedown the point.And the running i explained is worst case.And the range i explained before best
Message 5 of 6 , Jun 28, 2006
Hi Jens,
Last time i told u the small example.U didn't notedown the point.And the running i explained is worst case.And the range i explained before best case i got just 20000 calculations.I think its never ever possible(without special case and compact algorithm for every composite).Others pls respond to my prev mail.

Thank u.

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• ... I got down the library after that, and read up on Kac. Very interesting. But I remain very keen to get a *definitive* answer on the following, which as yet
Message 6 of 6 , Jun 25, 2007
Some weeks ago I asked a question, to get an answer from Phil Carmody:

> {on the subject of the shape of the skewed bell curve for number of
> factors beneath n):
>
> > The expected number of factors of x is ~ log(log(x)).
> > I think Knuth, Pardo, and perhaps Kac had some theories along those
> lines.
> > See Riesel's PN&CMfF.
> > I don't believe being squarefree changes anything significantly.

I got down the library after that, and read up on Kac. Very interesting.
But I remain very keen to get a *definitive* answer on the following,
which as yet has eluded me in my enquiries:
It is proven that, for any value of n, the histogram for number of
factors of squarefree integers lower than n, will conform to a normal
curve?

With thanks.
Tom
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