Synchronization is important to long range order, since it can effectively prevent patterns of destructive interference from introducing wrong configurations. While mathematically perfect growth functions may not ever occur in the quasicrystal lab, they can certainly give us insight to what may happen there. The biggest takeaway (which is hinted at in Socolar's "Growth Rules for Quasicrystals") is that frustration is an important part of growing Ulam structures. Snowflakes tend to grow on corners due to sharpness of the electromagnetic field, so perhaps a similar mechanism affects quasicrystals.

Now that we have a perfectly synchronized function up to time $t \sim 100$, we can calculate the growth sequences fairly easily:

Length /@ AStates[[All, 2, 2]]
Differences[%]
Divide[%% - 1, 5]
Divide[%%, 5]
Transpose[Partition[%, 4]]

(* editor's note: these sequences have been corrected on Oct. 26 *)
Out[]:= {1, 1, 1, 1, 1, 6, 6, 6, 6, 16, 16, 16, 16, 21, 21, 21, 21, 36, 36, 
36, 36, 61, 61, 61, 61, 71, 71, 71, 71, 101, 101, 101, 101, 121, 121, 
121, 121, 136, 136, 136, 136, 156, 156, 156, 156, 176, 176, 176, 176, 
191, 191, 191, 191, 236, 236, 236, 236, 306, 306, 306, 306, 331, 331, 
331, 331, 396, 396, 396, 396, 441, 441, 441, 441, 471, 471, 471, 471, 
511, 511, 511, 511, 551, 551, 551, 551, 581, 581, 581, 581, 671, 671, 
671, 671, 736, 736, 736, 736, 751, 751, 756}
Out[]:= {0, 0, 0, 0, 5, 0, 0, 0, 10, 0, 0, 0, 5, 0, 0, 0, 15, 0, 0, 0, 25, 0, 
0, 0, 10, 0, 0, 0, 30, 0, 0, 0, 20, 0, 0, 0, 15, 0, 0, 0, 20, 0, 0, 
0, 20, 0, 0, 0, 15, 0, 0, 0, 45, 0, 0, 0, 70, 0, 0, 0, 25, 0, 0, 0, 
65, 0, 0, 0, 45, 0, 0, 0, 30, 0, 0, 0, 40, 0, 0, 0, 40, 0, 0, 0, 30, 
0, 0, 0, 90, 0, 0, 0, 65, 0, 0, 0, 15, 0, 5} (* note sync breaking *)
Out[]:= {0, 0, 0, 0, 0, 1, 1, 1, 1, 3, 3, 3, 3, 4, 4, 4, 4, 7, 7, 7, 7, 12, 
12, 12, 12, 14, 14, 14, 14, 20, 20, 20, 20, 24, 24, 24, 24, 27, 27, 
27, 27, 31, 31, 31, 31, 35, 35, 35, 35, 38, 38, 38, 38, 47, 47, 47, 
47, 61, 61, 61, 61, 66, 66, 66, 66, 79, 79, 79, 79, 88, 88, 88, 88, 
94, 94, 94, 94, 102, 102, 102, 102, 110, 110, 110, 110, 116, 116, 
116, 116, 134, 134, 134, 134, 147, 147, 147, 147, 150, 150, 151}
Out[]:={0, 0, 0, 0, 1, 0, 0, 0, 2, 0, 0, 0, 1, 0, 0, 0, 3, 0, 0, 0, 5, 0, 0, 
0, 2, 0, 0, 0, 6, 0, 0, 0, 4, 0, 0, 0, 3, 0, 0, 0, 4, 0, 0, 0, 4, 0, 
0, 0, 3, 0, 0, 0, 9, 0, 0, 0, 14, 0, 0, 0, 5, 0, 0, 0, 13, 0, 0, 0, 
9, 0, 0, 0, 6, 0, 0, 0, 8, 0, 0, 0, 8, 0, 0, 0, 6, 0, 0, 0, 18, 0, 0, 
0, 13, 0, 0, 0, 3, 0, 1}
Out[]:= {
{0, 1, 2, 1, 3, 5, 2, 6, 4, 3, 4, 4, 3, 9, 14, 5, 13, 9, 6, 8, 8, 6, 18, 13}, 
{0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0}, 
{0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0}, 
{0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0}}

The first sequence is the number of suns turned on at time $t$. None of these integer sequences appear to be on OEIS, though many similar sequences are. Actually, the growth algorithm is fairly difficult to program, so it's more likely that OEIS would have something on the triality pattern itself, for example:

Length /@ NestList[Function[{verts}, Union@Join[Flatten[
      Nearest[Complement[AStates[[-1, 2, 2]] , verts ], #] & /@ verts,
       1], verts]], {{0, 0}}, 10]
Divide[% - 1, 5]
Differences[%]
Differences[%%%]

Out[276]= {1, 6, 16, 21, 36, 61, 81, 106, 136, 176, 211}
Out[277]= {0, 1, 3, 4, 7, 12, 16, 21, 27, 35, 42}
Out[278]= {1, 2, 1, 3, 5, 4, 5, 6, 8, 7}
Out[279]= {5, 10, 5, 15, 25, 20, 25, 30, 40, 35}

Neither do any of these appear in OEIS, so perhaps we are really doing something fun and original here.

Hey thanks!

Here's one more graph: Growing the Penrose tiling Triality pattern. Adding one of three colors to star-centers ultimately decides all binary choices. The two alternate vertex configurations do not occur in equal proportion, and the above graph marks the less frequent figure in lime green. A very nice feature of this pattern is that all green vertices fall on a thick edge of one of the three or four larger tiles. After about ~100 time steps, we have:

So yes, it appears the Penrose tiling can be grown by a deterministic local rule, but the neighborhood may need to be on distance scale $\phi^4$ (as seen in the background).