[seqfan] Re: nice new board game puzzle A337663
Robert Gerbicz
robert.gerbicz at gmail.com
Mon Apr 26 19:46:51 CEST 2021
Collapsed, with a somewhat easy proof a(n) is linear! See my submission:
a(n)<714*n. Proof:
As above assume k>1, since the square containing k is the sum of its
neighbors one neighbor will be at most k/2. Continuing this in at most d=11
steps we get a square not larger than max(1,k/2048).
This means that the n ones and the integers in [2,k/2048] cover all
integers from [2,k] within d=11 distance. A single square cover at most
(2*d+1)^2 squares, hence
23^2*(n+k/2048)>=k-1
from this k<714*n, so a(n) is finite and a(n)<714*n. ( more precisely we
got a(n)<=(1083392*n+2048)/1519 ). This idea works for any d>8 steps, but
gives the best upper bound for d=11.
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