Logo
All Random Solved Random Open
OPEN
Let $f(n,k)$ count the number of self-avoiding walks of $n$ steps (beginning at the origin) in $\mathbb{Z}^k$ (i.e. those walks which do not intersect themselves). Determine \[C_k=\lim_{n\to\infty}f(n,k)^{1/n}.\]
The constant $C_k$ is sometimes known as the connective constant. Hammersley and Morton [HM54] showed that this limit exists, and it is trivial that $k\leq C_k\leq 2k-1$.

Kesten [Ke63] proved that $C_k=2k-1-1/2k+O(1/k^2)$, and more precise asymptotics are given by Clisby, Liang, and Slade [CLS07].

Conway and Guttmann [CG93] showed that $C_2\geq 2.62$ and Alm [Al93] showed that $C_2\leq 2.696$. Jacobsen, Scullard, and Guttmann [JSG16] have computed the first few decimal places of $C_2$, showing that \[C_2 = 2.6381585303279\cdots.\]

Additional thanks to: Cedric Pilatte