548 Proceedings of Royal Society of Edinburgh. [sess. 
(n - 2p)x 1 + (n- 2p)x 2 + • • • 4- (n - 2p)x n — • • • • 
- 2x 1 + 2x p+1 = ... . 
2x 2 + 2x = • • • • 
2x p _ 1 - 2x n _ 1 = • • • • 
et nous aurons, par les memes considerations que ci-dessus, 
A' = n(- If- 1 A = Hn - 2p)2 n ~\ - 1 )‘ 2n ~ 2 
done 
A = (n- 2p)( - 2) n_1 . ” 
The final line here contains the incorrect result above referred 
to. That it is incorrect one instance will suffice to show, say the 
instance in which n = 6 and p = 2. The circulant in question is 
then 
-1-11111 
1-1-1 1 1 1 
1 1-1-1 1 1 
1 1 1-1-1 1 
1 1 1 1-1-1 
- 1 1 1 1 1 - 1 ; 
and, as the sum of its 1st, 3rd and 5th rows is the same as the 
sum of its 2nd, 4th and 6th rows, its value is 0 instead of 
(6 -4) (-2)5. 
(3) The error in the proof lies in saying at the close that after 
the manner of § 17 it can he shown that the determinant of the 
derived set of equations equals n{n - 2p)2 n ~ 1 * ( - 1) 2M-2 . This 
determinant, in fact, when we remove the factor n - 2p from the 
modified first row and 2 from each of the other rows as afterwards 
modified is, in the special case just adduced, equal to 
(6 - 4)-2 5 - 
1 1 
- 1 
. -1 
1 1 
1 
. - 1 
- 1 
. - 1 
1 1 
-1 
. -1 
- 1 
1 
