[seqfan] Re: Unexpected help by A052515
Rainer Rosenthal
r.rosenthal at web.de
Thu Feb 11 23:09:14 CET 2010
Rick Shepherd schrieb:
> Here is a very straightforward (using Ralf Stephan's claim) proof.
> [nice proof skipped, thanks!]
>
> Ralf Stephan stated that, for n>2, A052515(n) = 2^n-2n-2. Assuming
> that is true, then also a = 2*A052515(n).
I was able to understand Ralf's claim and prove it by simply
enumerating the pairs [t,complement(t)]. This yields 2^n, as there
are 2^n subsets t of {1,...,n}. We have to exclude the empty set
and its complement as well as the n singletons {1}, ..., {n}
and their complements. Thus we arrive at 2^n - 2n - 2 quite
easily.
It took a while until I realized that the sets in the pairs are
supposed to be complementary to each other. And I suggest to
put this piece of information into the definition:
A052515 Number of pairs of complementary sets
=============
of cardinality at least 2.
Best regards,
Rainer
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