Problem with Queens and Pawns
John Conway
conway at math.Princeton.EDU
Thu Jun 1 20:07:49 CEST 2000
On Thu, 1 Jun 2000, Erich Friedman quoted and wrote:
> >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...
[they are alternately 2 and 3, if we redefine f(1) = -1]
> the same thing happens when we do only the odd index terms...
In such circumstances there's often a single formula of the type
"nearest-integer-to-a-polynomial" that does all cases. Let me first
find the formulae for the separate residue classes modulo 4:
1 5 9 13 2 6 10 14 3 7 11 15 4 8 12
--------------- --------------- --------------- -----------
-1 4 19 44 0 7 24 51 0 10 30 60 2 14 36
5 15 25 7 17 27 10 20 30 12 22
10 10 10 10 10 10 10
Since the second differences of x^2 are 2 this means that
the leading term of f(4x+k) is 5x^2, so the leading term of
f(x) will be 5x^2/16. So I'll compute 5x^2 - 16f(x):
1 5 9 13 2 6 10 14 3 7 11 15 4 8 12
--------------- --------------- --------------- -----------
21 61 101 141 20 68 116 164 45 85 125 165 48 96 144
40 40 40 48 48 48 40 40 40 48 48
This tells us that the x-term in this difference is 10x
or 12x according as x is odd or even, giving the new
approximations (5x^2 - 10x)/16 or (5x^2 - 12x)/16
so I'll work out the nearest integer to 5(x-1)^2/16 for odd x:
1 3 5 7 9 11 13 15
------------------------
0 1 5 11 20 31 45 61 {5(x-1)^2/16}
-1 0 4 10 19 30 44 60 f(x)
So f(x) is 1 more than the nearest integer to 5(x-1)^2/16,
for odd x. I have to go to lunch now, but "I shall return".
John Conway
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