counting different sums of KSubsets over the Unit Circle

[wouter meeussen]

take the familiar KSubsets of an integer n (say 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}

The fool says "so, what?", the amateur says "oh? why?", the mathematician
thinks "how to prove this?", and the wise man says "there's surely an easy
way to prove this, but, not me, not now..."

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 numbers thrown together (conjecture).

W.
and Neil, ..in the end, the molds will nicely clean me up, I'm sure.







I posted this to a different mailing list, but since
Wouter referred to it here,. I will post it here too.
It IS important!

I just read a truly great - funny, scholarly and deeply scary -
book called "The Triumph of the Fungi - A Rotten History" by
Nicholas P. Money.
It's the kind of book you come across every ten years that
you want to buy copies of for all your friends.
It got a great review in Nature.
It's scary in the way that The Hot Zone was: rusts and molds
could easily destroy civilisation.
Neil