[seqfan] Re: nth cyclotomic polynomial values modulo n
Hugo Pfoertner
yae9911 at gmail.com
Sun Aug 7 14:37:24 CEST 2016
Up to n=29 your list is correct (checked with PARI/GP's polcyclo function).
Starting from n=30, the next entries should be
n=30
1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
n=31
1 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
n=32
1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2
n=33
1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
n=34
1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 17 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 17
n=35
1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
n=36
1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
n=37
1 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
n=38
1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 19 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
19
n=39
1 1 1 13 1 1 1 1 1 13 1 1 1 1 1 1 13 1 1 1 1 1 13 1 1 1 1 1 1 13 1 1 1 1 1
13 1 1 1
n=40
1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
1 1
n=41
1 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
1 1 1
n=42
1 1 1 7 1 7 1 1 1 1 7 1 7 1 1 1 1 7 1 7 1 1 1 1 7 1 7 1 1 1 1 7 1 7 1 1 1 1
7 1 7 1
Regards
Hugo Pfoertner
On Fri, Aug 5, 2016 at 7:20 AM, Peter Lawrence <peterl95124 at sbcglobal.net>
wrote:
>
> I was playing around with cyclotomic polynomials,
> in particular I was wondering how to verify my calculations
> of their coefficients without using floating-point arithmetic
> to evaluate their supposed roots
>
> and wondered about the values of Cn(x) modulo n
> evaluated for x in 0..n-1,
>
> I did not seem to find these values in OEIS,
> did I compute them incorrectly ?
>
> there are some obvious patterns in the numbers I computed with modulo n
> arithmetic
> Cp(x) ---> 1,0,1,1,1,1,.....
> Cp^e(x) : all 1's except Cn(1), Cn(1+p), Cn(1+2p), ..., Cn(1+p^e-p) ---> p
> Cn(x) with n = 2q with q odd: Cn(q-1), Cn(2q-1) ---> q
>
> but things seem to get wild around C30(x),
>
> would anyone else like to verify the triangle of values I came up with
> for n = 1,..., 30 ?
> 1
> 1 0
> 1 0 1
> 1 2 1 2
> 1 0 1 1 1
> 1 1 3 1 1 3
> 1 0 1 1 1 1 1
> 1 2 1 2 1 2 1 2
> 1 3 1 1 3 1 1 3 1
> 1 1 1 1 5 1 1 1 1 5
> 1 0 1 1 1 1 1 1 1 1 1
> 1 1 1 1 1 1 1 1 1 1 1 1
> 1 0 1 1 1 1 1 1 1 1 1 1 1
> 1 1 1 1 1 1 7 1 1 1 1 1 1 7
> 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
> 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2
> 1 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
> 1 1 3 1 1 3 1 1 3 1 1 3 1 1 3 1 1 3
> 1 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
> 1 1 5 5 1 1 1 5 5 1 1 1 5 5 1 1 1 5 5 1
> 1 1 7 1 7 1 1 1 1 7 1 7 1 1 1 1 7 1 7 1 1
> 1 1 1 1 1 1 1 1 1 1 11 1 1 1 1 1 1 1 1 1 1 11
> 1 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
> 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
> 1 5 1 1 1 1 5 1 1 1 1 5 1 1 1 1 5 1 1 1 1 5 1 1
> 1
> 1 1 1 1 1 1 1 1 1 1 1 1 13 1 1 1 1 1 1 1 1 1 1 1
> 1 13
> 1 3 1 1 3 1 1 3 1 1 3 1 1 3 1 1 3 1 1 3 1 1 3 1
> 1 3 1
> 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
> 1 1 1 1
> 1 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
> 1 1 1 1 1
> 1 17 1 1 1 21 1 1 1 25 1 27 1 1 15 1 1 1 1 5 21 1 1 1
> 25 1 1 1 1 15
>
> if these values are correct I'll go ahead and submit the sequence,
> then see if I can prove the observations,
> but the last line above for 30 seems without pattern,
>
>
> thanks,
> Peter Lawrence.
>
>
>
> --
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>
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