25520Re: [PrimeNumbers] Re: Small prime divisors of very large numbers
- Mar 23, 2014Dear 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.
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