A103314 conjecture

[wouter meeussen]

hi David,

good counterexample. Hmmpff.
Can you also do it without subsets (with abs values) that add to 1?

In your example:
{{14, 84}, {21, 56}, {42, 77}, {63, 98}}
{{14, 56, 77, 98}, {21, 42, 63, 84}}
{{10, 25, 40, 55, 85, 100}}

such extra condition would constrain you to avoid taking p-1 out of a p-star. I guess the remainder
of such p-star with 2 or more roots dropped, would need to be 'balanced' with 2 or more equally
defective rotated p-stars, and that it couldn't be done with defective rotated q-stars (q <> p).

Strange to see that so much references suddenly get found.

Thanks,

W.

----- Original Message ----- 
From: "David Wilson" <davidwwilson at comcast.net>
To: "wouter meeussen" <wouter.meeussen at pandora.be>
Cc: "Sequence Fans" <seqfan at ext.jussieu.fr>
Sent: Friday, May 20, 2005 12:22 PM
Subject: Re: A103314 conjecture


Wouter Meussen wrote:

> Note that n must be even in order to realise the 'blackness' of the
> complementary blue star.
> This means that, in my opinion, the least prime factor for the 'difficult'
> cases must be 2. No
> special cases for 3*5*7 for instance.

Is this a difficult case for 3*5*7?

10 14 21 25 40 42 55 56 63 77 84 85 98 100