This is a summary of the meeting of Wed. July 3, 2024:

• Three strategies for exact 3 copy Skolem:
  • Place the smallest remaining integer that can fit in the first available gap
  • Place the smallest remaining integer in the first gap that fits
  • If the next smallest integer cannot fit in the first available gap, drop it
  These strategies can be related to $1/x+1/(x+1)+1/(x+2)=1$. It is likely that $a_n/n$ does not converge to $x$, but it fluctuates around it (experimentally). In case of the third strategy, the equation becomes $(1-r)[1/x+1/(x+1)+1/(x+2)]=1$, where $r$ is the rate of dropping. Valentio worked on the implementation of the three strategies and shared them on slack.


• In terms of partitioning $\mathbb{N}$ into triplets $(a_n, b_n, c_n)$, it is not possible using the idea of a spectrum of an irrational number. This is probably why the exact 3 copy Skolem sequence is not easily captured by the idea of rate. For the same reason, a game of Nim with 3 piles might not be able to partition $\mathbb{N}$ into triplets by considering the cold positions. Jack worked on several variants of the game of Nim with 3 piles and confirmed that there is no partitioning. Here's a reference on the subject of spectrum in general (reference provided by Valentio).


• One idea to work with $1/p+1/q+1/r=1$ as a partitioning method is to consider whether these 3 numbers can be used to somehow partition a different space, something like $\mathbb{N}^2$, and study what that might mean in terms of a generalized Skolem sequence. Input from Saman.


• Ben Gildea suggested a way for constructing an infinite Skolem sequence by "switching" between the next number and some large number. Ben, Jack, Sambhu, and Valentio are considering ideas by which we can systematically construct an exact 3 copy Skolem sequence, not necessarily based on any of the 3 strategies mentioned above. The general theme relies on this "switching" idea, using large numbers whenever the next number does not fit, to fill existing gaps and create a stretch of gaps that is large enough for the next number to fit. This is similar to the first strategy where "large" means simply the next number that can fit in the first available gap.


• Restricted set of integers. One approach is to consider an exact 3 copy Skolem sequence for a subset of $\mathbb{N}$; for instance, the powers of 2. Sambhu was exploring this idea. Saman pointed out that this is equivalent to saying that $b_n-a_n=c_n-b_n=f(n)$, where $f(n)=2^{n-1}$. Other functions can be explored...


• Finite 3 copy Skolem sequences exist. I don't know what the necessary conditions are. We know we must have: $\sum a_i + \sum b_i + \sum c_i=1+2+\ldots +3n=3n(3n+1)/2$ and $\sum b_i - \sum a_i = \sum c_i - \sum b_i = 1+2+\ldots +n=n(n+1)/2$, but that did not lead to anything useful. We get $\sum a_i=n^2$, $\sum b_i=n(3n+1)/2$, and $\sum c_i=n(2n+1)$, all of which are integers. I shared on slack a C program that searches for a 3 copy Skolem sequence for a given $n$. It will be slow for higher values of $n$. It is possible that the program can be speeded up by not going through all permutations, but instead using a randomized strategy like in here. Here are examples found by search (the search favors lexicographic order):
  • $n=1$: 1 1 1
  • $n=9$: 2 4 2 8 2 4 6 7 9 4 3 8 6 3 7 5 3 9 6 8 5 7 1 1 1 5 9
  • $n=10$: 1 1 1 2 9 2 10 2 6 3 7 8 3 9 6 3 10 7 5 8 6 4 9 5 7 4 10 8 5 4
  • $n=11$: 1 1 1 2 4 2 11 2 4 5 10 7 4 9 5 6 8 11 7 5 10 6 9 3 8 7 3 6 11 3 10 9 8
  • $n=18$: 1 1 1 2 4 2 5 2 4 6 18 5 4 16 12 6 5 15 17 3 11 6 3 14 10 3 12 13 18 16 9 11 15 8 10 17 7 14 12 9 13 8 11 7 10 16 18 15 9 8 7 14 17 13
  • $n=19$: 1 1 1 2 4 2 5 2 4 6 17 5 4 18 14 6 5 19 7 11 15 6 12 10 16 7 13 17 14 9 11 18 7 10 12 15 19 8 9 13 16 11 14 10 17 8 12 9 3 18 15 3 13 8 3 19 16
  • $n=20$: 1 1 1 2 4 2 5 2 4 6 9 5 4 17 20 6 5 14 19 9 15 6 18 3 12 13 3 16 9 3 17 14 10 11 20 15 12 19 13 7 18 8 10 16 11 14 7 17 12 8 15 13 10 7 20 11 19 8 18 16
  Note: It seems there are no exact $k$ copy Skolem sequences for small $n>1$ when $k>3$ (I tested $4\leq k\leq 100$ and $n\leq 20$). For $k=4$, it seems a necessary condition is $n\equiv 0,1 \bmod 4$.


• Skolem placement to minimize gaps. Alexis is exploring some ideas of online algorithms. For offline algorithms, one idea is to compare the two greedy strategies below:
  • Greedy 1: Sort the numbers. Place the smallest number in the first available gap that works.
  • Greedy 2: Sort the numbers. Place the largest number in the first available gap that works.
  • The second strategy seems to perform better in terms of minimizing gaps for sets of the form $\{1, 2, \ldots, n\}$. Observe that the optimal number of gaps for such a set is either 0 (if $n\equiv 0,1 \bmod 4$) or 1 (if $n\equiv 2,3 \bmod 4$).


• Let me know if there is anything I missed, or if you have some idea, and I will add it...