Problem with Queens and Pawns

Joe Crump joecr at
Thu Jun 1 17:29:15 CEST 2000

I like the 'mod n' approach!

I'll love the day when we have a big set of polynomials
in terms of pieces & their positions to get a 'best' move
to play. :)

But I suppose you could devote an entire database by itself
to 'Chess' sequences!

Sequences like these though, notwithstanding leading
to a better understanding of the pieces, could lead
to a simplified way of calculating values needed by
chess programs. E.g. instead of having to find the
legal moves, iterate them looking for captures, pins,
etc, it would be astounding to be able to do these types
of operations with some sort of polynomial operations
on the pieces & positions.

-----Original Message-----
From: James A. Sellers [mailto:SELLERSJ at]
Sent: Thursday, June 01, 2000 11:18 AM
To: seqfan at; math-fun at optima.CS.Arizona.EDU;
efriedma at
Subject: Re: Problem with Queens and Pawns

Erich's comment is a very good one!  Rather than taking even index
terms and odd index terms, why not take the terms with index congruent
to i mod 4, i=0,1,2,3?  Then the second differences will be constant,
which should allow for a formula to be calculated (which will be in
four parts, based on n mod 4).  So, for example, if we only look at
0,7,24,51, ..., and write f(0)=0, f(1)=7, and f(2)=24, and assume a
quadratic function for these (since the second difference is
constant), then f(x) = 5*x^2+2x.  (Note that plugging in 3 for x in
this function does yield 51, which is nice to see!!!)  The other three
subsequences should have formulas that are nice as well.  

Ultimately, it would be nice to have more terms in the original
sequence to make sure that the pattern noted by Erich continues to
hold!!  (I have assumed that the pattern does continue to hold in the
above work.)

Hope this helps!


James A. Sellers
Associate Professor, Mathematics
Cedarville College

sellersj at

>>> Erich Friedman <efriedma at> 06/01/00 10:53AM >>>
>Consider an n X n chessboard. Place n queens in the cells of the
first row,
>namely in the cells (1,1), (2,1),..., (n,1), and [(n+1)/2] pawns in
>odd cells of the second row, namely in the cells (1,2), (3,2),
(5,2), ...
>Which is the number of the unattacked (by the queens) cells ?
>n |  1 2 3 4 5 6  7  8  9 10 11 12 13 14 15 .....
># |  0 0 0 2 4 7 10 14 19 24 30 36 44 51 60 .....
>Formula ??

look at only the even index terms, and compute second differences:

0   2   7  14  24  36  51
  2   5   7  10  12  15
    3   2   3   2   3

the same thing happens when we do only the odd index terms:

0   4  10  19  30  44  60
  4   6   9  11  14  16
    2   3   2   3   2

hope this helps.

erich friedman 

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