42 
Proceedings of the Royal Society 
where the numbers printed in darker type are inserted by the rule 
given above. This is, of course, in one sense a complete solution 
of the problem ; but the results may easily be put in an analytical 
form. 
Had we had zeros along the line 
V = x - 2 
we should have had the following scheme instead of that above : 
0 1 0 • . x 
0 1110 
0 1 2 3 2 1 0 . . ( 6 ) 
013676310 
0 1 4 10 16 19 16 10 4 1 0 
i 
&c. y &c. 
Hence the part added by the units along the line 
y = x - 2 
is 
0 . . x 
0 1 
0 112 
0 1 2 4 3 3 . . (c) 
01379 10 64 
1 
y &c. 
This, again, differs from (6) shifted one place downwards, by 
0 . . x 
0 0 
0 0 1 
0 112 . . (d) 
1 2 4 3 3 
y &c. 
But it is obvious that this is a repetition of the same one place 
diagonally downwards to the right. 
