

A355903


Variant of Stepping Stones problem: here the stone you place only needs to divide the sum of its 8 neighbors.


2




OFFSET

1,2


COMMENTS

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 k1 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 MurphyDavies 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.


LINKS

Skylark Xentha MurphyDavies and Hugo van der Sanden, Illustration for a(2) = 27 (Discovered by Skylark Xentha MurphyDavies and proved optimal by Hugo van der Sanden.) [There are two choices for the 27 stone, indicated by dashed lines. The 27 in the top left corner is not part of the arrangement.]


EXAMPLE

. . . . . . . .
. 21 20 . . . . .
23 13 9 18 . . . .
. 12 5 4 . 14 25 .
. 27 10 1 3 11 26 .
. . . . 2 6 . 22
. . . 19 . 1 7 15
. . . . 17 16 8 .
. . . . . . 24 .
. . . . . . . .


CROSSREFS



KEYWORD

nonn,bref,more


AUTHOR



STATUS

approved



