Here's how I think I would approach this problem (ignoring the "golf" issue):
monicN=: (<1) >@,@{@,~ (<i:1) #~ ]
primrootN=:3 :0
if. 0=y do. 1 return. end.
<.&.(%&(9!:18'')) j./2 1 o. 2p1%2^y
)
rootsN=:3 :0
(primrootN y)^i.2^y
)
polysN=:3 :0
maybe=. monicN y
keep=. +./"#.0=maybe p."1 (primrootN y)^i.2^y
keep#maybe
)
#@polysN"0 i. 10
0 2 4 6 16 45 126 357 1016 2908
Note that these numbers are different from Peter Taylor's results. But
looking at the first example in his list of a polynomial which is
supposed to have a root which is a root of unity:
(x^4) + (-x^3) + (-x^2) + (- x) + 1
I can't find any roots which make that one work.
<./|1 _1 _1 _1 1 p. rootsN 4
0.414214
That's not very close to zero, so I don't think I've got any precision
errors. But I could be wrong. I'm actually pretty good at being wrong.
Anyways, I'd like to know what the primitive root of 1 is that
satisfies that equation. I don't think I can get any farther without
that.
Thanks,
--
Raul
On Mon, Jun 2, 2014 at 4:12 PM, Dan Bron <[email protected]> wrote:
> Don wrote:
>> To the original problem.
>
> Here is the original problem:
>
> http://codegolf.stackexchange.com/questions/29901/find-number-of-polynomials-with-a-root-which-is-a-root-of-unity
>
>> Does it require only that the absolute value of the number be 1 or does it
>> require the actual definition of a primitive root of unity?
>
> How about you tell me? ;)
>
> -Dan
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