Timeline for Connective constant for self-avoiding walks on a skip-chain
Current License: CC BY-SA 2.5
9 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Nov 7, 2010 at 9:22 | comment | added | Yaroslav Bulatov | interesting, I'm planning on checking it against a transfer matrix method | |
| Nov 7, 2010 at 5:19 | comment | added | Yuval Filmus | I now have a complete proof that $\mu \approx 2.20556943040059$, following the steps of my answer below. | |
| Nov 2, 2010 at 9:36 | answer | added | Yuval Filmus | timeline score: 3 | |
| Nov 2, 2010 at 9:11 | comment | added | Yuval Filmus | Approximately, the process divides into "bricks" of sizes 1,2, and for $n\geq 3$, two bricks of size $n$. The generating series is then $(1-x)/(1-2x-x^3)$. The only real root of the denominator is 0.453397651516404, so your $\mu$ is approximately 2.20556943040059. | |
| Nov 2, 2010 at 8:34 | comment | added | Yuval Filmus | I got a recurrence relation for the number of such walks. The first few numbers are 4, 12, 30, 70, 160, 360, 802, 1778, 3932, 8684, 19166, 42286, 93280, 205752, 453818. With some perseverance, one could extract the asymptotics (perhaps even a closed form) from the recurrence relation, but it's a bit messy. | |
| Nov 2, 2010 at 3:39 | comment | added | Yaroslav Bulatov | yes, added clarification | |
| Nov 2, 2010 at 3:38 | history | edited | Yaroslav Bulatov | CC BY-SA 2.5 |
clarification
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| Nov 2, 2010 at 3:21 | comment | added | JBL | Since it's undirected, I guess you mean that each $i$ is connected to $i \pm 1$ and $i \pm 2$? | |
| Nov 2, 2010 at 2:04 | history | asked | Yaroslav Bulatov | CC BY-SA 2.5 |