10-direction clock

[Robert Israel]

I'd prefer to put the clock in the complex plane with
"0" at 1 and "1" to "9" counterclockwise. 
Then if d_j are the decimal digits of m,
the number "m" brings us to S(m) = sum_j w^(d_j)
where w = exp(pi i/5).  Let r_i(m) be the number
of occurrences of the digit i in m, i=0...9.
Then S(m) = P_m(w) where P_m(x) = sum_{i=0}^9 r_i(m) x^i.
The minimal polynomial of w over the rationals is
P(x) = x^4 - x^3 + x^2 - x + 1, so m and n lead to
different spots if and only if P_m(x) - P_n(x) is divisible
by P(x).  Thus one set of ways to list each achievable spot once
would correspond to enumerations of Z^4, where 
(z_1, z_2, z_3, z_4) in Z^4 corresponds to a number m
such that r_0(m)-r_5(m) = z_1, r_1(m)-r_6(m) = z_2,
r_2(m)-r_7(m) = z_3, r_3(m)-r_8(m) = z_4, r_4(m) = r_9(m) = 0.

Cheers,
Robert Israel

On Tue, 22 Apr 2008, Eric Angelini wrote:

>
>                          . 11
>
>               0
>        9      *      1
>         *          *
>
>     8 *                * 2
>               .                  . 23 & 32
>
>     7 *                * 3
>
>         *          *
>        6      *     4
>               5
>
> Hello SeqFans,
> Imagine we assign a direction to each digit, as above.
> If we start at the center of the circle, the number "16"
> would bring us back to the center (we move one step in
> direction "1" and, from there, one step in direction "6").
> The number "11" would drive us outside the clock, as shown,
> -- and so would "23" or "32" (sharing the same spot).
>
> What could be a sequence of integers a(n) where all a(n)s
> would lead us on _different_ spots? (so if 23 is in the
> seq, 32 would not be allowed to).
>
> This (monotonically increasing) seq would start like this,
> I guess:
>
> 0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,22,...
>
> Hope this is not old hat,
> Best,
> É.
>