Re: Small prime divisors of very large numbers
- --- In firstname.lastname@example.org,
"Jens Kruse Andersen" <jens.k.a@...> wrote:
> David Broadhurst wrote:As might have been expected, Jens was the first
> > 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.
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 numberLike Richard, I find this result to be "strange indeed".
> (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!
Can anyone tell Richard and me why we should not be surprised?
With thanks to all concerned
- Dear all,
I come back to this topic, looking for a working "pow_mod".
but 2^2^2 % 5 = 16 % 5 = 1.
(and idem for 2^2^2^2 - of course.).
David gave some other code, in
Below is a Pari-GP procedure "pmod(a,m)" to compute
a^(a^(a^ ... ^(a[k-1]^a[k]) ... ) modulo m
where the modulus "m" need not be prime.
Although this may work for prime m
(at least, pmod([2,2,2],5) = Mod(1,5) as should),
= Mod(1, 4)
which is most certainly wrong.
So, to put it short, has anyone a working pmod() in his "library" ?
Thanks in advance!
--- In email@example.com, "David Broadhurst"
I consider that his code should not be trusted.