[seqfan] A potentially new sequence based on a simple game

Andrew Plewe andrew at nevercenter.com
Fri Dec 26 09:33:58 CET 2008

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:


When you form a sum table, you get the even numbers starting with 6,  
or set B:


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  

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:


and the corresponding subsets of A are:
{3,5,7,11,13}  or  {3,5,9,11,13}

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-

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