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

Max Alekseyev maxale at gmail.com
Mon Jan 5 03:56:29 CET 2009

```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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>
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>

```