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

[Andrew Plewe]

I suppose the choice of starting with 3 is somewhat arbitrary; I  
believe the game doesn't change much (if at all) if you start with 1.  
In fact it shouldn't matter what set of "characters" you use to play  
the game -- the ordinal position of the values you can remove should  
be independent of what base/alphabet/numeral system you choose.

What I'd ultimately like to do is figure out if the game has any  
bearing on what happens at infinity -- can you generalize the game to  
make any statements about infinite sets?

	-Andrew Plewe-


On Jan 4, 2009, at 6:56 PM, Max Alekseyev wrote:

> Andrew,
>
> It is not clear what is exactly A in the definition of you sequence.
> Is it a finite subset of odd positive integers >=3? Does it have to  
> contain 3?
>
> In any case, there is another somewhat simpler sequence related to
> your construction - the maximum number of elements that can be removed
> from the n-element set A (whatever it is) without affecting the set of
> its pairwise sums. This sequence starts with
>
> b(n): 0, 0, 0, 0, 1, 1, 2, 2, 3, 3, 4, 4, 5, 6, 6, 7, 8, 8, 9, 10, 11,
> 11, 12, 12, 13
>
> There is a trivial connection between your sequence a(n)
>
> a(n):  
> 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
> (as computed by Peter)
>
> and the defined above b(n):
>
> a(n) = n + (n-1) + ... + (n-b(n)+1) = b(n) * (2n+1-b(n)) / 2
>
> Correspondingly, the reverse relation is:
>
> b(n) = ( (2n+1) - sqrt( (2n+1)^2 - 8a(n)) ) / 2.
>
> Regards,
> Max
>
> On Fri, Dec 26, 2008 at 3:33 AM, Andrew Plewe  
> <andrew at nevercenter.com> wrote:
>> 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-
>>
>>
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