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The possible numbers of initial 1's required to reach T(n,k) are not always consecutive. For n = 5 and k = 2, 2 or 4 initial 1's are required to reach T(5,2) = 6.

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.

Start with an infinite square grid. Each cell has eight neighbors. Place the numbers 1, 2, ..., n anywhere. Now place the numbers n+1, n+2, ..., 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 similar to the Stepping Stones problem discussed in A337663, but predates it by more than 20 years.

As can be seen in the El Acertijo (The Riddle) links and in Rodolfo Kurchan's webpage, there are at least six similar problems, for example when the numbers are restricted to an n X n square board. All of these are worthy of inclusion in the OEIS once enough terms are known.

This is a variant of the stepping stone puzzle sequence (A337663), where you start by placing n 1's on an infinite square grid. Then place the numbers 2,3,... in order on the grid, following the rule that the sum of the 8 surrounding cells has to be equal to the number placed on a cell. a(n) is the largest number which can be achieved starting with n 1's. Additionally, there are 2 different new rules in this "stacking stones" sequence:

1. You can "stack" numbers on top of already placed numbers when the sum of the surrounding 8 cells is equal to the new number. The number inside the cell is not added to the sum and is replaced as value of the cell with the new number.

2. The starting 1's have to be at least one cell apart (to avoid a trivial infinite solution).

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.

To find a(n), start by placing n stones labeled 1 on an infinite chessboard.

Set k=2. At step k, you must place a stone labeled k in a square where the sum of its neighbors is x*k for some x >= 1. That is, the sum of the neighbors must be an integral multiple of k.

If there is no way to do that, this game ends, and you win k-1 dollars. If you can do it, you increment k and repeat.

Then a(n) = maximum number of dollars you can win with optimal play for any initial placement of the n starting stones.

The sequence was proposed by Skylark Xentha Murphy-Davies on Sep 16 2022. She found lower bounds for a(2), a(3), and a(4). On Sep 17 2022, Hugo van der Sanden showed that her lower bound for a(2) was the correct value, and found the values of a(3) and a(4).

In the original version of the problem (see A337663) each stone that you place must equal the sum of its 8 neighbors.

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.

Start with an infinite square grid. Each cell has eight neighbors. Place one 1 and n 2's anywhere. Now place the prime numbers 3, 5, 7, ..., 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 n X n square grid. Each cell has up to eight neighbors. Place the numbers 1 and 2 in any two cells of the grid. Then place a number in each remaining cell, in increasing order (not necessarily using consecutive integers), and with the rule that when you place the number k in a cell, the sum of the numbers in its neighboring cells must equal k. The goal is to fill the grid in such a way as to minimize the largest number placed.

This is similar to the Stepping Stones problem discussed in A337663, but predates it by more than 20 years.

Computer solutions by Dmitry Kamenetsky.

a(7) = 292, a(8) = 502, a(9) = 787 and a(10) 1391 not yet confirmed to be optimal.