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a(n) is the number of climbable arrangements that exist for sets of n adjacent "broken ladders" with height n, where a broken ladder is an array of n steps with some number of the steps unusable, the rest usable; an arrangement is the configuration of the locations of the broken rung(s) on the n ladders of height n; and a climbable arrangement is a set of ladders such that with movement up, down, left, and right, there exists a path from the bottom to the top.

Also, a(n) is the sum of the coefficients of exact spanning probabilities in 2d lattices along the second dimension for an n X n square lattice.

This is a variant of the stepping stone sequence (A337663), where any cell has just 4 neighbors (Von Neumann neighborhood). The game works as follows:

Start with an infinite square grid. Each cell has four neighbors. Place n 1's anywhere. Now place the numbers 2, 3, ..., m in order, subject to the rule that when you place k, the sum of its neighbors must equal k. Then a(n) is the maximum m that can be achieved.

This is a variant of the stepping stone sequence (A337663), where now the board is a hexagonal grid.

Start with an infinite hexagonal grid. Each cell has six neighbors. Place n 1's anywhere. Now place the numbers 2, 3, ..., m in order, subject to the rule that when you place k, the sum of its neighbors must equal k. Then a(n) is the maximum m that can be achieved.

Start with an infinite square grid. Each cell has eight neighbors. Place n 1's anywhere. Now place the numbers 2, 3, ..., m in order, subject to the rule that when you place k, the sum of its neighbors must equal k. Then a(n) is the maximum m that can be achieved.

a(1) - a(4) were found by Thomas Ladouceur and Jeremy Rebenstock by considering all the possibilities (by computer - see link to Python program).

Note that a(n) always exists, by definition. But it is possible that it is infinity. One consequence of the following argument is that a(n) < oo. - N. J. A. Sloane, Oct 08 2020

From Robert Gerbicz, Oct 08 2020: (Start)

a(n) = O(n*log(n)^2). Proof:

Assume k > 1. Since the square containing k is the sum of its neighbors, one neighbor will be at most k/2. Continuing this (with the smallest term from the sum): if k < 2^(d+1) then one term within distance d from the square containing k will be at most 1, hence exactly one.

But n squares (containing ones) cover at most (2*d+1)^2*n squares within distance d. So for all d > 0, min(2^(d+1)-2, a(n)-1) <= (2*d+1)^2*n.

From this a(n) is finite because 2^d/d^2 is unbounded. Use the inequality for that d where 2^(d+1) < a(n) <= 2^(d+2), then (a(n)-4)/2 <= 2^(d+1)-2 = min(2^(d+1)-2, a(n)-1) <= (2*d+1)^2*n < 4*log_2(a(n))^2*n, and from a(n) < 4 + 8*log_2(a(n))^2*n it is easy to see that a(n) = O(n*log(n)^2). (End)

From Robert Gerbicz, Apr 26 2021: (Start)

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 a distance of d=11. A single square covers 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. (End)

From N. J. A. Sloane, Aug 26 2022: (Start)

This entry has grown too long, so I have moved some of the comments to attached text files. At present these are, in chronological order:

- Charlotte Darroch, Jan 11 2022: a(n) <= 278*n. (See link)

- Charlotte Darroch, Jan 11 2022, and Robert Gerbicz, Jan 12 2022: a(n) <= 183*n. (See link)

- Tejo Vrush, Jan 22 2022: a(n) <= 155*n. (See link)

- Jonathan F. Waldmann, Aug 17 2022: a(n) < 86*n + 32. (See link)

- Jonathan F. Waldmann, Oct 01 2022: a(n) < 79*n + C. (See link)

- Robert Gerbicz, Oct 05 2022: lim inf a(n)/n > 6 (Probably a(n) > 6.0128*n-5621 for all n.) See link

- Skylark Xentha Murphy-Davies, a(n) >= 6*n for n >= 3 (see link)

(End)

Al Zimmermann has informed me that he is running a computer-programming competition (see link) in which contestants try to improve the lower bounds on a(n). This has already produced many improvements. Several contestants (the first was Mark Beyleveld) have shown that a(7) >= 71. Other lower bounds are a(8) >= 79, a(9) >= 89, a(10) >= 99, a(11) >= 109, a(12) >= 115. The full results will be announced when the competition ends in November 2022, and at that time the contestants may reveal that they also have proofs that some of these lower bounds are in fact the exact values. - N. J. A. Sloane, Aug 26 2022

See A350627 for several older problems that are similar to this, such as the Forest of Numbers (Bosque de Números) puzzle. - N. J. A. Sloane, Feb 05 2022