[seqfan] Re: Coins puzzle and a sequence

[Max Alekseyev]

On Thu, Jul 2, 2009 at 11:41 PM, Tanya
Khovanova<mathoflove-seqfan at yahoo.com> wrote:
>
> Yep,
>
> Can you prove that for 7, 8, and 9 we need 3? (see my essay)

Exhaustive search tells that each a(7), a(8), a(9) >= 3.

btw, for 6 coins there is no "etc." below - there exist exactly two
essentially different solutions with two weightings:

>> For example:
>>
>> 6=1+2+3 and 1+6<3+5
>>
>> 3+6>1+2+5 and 1+3<5
>>
>> etc.

Regards,
Max