a(1)-a(5) = 1 as the only stable walk is a walk straight down from the first node.
a(6) = 5. There is one stable walk confined to a single plane:
.
X-----+
|
|
+-----+-----+
|
|
+-----+
.
where 'X' represents the hanging point first node at (0,0,0).
Assuming a mass of p for the nodes, q for the rods, and a length l for the rods, the total torque from the nodes to the right of the first node is 2*p*l, which equals that from the nodes to the left. The total torque for the rods to the right of the first node is 2*q*(1/2)*l + 1*q*1*l = 2ql, which equals that from the rods to the left. The center of mass is at coordinate (0,0,-1). This walk can be taken in 4 ways thus, with the straight down walk, the total number of stable walks is 4+1 = 5.
a(10) = 193. Other than the straight and single plane walks those using three dimensions now occur. An example of such a walk is:
.
+--------+ z y
/ / \ /
/ X-------+ \/
/ / \ +-----x
/ / \
+ + +
\ / /
\ / /
+--/----+ /
/ /
+--------+
.
The total rotational torque around the y-axis from the nodes with x>0 is 3*p*l, which equals that from the nodes with x<0. The total rotational torque around the x-axis from the nodes with y>0 is 2*p*l, which equals that from the nodes with y<0. The total rotational torque around the y-axis from the rods with x>0 is 2*q*(1/2)*l + 2*q*1*l = 3ql, which equals that from the rods with x<0. The total rotational torque around the x-axis from the rods with y>0 is 2*q*(1/2)*l + 1*q*1*l = 2ql, which equals that from the rods with x<0. The center of mass is at coordinate (0,0,-1).
a(11) = 581. An example of an 11 step stable walk where the final node is above the first node:
.
+ +
/ /| z
/ / | | y
/ / | | /
+ / | |/
+ | X---------+ | +-----x
/| | +----------+
/ | | /
/ | | /
/ | | /
+-----------+ /
+---------+
.
This particular walk would be counted twice as it is also stable if hung from the final node.
See the linked text file for the step directions for the stable walks for n=6 to n=12.
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