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The Hales-Jewett numbers $c_{n,k}$ are defined, essentially, to be the largest possible size of a subset of $\mathbb Z_k^n$ free of $k$-term arithmetic progressions with the difference in $\{0,1\}^n$ and satisfying the additional restriction that the support of the initial term of the progression is disjoint from the support of its difference. A significant part of the Polymath project Density Hales-Jewett and Moser Numbers consists in determining the numbers $c_{n,3}$. We know the first seven Hales-Jewett numbers (with $k=3$): $$ c_{0,3}=1,\ c_{1,3}=2,\ c_{2,3}=6,\ c_{3,3}=18,\ c_{4,3}=52,\ c_{5,3}=150,\ c_{6,3}=450; $$ see Theorem 1.4 of the aforementioned project, or OEIS sequence A156989.

What if we drop the support disjointness restriction and define, say, $\mu_n$ to be the largest possible size of a subset of $\mathbb F_3^n$ free of three-term arithmetic progressions with the difference in $\{0,1\}^n$? It is not particularly difficult to see that $$ \mu_0=1,\ \mu_1=2,\ \mu_2=6,\ \mu_3=14,\ \text{and}\ 36\le\mu_4\le 40. $$

Virtually nothing is known about the order of growth of the numbers $\mu_n$, and to develop some intuition, it would be extremely helpful to compute several more of them.

Added March 8, 2017

Robert Israel reports $\mu_4=36$ using cplex. A wild guess that I certainly disbelieve myself: $\mu_n=\lfloor 6^{n/2}\rfloor$.

The Hales-Jewett numbers $c_{n,k}$ are defined, essentially, to be the largest possible size of a subset of $\mathbb Z_k^n$ free of $k$-term arithmetic progressions with the difference in $\{0,1\}^n$ and satisfying the additional restriction that the support of the initial term of the progression is disjoint from the support of its difference. A significant part of the Polymath project Density Hales-Jewett and Moser Numbers consists in determining the numbers $c_{n,3}$. We know the first seven Hales-Jewett numbers (with $k=3$): $$ c_{0,3}=1,\ c_{1,3}=2,\ c_{2,3}=6,\ c_{3,3}=18,\ c_{4,3}=52,\ c_{5,3}=150,\ c_{6,3}=450; $$ see Theorem 1.4 of the aforementioned project, or OEIS sequence A156989.

What if we drop the support disjointness restriction and define, say, $\mu_n$ to be the largest possible size of a subset of $\mathbb F_3^n$ free of three-term arithmetic progressions with the difference in $\{0,1\}^n$? It is not particularly difficult to see that $$ \mu_0=1,\ \mu_1=2,\ \mu_2=6,\ \mu_3=14,\ \text{and}\ 36\le\mu_4\le 40. $$

Virtually nothing is known about the order of growth of the numbers $\mu_n$, and to develop some intuition, it would be extremely helpful to compute several more of them.

Added March 13, 2017

Robert Israel reports $\mu_4=36$ and $\mu_5\ge 102$, using cplex. Moreover, computations seem to suggest that, indeed, $\mu_5=102$ may hold true.

The Hales-Jewett numbers $c_{n,k}$ are defined, essentially, to be the largest possible size of a subset of $\mathbb Z_k^n$ free of $k$-term arithmetic progressions with the difference in $\{0,1\}^n$ and satisfying the additional restriction that the support of the initial term of the progression is disjoint from the support of its difference. A significant part of the Polymath project Density Hales-Jewett and Moser Numbers consists in determining the numbers $c_{n,3}$. We know the first seven Hales-Jewett numbers (with $k=3$): $$ c_{0,3}=1,\ c_{1,3}=2,\ c_{2,3}=6,\ c_{3,3}=18,\ c_{4,3}=52,\ c_{5,3}=150,\ c_{6,3}=450; $$ see Theorem 1.4 of the aforementioned project, or OEIS sequence A156989.

What if we drop the support disjointness restriction and define, say, $\mu_n$ to be the largest possible size of a subset of $\mathbb F_3^n$ free of three-term arithmetic progressions with the difference in $\{0,1\}^n$? It is not particularly difficult to see that $$ \mu_0=1,\ \mu_1=2,\ \mu_2=6,\ \mu_3=14,\ \text{and}\ 36\le\mu_4\le 40. $$

Virtually nothing is known about the order of growth of the numbers $\mu_n$, and to develop some intuition, it would be extremely helpful to compute several more of them.

