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

Cool findings, thanks David and others.

> --- In primenumbers@yahoogroups.com,

> "Jens Kruse Andersen" <jens.k.a@> wrote:

>

> > David Broadhurst wrote:

> > > The first 7 primes that divide

> > > 137^(137^(137^(137^137))) + 73

> > > are 2, 3, 5, 29, 821, 23339, 67525153.

> > >

> > > Puzzle: Find the first 7 primes that divide

> > > 137^(137^(137^(137^(137^137)))) + 73

> >

> > The same.

>

> As might have been expected, Jens was the first

> to post what I believe to be the correct solution.

>

> Almost at the same time, Richard FitzHugh

> sent me, privately, the same answer:

>

> > I get the same first 7 primes to divide your larger number

> > (namely 2, 3, 5, 29, 821, 23339, 67525153). This seems very

> > strange indeed! It took about 6 minutes of actual processor

> > time, after about 20 minutes of trying to accurately place

> > all the brackets in the nested eulerphi calls!

>

> Like Richard, I find this result to be "strange indeed".

>

> Can anyone tell Richard and me why we should not be surprised?

>

> With thanks to all concerned

>

> David

>

After some reflection, I could tell everyone why we shouldn't be surprised. It's from a very simple and more general result. And I don't even know what eulerphi means.

Here's a hint: (n-1) is a factor of (n^n-n).

A second and a third simple hint will be forthcoming if the above isn't enough.

Mark

PS I apologize in advance if it turns out my very simple reasoning turns out to be delusional... :) - 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