[seqfan] Re: nth cyclotomic polynomial values modulo n

[Neil Sloane]

That's a nice triangle - please go ahead and submit it. When you have an
A-number for it,
you might send a follow-up message here so people can look at it.

Best regards
Neil

Neil J. A. Sloane, President, OEIS Foundation.
11 South Adelaide Avenue, Highland Park, NJ 08904, USA.
Also Visiting Scientist, Math. Dept., Rutgers University, Piscataway, NJ.
Phone: 732 828 6098; home page: http://NeilSloane.com
Email: njasloane at gmail.com


On Sun, Aug 7, 2016 at 8:37 AM, Hugo Pfoertner <yae9911 at gmail.com> wrote:

> 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.
> >
> >
> >
> > --
> > Seqfan Mailing list - http://list.seqfan.eu/
> >
>
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