[seqfan] Re: The weight puzzle sequence
Dan Dima
dimad72 at gmail.com
Tue Aug 24 22:39:55 CEST 2010
Hi,
Just note that lower bound for *a(n)* can be slightly improved by
considering instead of different powers of two the Fibonacci numbers (any
integer can be represented in unique way as sum of Fibonacci numbers -
Zeckendorf representation).
Instead of:
*a(n) ≥ log2n
one gets
**a(n) ≥ loggn*, g - golden ratio
Best Regards,
Dan
On Tue, Aug 24, 2010 at 7:57 PM, Tanya Khovanova <
mathoflove-seqfan at yahoo.com> wrote:
> Dear SeqFans,
>
> I just wrote a piece for my blog with a problem from the 1966 Moscow
> Olympiad.
> http://blog.tanyakhovanova.com/?p=269
>
> There is a sequence there. I didn't compute enough terms to check if it is
> in the database. I would like if someone can compute more terms of the
> sequence.
>
> Here is the relevant part from the blog:
> From the 1966 Moscow Math Olympiad:
>
> Prove that you can choose six weights from a set of weights weighing 1,
> 2, ..., 26 grams such that any two subsets of the six have different total
> weights. Prove that you can’t choose seven weights with this property.
>
> Let us define the sequence a(n) to be the largest size of a subset of the
> set of weights weighing 1, 2, ..., n grams such that any subset of it is
> uniquely determined by its total weight. I hope that you agree with me that
> a(1) = 1, a(2) = 2, a(3) = 2, a(4) = 3, and a(5) = 3. The next few terms are
> more difficult to calculate, but if I am not mistaken, a(6) = 3 and a(7) =
> 4.
>
> Best, Tanya
>
>
>
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