PROFESSOR CAYLEY’S NINTH MEMOIR ON QUANTICS. 
27 
Table No. 88 (concluded). 
Ind. a 
Ind. x. 
Coeff. 
8 
40 
i 
A 8 
1 
38 
0 
36 
i 
A 6 C 
34 
i 
A 6 F 
32 
2 
A 6 B, A 4 C 2 
30 
2 
A 5 E, A 3 CF 
28 
3 
AT, A 4 BC, A 2 C 3 , AT 2 
V' 
26 
3 
A 4 I, A 3 BF, ATE, ACT 
S' 
24 
5 
A 4 B 3 , A 4 H, A 3 CD, A 2 BC 2 , A 2 EF, CF 2 
2' 
22 
4 
A 3 BE, A 3 L, ATI, A 2 DF, ABCF, ACT 
2S' 
20 
6 
A 4 G, A 3 BD, A 2 B 2 C, A 2 CH, AT 2 , ACT, AFI, BC 3 , BF 2 , CEF 
<r 
18 
5 
A 3 K, A 2 BI, ATE, ABT, ABCE, ACL, AFII, C 2 I, CDF 
4S' 
16 
7 
A 3 J, A 2 B 3 , A 2 BH, A 2 CG, AT 2 , ABCD, AEI, B 2 C 2 , BEF, C 2 H, CE 2 , FI . 
5S' 
14 
5 
A 2 N, AB 2 E, ABL, ACK, ADI, A EII, AFG, BCI, BDF, CDE 
(T 
12 
7 
A 2 BG, A 2 M, AB 2 D, ACJ, ADH, B 3 C, BCH, BE 2 , C 2 G, CD 2 , EL, FK, l 2 . 
3S 
10 
5 
ABK, AEG, AP, B 2 I, BDE, CN, DL, FJ, HI 
3S 
8 
6 
ABJ, ADG, B 4 , B 2 H, BCG, BD 2 , CM, EK, H 2 
2S 
6 
3 
AO, BN, DK, EJ, GI 
2S 
4 
4 
B 2 G, BM, DJ, DH 
2 
1 
R 
* 
0 
2 
G 2 , Q 
* 
341. The syzygies and interconnexions of syzygies are given in 
Table No. 89. 
(5, 
11) 
AI + BF— CE = 0 
(6, 
18) 
AT- A 2 BC + 4C 3 +F 2 
= 0 
(6, 
14) 
A 2 H— 6ACD — 4BC 2 — EF 
= 0 
(6, 
12) 
AL - 2CI + 3DF 
= 0 
(6, 
10) 
A 2 G — 12ABD — 4B 2 C — E 2 
= 0 
(6, 
8) 
AK + 2BI — 3DE 
= 0 
(6, 
6) 
AJ - B 3 +2BH-CG-9D 
’=0 
(7, 
15) 
A 2 BD-ABC 2 + ACII-6CT- 
- FI =0 
(7, 
13) 
A 2 K - ABI— 3B 2 F +6CL +3FH =0 
(7. 
11) 
A 2 J - AB :i + ABII — 9 AD 2 — 
6BCD-E1 =0 
(7, 
9) 
AN - B 2 E — 6DI + 2EH - 
FG =0 
2BL + 6DI - EH + 
FG =0 
2CK -12DI + EH- 
FG =0 
(7, 
7) 
AM + 2BT + CJ — 3DII 
= 0 
tr, (8, 
20) 
0 . A 2 ( A 2 G -12ABD -4B 2 C - 
- E 2 ) 
supra 
(6, 
10) 
— A (A 2 BD— ABC 2 + ACII 
-6CT-FI) 
55 
(7, 
15) 
+ B(AT - A 2 BC +4C 3 + F 2 ) 
5^ 
(6, 
18) 
+ C(A 2 H - 6ACD -4BC 2 - 
- EF) 
55 
(6 
14) 
-F (AI + BF - CE) 
= 0 
( 5 , 
11) 
cr, (8, 
14) 
0. A(AN— B 2 E- 6DI + 2EH- 
-FG) 
supra 
(7, 
9) 
+ A( 2BL + 6DI — EH + FG) 
55 
( 
> ) 
+ A( 2CK-12DI+ EH- 
-FG) 
55 
( 
— 2B(AL — 2CI + 3DF) 
„ 
(6, 12) 
-2C(AK + 2BI - 3DE) 
55 
(6, 
8) 
+ 6D(AI + BF - CE) 
= 0 
” 
(5, 
U) 
e 2 
