mechanism for submitting new sequences is broken

[N. J. A. Sloane]

Hi Tony,

so you received my first mail of Saturday night.
I just reposted, because I saw no copy coming into my inbox. Shouldn't have.

Now that you mention it,
I think I've fallen into the trap and over-counted, due to 'Union' not
eliminating
close pairs (without the proper SameTest setting).

By definition, my proposal should equal yours.
What to make of the 'close pairs' count, standing out against the formula I
gave?
(that erroneously counts 'close pairs' as different).
Any new insights?

Wouter.

----- Original Message ----- 
From: "T. D. Noe" <noe at sspectra.com>
To: "wouter meeussen" <wouter.meeussen at pandora.be>
Sent: Sunday, June 24, 2007 7:00 PM
Subject: Re: counting different sums of KSubsets over the Unit Circle


> >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}
>
> You might reference A107861.
>
> Best regards,
>
> Tony
>
>
>
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