this is clearly the wrong rabbit hole to have fallen down but at this point i'm all out of ideas. previous bests were a 3-arm at 1179 (min cycles at pseudoperiod 5) and a 1-arm at 1138.

this solve is essentially two independent 1-arm machines which share a common output--the one adjacent to the salt. the tape loop is 2P, in which each arm delivers 13 times: twice to each of their unique outputs, and once to the common output. 

if the six unique outputs for each arm are A through F, and the shared output is G, then the delivery pattern is 

Arm 1: (A,B) (C,D) (E,F) (G)   (A,B) (C,D) (E,F)
Arm 2: (A,B) (C,D) (G,C) (A,B) (D,E) (F)   (E,F)

the peculiar pattern on arm 2 allows for more efficient use of dead cycles, leading to a reduction in the tape loop.