- --- In primenumbers@yahoogroups.com, "mikeoakes2" <mikeoakes2@...> wrote:
>

New experimentally-determined data point supports the picture:-

> --- In primenumbers@yahoogroups.com, "djbroadhurst" <d.broadhurst@> wrote:

> >

> > Puzzle 2: Find a triplet of primes (b,p,q) with b < p < q,

> > b^p = 1 mod q^2, and q^2 having precisely 21 decimal digits.

>

> I reckon there should be about 208952 solutions to this puzzle.

> Rationale:-

>

> max_q solutions ratio

> 10^1 0 -

> 10^1.5 1 -

> 10^2 1 1.0

> 10^3 1 1.0

> 10^3.5 1 1.0

> 10^4 3 3.0

> 10^4.5 6 2.0

> 10^5 22 3.667

> 10^5.5 58 2.636

> 10^6 125 2.155

> 10^6.5 322 2.576

> 10^7 781 2.435

> Taking the ratio to be about 2.4 then gives these estimates:-

> 10^7.5 1874 2.4

>...

10^7.5 1944 2.489

Mike - Hi , all ,

Not even draft yet , let alone published :

http://upforthecount.com/math/nnp.html

still needs a lot of work .

But , I have updated the top of the main table :

http://upforthecount.com/math/nnp2.txt

with complete factorizations now extending through

np ( 90 ) .

When is np ( n ) prime ?

So far , only for 1 , 2 and 3 .

When n mod 6 = 4 , np ( n ) has at least 3 algebraic

factors , 3 and n^2 + n + 1 ( twice ) .

But what about the residue ( primitive part ? ) ?

When is this prime ?

n t n^n + (n+1)^(n+1)

4 4 3 * 7 * 7 * 23

16 6 3 * 7 * 7 * 13 * 13 * 34041259347101651

Any others ?

Cheers ,

Walter