[seqfan] Re: A potentially new sequence based on a simple game
Peter Pein
petsie at dordos.net
Tue Dec 30 19:15:01 CET 2008
Andrew Plewe schrieb:
> I have a game for the list; I'd be somewhat surprised if it hasn't
> been "solved" before as it's a pretty simple game. Let set A be the
> set of odd numbers starting with 3:
>
> {3,5,7,etc..}
>
> When you form a sum table, you get the even numbers starting with 6,
> or set B:
>
> {6,8,10,etc..}
>
> Like this:
>
> 3 5 7
> 3 6 8 10
> 5 10 12
> 7 14
>
> The "complete" expression of B shown here is {6,8,10,12,14}; in other
> words, when you take the first 3 characters from set A, {3,5,7}, and
> use that subset to form a sum table you get the first 5 characters of
> set B. The game is, can we remove any characters from a subset of A
> and still get a "complete" expression of set B? In this case you
> can't; removing 3 or 5 or 7 would render the resulting sum table
> incomplete.
>
> I believe the first subset of A that allows for a deletion is
> {3,5,7,9,11}, which in turn "expresses" {6,8,10,12,14,16,18,20,22}.
> You can remove 7 and you'll still get all of those values when you
> compute the sum table:
>
> 3 5 9 11
> 3 6 8 12 14
> 5 10 14 16
> 9 18 20
> 11 22
>
> Now, the sequence is formed by counting the maximum number of sums
> that can be removed from the sum table when you play this game for a
> subset of A of length n. Here are the first few values of the sequence
> (based on by-hand computation), starting with n = 1:
>
> 0,0,0,0,5,6,13,15,22
>
> and the corresponding subsets of A are:
> {3}
> {3,5}
> {3,5,7}
> {3,5,7,9}
> {3,5,9,11}
> {3,5,7,11,13} or {3,5,9,11,13}
> {3,5,9,13,15}
> {3,5,9,13,15,17}
> {3,5,9,13,17,19}
>
> Thus in the example above n = 5 and when we removed the number 7 from
> the subset we removed 5 sums from the resulting sum table. In other
> words, imagine writing out the sum table for {3,5,7,9,11} and deleting
> the row and the column starting with 7 and counting the deleted sums.
>
> Is there an algorithmic way to determine this number? It would appear
> to be so (I think there's a fairly simple pattern, but I'm not quite
> sure about it yet), but if that's the case then odd things happen if
> you let your subset equal A and thus extend to infinity.
>
> -Andrew Plewe-
>
Hi Andrew,
It would be nice if there was a _simple_ pattern, but I can't find one.
Using a brute-force backtracking method [1] I get the following start of this
sequence:
0,0,0,0,5,6,13,15,24,27,38,42,55,69,75,91,108,116,135,155,176,187,210,222,247...
which differs at a(9)=24 from your value 22 (maybe a typo?). This value is
reached when you take {7, 11, 15} out of A_9 ={3,5,...,19}.
Peter.
[1] http://home.arcor.de/petsie/seqfans/AndrewsGame.nb (Mathematica notebook)
and http://home.arcor.de/petsie/seqfans/AndrewsGame.pdf (the same as pdf )
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