The Hales-Jewett numbers $c_{n,k}$ are defined, essentially, to be the largest possible size of a subset of $\mathbb Z_k^n$ free of $k$-term arithmetic progressions with the difference in $\{0,1\}^n$ and satisfying the additional restriction that the support of the initial term of the progression is disjoint from the support of its difference. A significant part of the Polymath project Density Hales-Jewett and Moser Numbers consists in determining the numbers $c_{n,3}$. We know the first seven Hales-Jewett numbers (with $k=3$): $$ c_{0,3}=1,\ c_{1,3}=2,\ c_{2,3}=6,\ c_{3,3}=18,\ c_{4,3}=52,\ c_{5,3}=150,\ c_{6,3}=450; $$ see Theorem 1.4 of the aforementioned project, or OEIS sequence A156989.

What if we drop the support disjointness restriction and define, say, $\mu_n$ to be the largest possible size of a subset of $\mathbb F_3^n$ free of three-term arithmetic progressions with the difference in $\{0,1\}^n$? It is not particularly difficult to see that $$ \mu_0=1,\ \mu_1=2,\ \mu_2=6,\ \mu_3=14,\ \text{and}\ 36\le\mu_4\le 40. $$

Virtually nothing is known about the order of growth of the numbers $\mu_n$, and to develop some intuition, it would be extremely helpful to compute several more of them.

Added March 8, 2017

Robert Israel reports $\mu_4=36$ using cplex. A wild guess that I certainly disbelieve myself: $\mu_n=\lfloor 6^{n/2}\rfloor$.

The Hales-Jewett numbers $c_{n,k}$ are defined, essentially, to be the largest possible size of a subset of $\mathbb Z_k^n$ free of arithmetic progressions with the difference in $\{0,1\}^n$ and satisfying the additional restriction that the support of the initial term of the progression is disjoint from the support of its difference. A significant part of the Polymath project Density Hales-Jewett and Moser Numbers consists in determining the numbers $c_{n,3}$. We know the first seven Hales-Jewett numbers (with $k=3$): $$ c_{0,3}=1,\ c_{1,3}=2,\ c_{2,3}=6,\ c_{3,3}=18,\ c_{4,3}=52,\ c_{5,3}=150,\ c_{6,3}=450; $$ see Theorem 1.4 of the aforementioned project, or OEIS sequence A156989.

What if we drop the support disjointness restriction and define, say, $\mu_n$ to be the largest possible size of a subset of $\mathbb F_3^n$ free of arithmetic progressions with the difference in $\{0,1\}^n$? I know that $$ \mu_0=1,\ \mu_1=2,\ \mu_2=6,\ \mu_3=14,\ \text{and}\ 36\le\mu_4\le 40. $$

Virtually nothing is known about the order of growth of the numbers $\mu_n$, and to develop some intuition, it would be extremely helpful to compute several more of them.

The Hales-Jewett numbers $c_{n,k}$ are defined, essentially, to be the largest possible size of a subset of $\mathbb Z_k^n$ free of $k$-term arithmetic progressions with the difference in $\{0,1\}^n$ and satisfying the additional restriction that the support of the initial term of the progression is disjoint from the support of its difference. A significant part of the Polymath project Density Hales-Jewett and Moser Numbers consists in determining the numbers $c_{n,3}$. We know the first seven Hales-Jewett numbers (with $k=3$): $$ c_{0,3}=1,\ c_{1,3}=2,\ c_{2,3}=6,\ c_{3,3}=18,\ c_{4,3}=52,\ c_{5,3}=150,\ c_{6,3}=450; $$ see Theorem 1.4 of the aforementioned project, or OEIS sequence A156989.

What if we drop the support disjointness restriction and define, say, $\mu_n$ to be the largest possible size of a subset of $\mathbb F_3^n$ free of three-term arithmetic progressions with the difference in $\{0,1\}^n$? It is not particularly difficult to see that $$ \mu_0=1,\ \mu_1=2,\ \mu_2=6,\ \mu_3=14,\ \text{and}\ 36\le\mu_4\le 40. $$

Virtually nothing is known about the order of growth of the numbers $\mu_n$, and to develop some intuition, it would be extremely helpful to compute several more of them.