counting different sums of KSubsets over the Unit Circle

[wouter meeussen]

take the familiar subsets of  integers 1..n (say 1..5) in k parts (say 3):
{{1,2,3},{1,2,4},{1,2,5},{1,3,4},{1,3,5},{1,4,5},{2,3,4},{2,3,5},
{2,4,5},{3,4,5}}
now, place them on the unit circle as:

{{{E^((2*I)/5*Pi), E^((4*I)/5*Pi), E^((-4*I)/5*Pi)},
{E^((2*I)/5*Pi), E^((4*I)/5*Pi), E^((-2*I)/5*Pi)}, ...

and add each subset of k complex numbers.
Now, scan k from 1 to n, and check the count of *different* values you get
as function of n.
(Needless to say that the sum of sums adds to zero.)

In 'linguae mathematicae', you get:
Table[r=Table[Apply[Plus,Map[E^(2*I*Pi#/n)&,KSubsets[Range at n,k],{2}],{1}],{k
,n}];Length[Union at Flatten[r]],{n,17}]

{1,3,7,9,31,48,127,144,511,768,2047,2304,8191,12288,32767,36864,131071}

and now, it gets slightly less boring:
odd terms are
2^(2Range[8]-1)-1   equals Paul Barry's  A083420
{1,7,31,127,511,2047,8191,32767,131071}
and, a bit silly, but sooo cute: even terms are:
2^{0,0,4,4,8,8,12,12}  3^{1,2,1,2,1,2,1,2} equals
{3,9,48,144,768,2304,12288,36864}

we end up with:
Table[If[OddQ[n],2*4^((n+1)/2-1)-1,2^(4*Floor[(n/2-1)/2])*
3^(1+Mod[n/2+1,2])],{n,17}]
and beyond n=17, this sequence is just conjecture.

W.





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