Theorem: For the Stepping Stones Sequence, A337663, liminf a(n)/n > 6 (The argument gives something like a(n) > 6.0128*n-5621 for all n.) Robert Gerbicz From a post to the Sequence Fans Mailing List, October 5 2022 (Edited by N. J. A. Slaone, Jan 03 2023) This message gives an explicit construction that shows that Liminf a(n)/n>6 ! The computer prgram mentioned in this document is available from the adjacent link in the OEIS entry A337663. Theorem: for every n: a(n) > 6.0128*n-5621 Proof: Of course a(n) is an increasing sequence: for m>n to see that a(m)>=a(n), use the construction for n, and place the m-n additional 1s in cells that have no neighbour with numbers from [2,a(n)]. We will actually show that a(934*t+528)>=5616*t+3164 for every integer t>=0, and then from the increasing condition it follows that: a(n)>=(2808*n-2624900)/467 so a(n)>6.0128*n-5621 Start with the basic 6n-1 construction: 9 8 1 10 7 11 6 12 5 1 13 4 14 3 15 2 1 16 1 17 Call this a tree (or house). The first idea is that we can build (main) branches on this: 9 8 1 10 7 11 33 34 35 36 37 38 39 40 41 6 12 1 1 1 42 5 1 13 50 49 48 47 46 45 44 43 4 14 3 15 2 1 16 1 17 (This is not a construction for n=50, it just shows how to place the interval [33,50]). This places p additional 1s and gains 6p numbers. In some cases it is better to push down almost the whole branch: In some cases it is better to push down almost the whole branch by one notch: 9 8 1 10 7 11 33 6 12 34 35 36 37 38 39 40 41 5 1 13 1 1 1 42 4 14 50 49 48 47 46 45 44 43 3 15 2 1 16 1 17 Notice we cannot start these branches with every number, since in general we see t t+1 3t+3 t+2 So we can start only with a number that is divisible by 3. If somehow we could freely place (without using more 1s) 3 consecutive numbers, then just start a branch at t+2 instead of t+1 and saved a "half" one. And this works: use the places marked at A,B and W. 9 8 1 10 7 11 33 6 12 34 35 36 37 38 39 40 41 A 5 1 13 W 1 1 1 42 4 14 50 49 48 47 46 45 44 43 B 3 15 2 1 16 1 17 (Now you can see why we pushed down the branch, otherwise 33 would be also included in the sum for W). Why does this work? Because even these are of course not 3 consecutive numbers. But gives a complete residue system mod 3, it is also true that B=A+2, but that isn't interesting. Drawing enough const*n branches, we will see const*n with a complete residue system. Hence we get a (6+c)*n construction for fixed c>0. At W the number is larger than A or B, but this is not a problem, we can use that independently of A and B. We are almost done. Since A,B,W are too large, we need also to build branches on the main branches. And also there is a small complication because on the (main) branch we would need to find 6 consecutive numbers and not three(!) that we could place freely. To avoid this use also the following ladder construction between the main branches: (The numbers marked with x are on the branch.) xxxxxxxxxxxxxxxx h h+1 g+8 1 h+2 g+7 h+3 g+6 h+4 g+5 1 h+5 g+4 h+6 g+3 h+7 g+2 1 h+8 g+1 g xxxxxxxxxxxxxxx (Yes, we could also freely place g+9 and h+9, but we do not need to use that) To get a solution for n=934*t+528 we call the computer program for t (without any switch). Do not use line breaking in the file to see the construction. The grid is checked to be correct, and the coordinates are printed out in another file. (End) ---------------------------------------------------------------------------- Neil Sloane replied as follows: Oct 6, 2022, 12:41 AM Robert, that is really amazing, I did not think it was possible by that approach. Congratulations! I see the general idea, but I got lost around the point where you mentioned A, B, and W. When you have time, could you give more details starting at that point? Best regards Neil ------------------------------------------------------------------------------ Robert's reply, Oct 6, 2022 (lightly edited) OK, but it was not my intention to give a very strict proof with all computed constants, only to show the ideas. I have to say it is not really proved that it gives n=934*t+528, I just calculated it from two t values, assuming that n is linear in t. For the input t number: p=156t+97 and the given construction will show that a(934*t+528)>=5616*t+3164 So first use the basic construction using p 1s (house/tree construction), an additional one to push down 6p,6p+1,6p+2. Really don't know why we needed these 3 extra numbers. What I have not said, but is clear from the construction, is that the (horizontal) main branches contain an equal number of 1s, that is H=13. This also means that 6H numbers (bigger than 1) are on each main branch. And the distance of these consecutive main branches is also constant, 6*H/3=26. Then the first missing number is 6p+3, so the first main branch starts at 6p+3. The k-th main branch starts at 6p+78k+3. It is easily seen that A=12p+156k+84, B=A+2, W=24p+312k+169. If for k=k' at W we would see A+1=W then we have 3 consecutive numbers A,W,B. A+1=W gives that: 12*p+156*k+84+1=24*p+312*k'+169 Solving for k' gives k'=(-12*p+156*k-84)/312 it turns out that if for example p=97 mod 156 and k is even then k' is an integer. For A,W,B we do not use additional 1s, so for each such triple saving a half one over the classical 6m construction, if we can place down linear number of triplets, and that is the case, notice that (roughly): 12p