I believe I have identified the 1.264 mentioned in the formula for A005186. Let c be the Collatz function, that is c(n) = n/2 if n even 3n+1 if n odd We then have c^-1(n) = {2n} if n == 0, 1, 2, 3, or 5 (mod 6) {2n, (n-1)/3} if n == 4 (mod 6) Assume that in a large set of positive integers == 4 (mod 6), (n-1)/3 will be congruent to 1, 3, or 5 (mod 6) with equal likelihood. Then, if S is a large set of positive integers, and if v_k (0 <= k <= 5) is the number of integers in S congruent to k (mod 6), and M is the 6x6 matrix 1 0 0 1 0 0 0 0 0 0 1/3 0 0 1 0 0 1 0 0 0 0 0 1/3 0 0 0 1 0 0 1 0 0 0 0 1/3 0 and w = M v, then w_k should give a reasonable approximation to the number of integers == k (mod 6) in c^-1(S). This means that for large n, we might expect A005186(n+1)/A005186(n) to be near the large eigenvalue of M. By brute force, I found the large eigenvalue of M is about 1.26376261582, close to the reported 1.264. I believe that this is in truth (3+sqrt(21)/6 (thanks to Plouffe's inverter), but I don't have the matrix weaponry to prove it.