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

Indeed there is a nonsequitor, David :-(

> --- In primenumbers@yahoogroups.com,

> "Mike Oakes" <mikeoakes2@> wrote:

>

> > Then b(q,n+1) = b(q,n) mod p,

> > so q^b(q,n+1) = q^b(q,n) mod p,

>

> Err, Mike, how does the second line follow from the first?

>

> If x = y mod (p-1)

> then q^x = q^y mod p,

> in my book.

>

> Has one of us made a big boob?

While I try and mend the proof (I still believe the Theorem to be true - HELP, anyone?), here are the results of a nontrivial run that has just finished (17.2 hrs @3.79GHz).

I used Jens's pari script, with the prime limit upped from 10^8 to 4*10^9, which is just about as high as you can go with 32-bit gp.exe, on your original puzzle form of 137^^n+73.

Here are the sets of prime divisors up to that limit:-

n=1: 2, 3, 5, 7 [done manually, since 137+73=210]

n=2: 2, 3, 5, 821, 71757331, 152555243

n=3: 2, 3, 5, 29, 821

n=4: 2, 3, 5, 29, 821

n=5: 2, 3, 5, 29, 821, 23339, 67525153, 224354393

6<=n<=15: 2, 3, 5, 29, 821, 23339, 67525153

They still support my Theorem.

Can anyone find a counterexample to it, for any (q^^n+k sequence)?

Mike - Dear all,

I come back to this topic, looking for a working "pow_mod".

This :

{mypmod(b,n,p)=local(m=[p],f=0);

while(n=n-1,m=concat(eulerphi(m[1]),m));

for(p=n=1,#m,n=lift(Mod(b,m[p])^n);

if(f=f+(n*log(b)>=log(m[p])),n=n+m[p]));n%m[#m];}

yields:

mypmod(2,3,5)

= 2

but 2^2^2 % 5 = 16 % 5 = 1.

(and idem for 2^2^2^2 - of course.).

David gave some other code, in

https://groups.yahoo.com/neo/groups/primenumbers/conversations/topics/20526

:

"

Below is a Pari-GP procedure "pmod(a,m)" to compute

a[1]^(a[2]^(a[3]^ ... ^(a[k-1]^a[k]) ... ) modulo m

where the modulus "m" need not be prime.

(...)

{pmod(a,m)=local(k,q,t);k=#a;q=[m];t=a[k];

for(j=2,k-1,q=concat(eulerphi(q[1]),q));

for(j=1,k-1,t=Mod(a[k-j],q[j])^lift(t));t}

"

Although this may work for prime m

(at least, pmod([2,2,2],5) = Mod(1,5) as should),

it gives:

pmod([2,2,2],4)

= Mod(1, 4)

which is most certainly wrong.

So, to put it short, has anyone a working pmod() in his "library" ?

Thanks in advance!

Maximilian

--- In primenumbers@yahoogroups.com, "David Broadhurst"

<d.broadhurst@...> wrote:

> fail

(...)

I consider that his code should not be trusted.

David