The bishops problem asks for the maximum number of bishops 
that can be placed on an 
chessboard such that
no two attack each other, where 
is a positive integer.
The answer is 
and 
for 
(Dudeney 1970, Madachy 1979), giving the sequence 2, 4, 6, 8, ... (the even
numbers) for 
,
3, .... One maximum solution for 
is illustrated above. The numbers of placements of 
indistinguishable nonattacking bishops
on a fixed 
chessboard, counting rotations
and reflections separately, are 1, 4, 26, 260, 3368,
... for 
,
2, ... (OEIS A002465, with its empty-board
term omitted).
Let 
count maximum arrangements on an 
chessboard, with rotations and reflections
identified. Then 
,
and for 
,
![M(n)={2^((n-4)/2)[2^((n-2)/2)+1] for n even; 2^((n-3)/2)[2^((n-3)/2)+1] for n odd](/images/equations/BishopsProblem/NumberedEquation1.svg) |
(1)
|
(Dudeney 1970, p. 96; Madachy 1979, p. 45; Pickover 1995). An equivalent formula for 
is
 |
(2)
|
where 
is the floor function. The values 
for 
, 2, ... are 1, 1, 2, 3, 6, 10, 20, 36, .... For 
, 
for OEIS A005418;
the one-square value is prepended.
The minimum number of bishops needed to occupy or attack all squares on an 
chessboard is 
(Cockayne et al. 1986). A minimum arrangement places
one bishop on each square of a central row, as illustrated above.
Queens Problem." §C18 in