The Total RFEP Modulus and the RFEP Trichotomy

For an arbitrary sequence A = (a_n) with values in the profinite completion of the integers, let M(A) be the set of moduli m for which A mod m is eventually periodic. We show that M(A) is determined by a single object. With h_A(p) = sup{k : p^k in M(A)}, allowed to be infinite, and the formal product m(A) = prod_p p^(h_A(p)), we prove that m lies in M(A) if and only if m divides m(A), equivalently v_p(m) <= h_A(p) for every prime p. The proof uses only downward closure under divisibility, closure under least common multiples via the Chinese remainder theorem, and the resulting first-obstruction property. Consequently the three cases RFEP, partially RFEP and anti-RFEP correspond exactly to m(A) being maximal, strictly intermediate, or equal to 1; and A is RFEP if and only if it is p-complete RFEP for every prime p. We also record the prime-power-layer formulas for the preperiod and period spectra. As an application we consider the power-exponent suffix sequence R_n = [a^(b^n)]_(10^n), the last n decimal digits of a^(b^n). Three statements are recorded: the original sequence X_n = a^(b^n) is always RFEP; the heights h_R(2) and h_R(5) are infinite, so every modulus 2^alpha 5^beta is an RFEP modulus for R; and R is RFEP when 10 | a, and also when gcd(a,10) = 1 and 10 | b. No classification outside these regions is claimed. In the crossing regions and in the range where 10 does not divide ab, the question reduces to whether h_R(q) is finite for some prime q coprime to 10. We state three conjectures: (i) if 10 does not divide ab there is such a q with h_R(q) < infinity, equivalently R is not RFEP; (ii) q = 3 always suffices, i.e. h_R(3) = 0; (iii) for every q coprime to 10 one has h_R(q) in {0, infinity}, so no intermediate height occurs.