Let "k" be the numerousness of these "stable" digits, which satisfy the congruence relation G(mod 10"k")≡[GG](mod 10"k").  "k"="t"-1, where G("t"):=3↑↑"t". It follows that, . The algorithm above produces the following 500 rightmost decimal digits of Graham's number (or of any tower of more than 500 3s):