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A335780 The number of hanging vertically stable self-avoiding walks of length n on a 2D square lattice where both the nodes and connecting rods have mass.

%I A335780 #25 Sep 22 2020 01:47:10
%S A335780 1,1,1,1,1,3,7,15,37,65,115,223,503,1127,2761,6225,15393,34915,84399,
%T A335780 193489,477727,1113059,2753799,6486011,16181965,38447093,95995579
%N A335780 The number of hanging vertically stable self-avoiding walks of length n on a 2D square lattice where both the nodes and connecting rods have mass.
%C A335780 Consider a self-avoiding walk on a 2D square lattice where each visited node is given a fixed mass and each node is connected by a rod of another fixed mass. Hang the resulting lattice structure from a string at the first node. This sequence gives the number of walks of length n such that the structure will hang perfectly vertically, and will return to this position if perturbed.
%C A335780 For a walk to be stable requires the torque around the first node to be zero for both the node and rod masses, and that the overall center of mass of the structure is lower than the first node. As n increases the number of walks satisfying these conditions decreases rapidly. For example the total number of 2D self-avoiding walks on a square lattice in the lower two quadrants for n=27 is A116903(27) = 227399388019. The total number of hanging stable walks for n=27 is 95995579, indicating only one in about 2370 walks is stable.
%C A335780 For all stable walks it is found that the final node is always directly underneath the starting node. This is not the case if only the node or rod masses are considered.
%C A335780 See A337761 for the equalivalent sequence on a 3D cubic lattice.
%H A335780 Scott R. Shannon, <a href="/A335780/a335780.txt">Stable walks for n=6 to n=15</a>.
%e A335780 a(1)-a(5) = 1 as the only stable walk is a walk straight down from the first node.
%e A335780 a(6) = 3. There is one stable walk with a first step to the right:
%e A335780 .
%e A335780       X-----+
%e A335780             |
%e A335780             |
%e A335780 +-----+-----+
%e A335780 |
%e A335780 |
%e A335780 +-----+
%e A335780 .
%e A335780 where 'X' represents the hanging point first node at (0,0).
%e A335780 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,-1). This walk can be taken in 2 ways thus, with the straight down walk, the total number of stable walks is 2+1 = 3.
%e A335780 a(20) = 193489. An example of a 20-step stable walk is:
%e A335780 .
%e A335780             X---+
%e A335780                 |
%e A335780     +---+       +---+---+
%e A335780     |   |               |
%e A335780     +   +---+---+       +
%e A335780     |           |       |
%e A335780 +---+           +---+---+
%e A335780 |
%e A335780 +---+---+---+
%e A335780 .
%e A335780 The total torque from the nodes to the right of the first node is 4*p*1*l + 2*p*2*l + 3*p*3*l = 17pl. The torque from the left nodes is 3*p*1*l + 4*p*2*l + 2*p*3*l = 17pl. The total torque from the rods to the right of the first node is 2*q*(l/2)*l + 2*q*1*l + 2*q*(3/2)*l + 2*q*(5/2)*l + 2*q*3*l = 17ql. The torque from the rods on the left is 2*q*(l/2)*l + 1*q*1*l + 2*q*(3/2)*l + 2*q*2*l + 2*q*(5/2)*l + 1*q*3*l = 17ql. This shows the configuration does not have to be symmetrical to be balanced.
%e A335780 See the linked text file for the step directions for the stable walks for n=6 to n=15.
%Y A335780 Cf. A116903, A337761, A001411, A001412.
%K A335780 nonn,more,walk
%O A335780 1,6
%A A335780 _Scott R. Shannon_, Sep 13 2020