Re: Small prime divisors of very large numbers
- --- In email@example.com,
"Mark Underwood" <mark.underwood@...> wrote:
> if a factor appears twice in a row at two successive powerings,It would be good to see your demonstration, Mark.
> it will remain from then on
Note however that any such theorem would not provide,
of itself, a solution to my puzzle:
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
which Jens, like I, solved by exhaustive computation,
to arrive at the answer: "the same 7 primes as before".
- 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 firstname.lastname@example.org, "David Broadhurst"
I consider that his code should not be trusted.