Colin Mallows's problems

[Max]

Max wrote:

>> %S A000001 1,2,2,4,2,6,2,8,4,6,2,16,2,6,6,16,2,16,2,16,
>> %N A000001 Number of polynomials p with coefficients in {0,1} that 
>> divide x^n-1 and such that (x^n-1)/{p(x-1)} has all coefficients in 
>> {0,1}.
>>  
>> with differences at n = 12, 18, 20.
>>
> 
>  1 2 2 4 2 6 2 8 4 6 2 16 2 6 6 16 2 16 2 16 6 6 2 40 4 6 8 16 2 26 2 32 
> 6 6 6 52 2 6 6 40 2 26 2 16 16 6 2 96 4 16 6 16 2 40 6 40 6 6 2 88 2 6 
> 16 64 6 26 2 16 6 26 2 152 2 6 16 16 6 26 2 96 16 6 2 88 6 6 6 40 2 88 6 
> 16 6 6 6 224 2 16 16 52

As of the original problem:

N. J. A. Sloane wrote:
>   For each m,n, how many of these arrangments can be realised as the ordering 
> of numbers of the form x_i + y_j?

Assume x_1 = y_1 = 0.
For prime m,n the number of such arrangments is a(mn)/2-1 where a() is the sequence above.

E.g., m=2, n=3, there are a(6)/2-1=2 such rearrangements:
1) x_1=0, x_2=1, y_1=0, y_2=2, y_3=4
2) x_1=0, x_2=3, y_1=0, y_2=1, y_3=2

Max