672 states * 2 symbols = 1344

I rewrote Roman Czyborra's Haskell code in their comment into NotQuiteLaconic:

proc main() {
    digits = 3*2;
    q = 1;
    r = 180;
    t = 60;
    i = 2;
    
    while (true) {
        u = 3*(3*i+1)*(3*i+2);
        
        /* Inline div */
        a = q*(27*i-12)+5*r;
        b = 5*t;
        y = 0;
        while (a >= b) {
            a = a - b;
            y = y + 1;
        }

        if (i > 2) {
            digits = digits*2 + y;
        }
        /* Order of statements important here! */
        r = 2*u*(q*(5*i-2)+r-y*t);
        t = t*u;
        q = 2*q*i*(2*i-1);
        i = i+1;
    }
}

and then compiled it to a 2-symbol turing machine (this unfortunately makes it not exactly fit the requirements in the question, and it also does not write the digits on the tape in the way you want). But at least I wanted to show this "upper bound" of some sort.

Here is the list of states in the format you specified and here is the slightly more meaningful NQL listing.