--- In

primenumbers@yahoogroups.com, "Christ van Willegen"

<cvwillegen@...> wrote:

>

> Bill,

>

> just to clarify something for myself..

>

> For even n:

> T(n) = T(n/2)^2 - 2 (MOD 47)

yes!, if n is even...

> T(n + 1) = T(n/2) * T((n+2)/2) - 3 (MOD 47)

don't know about that complicated mess,

but if m IS odd, say 15... then T15 = T8 * T7 - 3 mod N

it's that simple!; the author described it easily...

here's my newest modification... but it's a bit more profound than

the original article; it combines the T3 sequence w/ sqrt N

it'll work with any number; take N = 207

find sqrt N = 14.3 {roughly};

then compute terms T14, T12, T7, T6, T4, T3, T2, and T0 = 2, T1 = 3;

so

T2 = T1^2-2 mod 207 = 7,

T3 = T1*T2-3 mod 207 = 18,

T4 = T2^2-2 mod 207 = 47,

T6 = T3^2-2 mod 207 = 115,

T7 = T4*T3-3 mod 207 = 15,

T12 = T6^2-2 mod 207 = 182,

& T14 = T7^2-2 mod 207 = 16;

Since 182 + 16 < 207, then 207 is composite;

if the sum were > 207, then it would've been prime;

this technique combines SQRT and the T3 sequence.

Try it! or wait 'til you understand the T3 sequence alone.

>

> I was figuring out where the -2 and -3 came from. Is this it?

>

> On Wed, Feb 13, 2008 at 3:57 PM, Bill Bouris <leavemsg1@...> wrote:

> > > take the number 47, for example...

> > > the program calculates the terms, ...

> > >

> > > T47

> > > T24,T23

> > > T12,T11

> > > T6,T5

> > > T3,T2

> > > T1,and T0

> > >

> > > so...

> > > every term is taken MOD 47,

> > > to keep them small...

> > >

> > > T0 = 2 mod 47

> > > T1 = 3 mod 47

> > > T2 is T1^2 -2 mod 47

> > > T3 is T1*T2 -3 ...

> > > T5 is T2*T3 -3 ... etc

> > > T6 is T3^2 -2

> > > T11 is T5*T6 -3

> > > T12 is T6^2 -2

> > > T23 is T11*T12 -3

> > > T24 is T12^2 -2

> > > and T47 is T23*T24 -3

> > >

> > > if the result is T47 = 3, then 47 is prime!

> > > neat, huh?

>

> I can't comment on this, though...

>

> Christ van Willegen

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