Let s for scale = .182803532968295464385265406184520048. And let tc for the tribonacci constant = (1 + (19-3sqrt(33))¹⸍³ + (19+3sqrt(33))¹⸍³)/3 = 1.839286755214161132551852564653... Then trib(n) = round(s*tcⁿ) gives the correct integer for n from 0 to 146.

If you can only represent s to half as many digits, it only works for n about half that far, and so on.

I would be very interested to see a closed-form formula for s that works for all n in an arbitrary precision calculator. I used Unix's bc for the above, and obtained s as trib(91)/tc^91 where 91 is 146 divided by the Golden Ratio (I have no idea why, it just works!).

As others have pointed out in similar languages, trib(n) in bc can be defined as define trib(n) {z=0; y=0; x=1; for (i=2; i < n; i++) {zz = z; z = y; y = x; x = y+z+zz;}; return x*(n>1)} Without the *(n>1), trib(0) would be 1 instead of 0.

I don't know how to compute "round" in bc, I just eyeball it.