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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

The puzzle is as described in A337663, but the numbers are placed on the nodes of a graph. Initially, the number 1 is placed on some of the nodes. After that, the numbers 2, 3, ... are placed, in order, on unused nodes. When the number m is placed on a node, the sum of the numbers already placed on the neighbors of that node must equal m.

The maximum reachable number for a graph G is 2 if and only if G is a path, a cycle, a star, or a complete graph, with at least three nodes. As a result, T(n,2) = 4 for n >= 4. (For graphs with three nodes, the path and the star are isomorphic, and the cycle and the complete graph are isomorphic, so T(3,2) = 2.) Proof: It is easy to see that the maximum reachable number for paths, cycles, stars, and complete graphs is 2 if the graph has at least three nodes, and 1 if it has one or two nodes. Assume that G is a connected graph which is not a path, cycle, star, or a complete graph. We must show that the number 3 can be placed on one of its nodes. Let C be a maximum clique of G, and first assume that it has size at least three. Since G is not complete, there is a node u adjacent to some, but not all, nodes in C. Let x, y, and z be nodes in C such that u is adjacent to x but not to y. We can then put 1's on u and z, 2 on x, and 3 on y. Next, assume that the size of C is less than three, i.e., that G is triangle-free. Since G is not a path or a cycle, G has a node x of degree at least three. Since G is triangle-free, there are no edges between the neighbors of x. Since G is not a star, x has a neighbor y which is adjacent to another node z. We can then put 1's on z and two of the neighbors of x other than y, 2 on x, and 3 on y.