Notes on sequence A306211 N. J. A. Sloane, Feb 01, 2019 Definitions: Let R(S) denote the RUNS transform of a finite or infinite string S. The sequence A306211 will be denoted by W. W is defined to be the limit of W_k as k -> oo, where W_1 = 1, and for k>1, W_k = W_{k-1} R(W_{k-1}). Let S_k = R(W_{k-1}) for k >= 2. The definition says that W_k = W_{k-1} S_k for k>=2. We call W_k the k-th generation of W. We have: W_1 = 1 S_2 = 1 W_2 = 11 S_3 = 2 W_3 = 112 S_4 = 21 W_4 = 11221 S_5 = 221 W_5 = 11221221 S_6 = 22121 W_6 = 1122122122121 S_7 = 221212111 W_7 = 1122122122121221212111 S_8 = 221212111211113 W_8 = 1122122122121221212111221212111211113 Of course in general, for strings S and T, R(ST) != R(S) R(T), because there can be runs at the end of S that get amalgamated with runs at the start of T. E.g. R(W_4) = R(11221) = 221 != R(W_3) R(S_4) = 21.11. Let's describe this by saying that there is an "amalgamation at generation 4". There are amalgamations at generations 2 and 4. But I don't know of any other amalgamations. So let's make a weak conjecture: Conjecture 1. There are no amalgamations after generation 4. If this is true, it simplifies the analysis. For it implies: Conjecture 2: S_k = S_{k-1} R(S_{k-1}) for k > 4. There are two conflicting conjectures in A306211: Conjecture 3a: (Peter Kagey) The terms in W do not exceed 5. Conjecture 3b: (me) Every positive integer appears in W. What we know at present is that 1, 2, 3, 4, 5 first appear at positions 1, 3, 37, 60, 255 in W (A323829). And, thanks to Ben Chaffin, if 6 does appear this happens after position 62350575902. Note that amalgamations of S_k and R(S_k) are common: that is why we see 3, 4, 5 (and, I believe, 6, 7, 8, ...) in W. Assuming Conj. 2 holds, we can forget about the W_k, and just look at the S_k. Let me illustrate by explaining how the first 5 appears. Start from (see A323478 for the S_k's): S_9 = 221212111211113211113141 *** We see that R(S_9) = 211113..., so when we concatenate S_9 R(S_9) to get S_10 we see S_10 = .......11113141 211113... and so S_11 = S_10 R(S_10) contains .....4111114... and so R(S_11) contains .....5.... and so S_12 = S_11 R(S_11) contains ....5.... In fact if you keep track of where these things are happening, you see that S_12, which has length 98, ends with ...513111, and in W, S_12 occupies positions 163 to 260, so the 5 is in position 255. Conclusions: 1. We can probably forget about the W_k's and just look at the S_k's, and 2. what we hope for is an S_k like S_9 (see ***) which ends with a bunch of runs of length 1.