551 
1902-3.] Dr Muir on a Special Circulant. 
and this when the a’s are 
-1 , -1 , 1 , 1 , 1 , 1 
vanishes. 
(7) Catalan’s oversight is the less easily accounted for because of 
the form of his next proposition, which is to the effect that the 
circulant C(1 , 1 , 1 , ... ,0,0) whose elements are p positive units 
followed by n — p zeros has the value p or 0 according as p and n 
are prime to one another or not. 
In support of one part of this we note that when p and n have 
a common factor h greater • than 1 the left-hand sides of the two 
equations which originate the circulant have a common factor 
1 +£c + ce 2 -f • • • +x h ~ 1 exactly as in the previous case. In fact the 
theorem of § 5 may be generalised into the following without 
altering the wording of the proof : — 
The circulant C(a , a , ... , b , b) whose elements are p a’s 
followed by n - p b’s has the value 0 when p is not prime to n. 
(8) There is thus suggested the problem of finding the value of 
C(a , a , . . . , b , b) in the remaining case, viz., when the number 
of a’s is prime to the number of U s. Instead, however, of dealing 
with this by itself and by a different method from that followed 
in the other case, it will be more instructive to show that the 
value of C(a , a , . . . , b , b) for any number of a’s or 6’s can be 
found without going outside the theory of determinants. As a 
first step towards this we have the following theorem : — 
The value of the circulant C(a , a , ... , b , b) whose elements are 
p a ’ 8 followed by v b’s is (/xa + j/b) (a-b)^+ v_1 multiplied by a 
determinant derivable from the said circulant by putting in the first 
row a = 1 , b = 1 , and in all the other rows a = 1 , b = 0 . 
To prove this we note that by increasing the first row of the 
given circulant by all the other rows there is obtained the row 
pa + vb , pa + vb , , pa + vb , 
so that we can remove the factor pa + vb , and leave a first row 
whose every element is 1. Next by subtracting b times this new 
first row from each of the other rows we obtain in every one of 
these rows an a — b wherever an a formerly was and a 0 wherever 
there was formerly a b. From each row therefore we can remove 
