From Charlotte Darroch, Jan 11 2022, a comment in the "Stones on an Infinite Chessboard" video: (Start)

The upper bound can be improved to a(n) <= 278*n. Suppose k > 1. Then k has a neighbor which is at most k/2. Suppose we continue this for d steps. Then there is a square no larger than max(1,k/2^d) within distance d of the number k.

Then the n ones along with the numbers in [2,k/2^d] cover all numbers in [2,k] within a distance of d. The argument for a(n) < 714*n continues by saying that each square covers at most (2d+1)^2 numbers. However, this can be improved by observing that any number x in [2,k/2^d] is within distance 1 of a smaller number y in that same set, or a 1. So x's square heavily overlaps y's square. In fact, there can be at most 4d+1 numbers covered by x which aren't covered by y, which occurs when x and y are diagonally adjacent.

Therefore the n ones cover at most (2d+1)^2 numbers each and the numbers in [2,k/2^d] cover at most 4d+1 numbers each. These cover all of [2,k]. So n(2d+1)^2+(k/2^d-1)(4d+1) >= k-1.

This becomes k <= (n(2d+1)^2-4d)/(1-(4d+1)/2^d), provided that 1-(4d+1)/2^d > 0, so provided 2^d > 4d+1, which is true for all d >= 5. Then to obtain the best bound with this method, we minimize the linear coefficient of n, subject to the constraint that d >= 5. This is minimized for d = 6, so we obtain k <= (10816n-1536)/39, or a(n) <= 278*n.

PS When I mention that any number x in [2,k/2^d] is within distance 1 of a smaller number y, I should have realized that x is in fact within distance 1 of at least two numbers smaller than it (the numbers which sum to x), y and z. And the overlap is slightly higher as a result, so x only has at most 2d+1 squares which aren't covered by smaller numbers, for any x in [2,k/2^d].

This involves checking all the ways that both y and z can be adjacent to x, but you'll find a maximum non-overlap of 2d+1 cells for x. So in fact we obtain that each of the n ones cover at most (2d+1)^2 numbers, each of the numbers in [2,k/2^d] cover at most 2d+1 numbers, yet all numbers in [2,k] are covered. Hence n(2d+1)^2+(k/2^d-1)(2d+1) >= k-1. This becomes k <= (n(2d+1)^2-2d)/(1-(2d+1)/2^d), provided that 1-(2d+1)/2^d > 0, which happens for d >= 3. Then the optimal choice of d is d = 5, which gives k <= (3872n-320)/21, or a(n) < 185*n.

(End)

Comment on the previous bound from _Robert Gerbicz_, Jan 12 2022: (Start)

A minor improvement: When we count the integers from [2,k] we can write: n*((2*d+1)^2-1)+(2*d+1)*(k/2^d-1) >= k-1.

This saves n on the left side, because we don't count the n ones (the ones in the center of squares of side length=2*d+1).

For d >= 4 this gives a(n) = k <= (1280*n-128)/7 < 183*n. (End)