Problem with Queens and Pawns

[James A. Sellers]

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



****************************************
James A. Sellers
Associate Professor, Mathematics
Cedarville College

sellersj at cedarville.edu
http://www.cedarville.edu/dept/sm/personalpages/jas_www.htm


>>> Erich Friedman <efriedma at stetson.edu> 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
the
>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
http://www.stetson.edu/~efriedma