

A337663


Solution to stepping stone puzzle if we start with n 1's (see Comments).


11




OFFSET

1,2


COMMENTS

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.
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
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)
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) >= k1.
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)
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:
 Daniel Darroch, Jan 11 2022: a(n) <= 278*n. (See link)
 Daniel 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)
 Robert Gerbicz, Oct 05 2022: lim inf a(n)/n > 6 (Probably a(n) > 6.0128*n5621 for all n.) See link
(End)
Al Zimmermann has informed me that he is running a computerprogramming 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


LINKS

N. J. A. Sloane, A lower bound of 6*n+3 for n >= 3, based on the optimal construction for n=2 (black) and continuing the obvious pattern of the red squares. Discovered by Menno Verhoeven, Jan 10 2022, and mentioned in a comment on the "Stones on an Infinite Chessboard" video.


FORMULA

a(n) >= 5*n  4.
Proof: This follows by continuing the following simple construction:
+++++++++++
   4       14  
+++++++++++
  3  1  5     13  1  15 
+++++++++++
  2   6     12   16 
+++++++++++
 1     7   11    
+++++++++++
     8  1  10    
+++++++++++
      9     
+++++++++++
(End)
a(n) >= 6*n  6. [This has since been strengthened to a(n) >= 6*n for n >= 3  see Comments and Link.  N. J. A. Sloane, Sep 14 2022]
Proof: This follows by continuing the following simple construction:
+++++++++++
 1          
+++++++++++
  2  3  4  5  6  7  8  9  
+++++++++++
   1    1    1  10 
+++++++++++
  18  17  16  15  14  13  12  11  
+++++++++++
(End)
Menno Verhoeven, Jan 10 2022, has shown that a(n) >= 6*n+3 for n >= 3. See my "Lower bound of 6n+3" link. This is obtained by starting from the a(2)=16 configuration. He remarks that by starting from a larger configuration one can improve the constant term 3, although not the multiplier 6. For example, starting from the a(6) configuration gives a(n) >= 6n+24 for n >= 6.  N. J. A. Sloane, Jan 10 2022


EXAMPLE

Starting with n = 2 cells containing 1's the following strategy achieves a(2) = 16 (this is also shown in the link):
+++++++
 9  5  10  11   
+++++++
  4  1    
+++++++
 12  8  3  2   16 
+++++++
    6  1  15 
+++++++
   13  7  14  
+++++++
. 24 . . . . .
. 8 16 17 . . .
15 7 1 . 19 . .
22 . 6 2 . 20 .
. 28 . 3 1 . 25
. . . . 4 5 .
. . 21 . 9 18 23
. . 11 10 . . .
. 12 1 . . . .
. 26 13 14 . . .
. . . 27 . . . (End)
Illustration for a(4) = 38 from Arnauld Chevallier:
. . . . . . . . . . . . . . .
. 35 18 36 . 23 . 21 . 32 . . . . .
. . 17 1 . 14 9 . 12 20 . . . . .
. . 34 16 15 . 5 4 8 . . 26 27 . .
. . . . 31 . 10 1 3 19 25 . 1 28 .
. . . . . . 11 . 2 6 . 33 . 29 .
. . . . . . 24 13 22 1 7 . . . .
. . . . . . 37 . . 30 38 . . . .
. . . . . . . . . . . . . . .
Illustration for a(6) = 60:
. . . . . . . . . . . . . 47 24 48 .
. . . . . . . . . . . . . . 23 1 49
. . . . . . . . . . . . 41 . 22 . 50
. . . . . . . . 51 . 36 . 20 21 43 . .
. . . . . . . . 34 17 . 19 1 . . . .
. . . . . . . . 16 1 18 38 58 59 . . .
. . . . . 37 30 15 40 . 57 . . . . . .
. . . . . . 7 8 . . . . . . . . .
. . . 35 46 6 1 25 33 . . . . . . . .
. 60 32 . 3 2 9 . . . . . . . . . .
. . 28 4 1 31 11 45 . 52 . . . . . . .
. 42 14 10 5 . . 12 13 39 . . . . . . .
. 56 . 29 44 . . 1 26 . . . . . . . .
. . . . . . 55 54 27 53 . . . . . . .
(End)


CROSSREFS

See A355903 for another version of the problem.


KEYWORD

nonn,more,nice


AUTHOR



EXTENSIONS

a(1)a(4) confirmed by Arnauld Chevallier
a(5) from Code Golf user xash (see Code Golf Stack Exchange link).  Peter Kagey, Oct 08 2020
a(7) >= 71 was found by Mark Beyleveld, Aug 07 2023 (see Links).  Al Zimmermann, Jan 02 2023
The programming contest also produced the lower bounds 80, 90, 99, 109, 118 for n = 8, ..., 12, respectively (see Links). Al Zimmermann, Jan 05 2023


STATUS

approved



