[seqfan] Re: In the vein of 103314

Allan Wechsler acwacw at gmail.com
Thu Dec 21 18:17:23 CET 2017


We can create a similar table by counting the number of distinct sums of k
(not necessarily distinct) nth roots of 1; this makes sense for any
positive integer n and nonnegative integer k. The

The first row (n=1) is 1,1,1,1,1,1 ... (A000012)

The second row is 1,2,3,4,5,6 ... (A000027)

The third row is 1,3,6,10,15,21 ... (A000217, beheaded)

The fourth row is 1,4,9,16,25,36 ... (A000290, beheaded)

The fifth row is 1,5,15,35,70,126 ... (I think this is a tail of A000332)

The sequence we have been calculating is the diagonal n=k, which is off the
main diagonal in the table above because n is 1-origin and k is 0-origin. I
believe that each row is polynomial with degree A000010(n) (Euler's totient
function phi(n)).

On Thu, Dec 21, 2017 at 1:57 AM, Max Alekseyev <maxale at gmail.com> wrote:

> Robert - yes, I agree.
> It may be worth also to consider other number of terms in the sum, giving
> raise to the table:
> 1,
> 1, 3,
> 1, 4, 10,
> 1, 5, 15, 25,
> 1, 6, 21, 56, 126,
> 1, 7, 28, 64, 100, 127,
> 1, 8, 36, 120, 330, 792, 1716,
> 1, 9, 45, 165, 495, 825, 1365, 2241,
> 1, 10, 55, 220, 715, 2002, 5005, 9724, 18469,
>
> The proposed sequence forms the main diagonal in this table.
>
> Regards,
> Max
>
> On Wed, Dec 20, 2017 at 1:06 PM, <israel at math.ubc.ca> wrote:
>
> > But the primitive roots are zeros of a polynomial of degree phi(n),
> namely
> > the n'th cyclotomic polynomial. So they are not all distinct.
> >
> > Cheers,
> > Robert
> >
> >
> > On Dec 20 2017, Max Alekseyev wrote:
> >
> > They all must be distinct as otherwise the n-th primitive root would be a
> >> zero of a polynomial of degree n-1.
> >> Hence, there are binomial(k+n-1,n-1) distinct values of k-term sums of
> nth
> >> roots of 1, for any k>=1.
> >>
> >> Regards,
> >> Max
> >>
> >> On Tue, Dec 19, 2017 at 4:24 PM, David Wilson <davidwwilson at comcast.net
> >
> >> wrote:
> >>
> >> How many distinct values are taken on by a sum of n nth roots of 1?
> >>>
> >>>
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