436 
Proceedings of Royal Society of Edinburgh, [july 18 , 
whose sum is 8, that is, 1 + 2 more than the sum of the last three 
indices of ^a 4 6 2 c 2 <i : this gives 
0, 2, 6; 0, 3, 5; 1, 2, 5; 1, 3, 4. 
Bring then each of the pairs to each of the triads, thus 
h, k 
l, m t n, 
and inquire if they dovetail, that is to say, if k<n<fm, and 
h<m<^l. It is seen that 0, 4 dovetails two times, and 1, 3 four 
times; that is, altogether, six times. The coefficient required is (3. 
I had never attempted to deal with expansions of so high a 
degree as a(0, 5, 13, 17) which is Professor Johnson’s example. 
But the following will show that the work necessary for obtaining 
three consecutive coefficients of this expansion is not heavy. 
Second Example. — Required the coefficients of ^a 10 5 9 cbi 3 , 
3a 10 6 9 c 6 ^ 4 , 5« 10 6 9 c 5 c7 5 in the expansion of cc(0, 5, 13, 17). 
Solution . — The groups are — 
0 1 2 3 4 
5 6 7 8 9 10 11 12 
13 14 15 16 
A 
B 
C 
Pairs with sum 11, i.e. 1 + (3 + 7), and lowest member ^>3. 
0,11; 1,10; 2,9; 3,8. 
Triads from A, B and C with sum 22, i.e. 1 + 2 + (3 + 7 + 9). 
0,6,16; 0,7,15; 0,8,14; 0,9,13; 
1,5,16; 1,6,15; 1,7,14; 1,8,13; 
2,5,15; 2,6,14; 2,7,13; 
3, 5, 14; 3, 6, 13. 
Number of dovetailings in the case of 0, 11 4, 
)> )•> 
55 55 55 
55 55 55 
which is the first coefficient required. 
In finding the coefficient of 5a 10 6 9 c 6 <i 4 the integer 4 would not 
be deleted in group A, and there would therefore be another triad 
4, 5, 13. There would also be another pair 4, 17, and consequently 
1, 10 
8, 
2, 9 
11, 
3, 8 
12: 
Total, 
35, 
