A079157
Sum of square displacements over all self-avoiding walks on cubic lattice trapped after n steps. Numerator of mean square displacement a(n)/A077817(n).
Original entry on oeis.org
5, 50, 529, 3870, 28900, 185014, 1191698, 7080332, 42072344, 238337862
Offset: 11
a(12)=50 because the A077817(12)=20 trapped walks stop at 5*(1,1,0)->d^2=2, 5*(2,0,0)->d^2=4, 10*(1,0,1)->d^2=2. So, a(12)=5*2+5*4+10*2=50. See "Enumeration of all self-trapping walks of length 12" at link.
A377161
a(n) is the numerator of the probability that a self-avoiding random walk on the cubic lattice is trapped after n steps.
Original entry on oeis.org
8, 38, 637, 9759, 86221819, 28522360751, 583791967829, 1801511107253, 6467456149881773
Offset: 11
8/1953125, 38/9765625, 637/58593750, 9759/976562500, 86221819/4687500000000, 28522360751/1687500000000000, 583791967829/22500000000000000, ...
A377162 are the corresponding denominators.
A077818
a(n) is the numerator of the probability P(n) of the occurrence of a 3-dimensional self-trapping walk of length n.
Original entry on oeis.org
40, 190, 15925, 48795, 86221819, 28522360751, 583791967829, 1801511107253, 32337280749408865
Offset: 11
a(13)=15925, A077819(13)=A077820(13)=1 because there are 5 different probabilities for the 1832 (=8*A077817(13)) walks: 256 walks with probability p1=1/125000000, 88 with p2=1/146484375, 600 with p3=1/156250000, 728 with p4=1/146484375 and 160 with p5=1/244140625. P(13)=256*p1+88*p2+600*p3+728*p4+160*p5=637/(6*5^10)=25*637/(5^12*6)= 15295/(5^(13-1)*3^1*2^1)
- See under A001412.
- More references are given in the sci.math NG posting in the second link.
A378903
Decimal expansion of the expected number of steps to termination by self-trapping of a self-avoiding random walk on the cubic lattice.
Original entry on oeis.org
- Martin Z. Bazant, Topics in Random Walks and Diffusion, Graduate course 18.325, Spring 2001 at the Massachusetts Institute for Technology.
- Martin Z. Bazant, Topics in Random Walks and Diffusion, Problem Sets for Spring 2001. In the no longer available solutions to Problem Set 2b, Dion Harmon gave 3960 using 10^5 walks, and E.C.Silva gave 3676 using 1.5*10^4 walks.
- Martin Z. Bazant, Problem Set 2 for Graduate course 18.325, local pdf version of postscript file. Problem 5. Self-Trapping Walk.
- Hugo Pfoertner, Probability density for the number of steps before trapping occurs, based on 27*10^9 simulated walks (2024).
A079158
Sum of end-to-end Manhattan distances over all self-avoiding walks on cubic lattice trapped after n steps.
Original entry on oeis.org
5, 40, 399, 2472, 17436, 98400, 601626, 3238694, 18355742, 96020478
Offset: 11
a(12)=40 because the A077817(12)=20 trapped walks stop at 5*(1,1,0)->d=2, 5*(2,0,0)->d=2, 10*(1,0,1)->d=2, so a(12)=5*2+5*2+10*2=40. See "Enumeration of all self-trapping walks of length 12" at link.
A107069
Number of self-avoiding walks of length n on an infinite triangular prism starting at the origin.
Original entry on oeis.org
1, 4, 12, 34, 90, 222, 542, 1302, 3058, 7186, 16714, 38670, 89358, 205710, 472906, 1086138, 2491666, 5713318, 13094950, 30003190, 68731010, 157423986, 360530346, 825626942, 1890615518, 4329196974, 9912914314, 22698017834, 51972012258, 119000208806
Offset: 0
a(0) = 1, as there is one self-avoiding walk of length 0, namely the null-walk (the walk whose steps are the null set).
a(1) = 4 because (using the terminology in the Comment), the 4 possible 1-step walks are W_1 = {l,r,c,c-}.
a(2) = 12 because the set of legal 2-step walks are {l^2, lc, lc-, r^2, rc, rc-, c^2, cl, cr, c^-2, c-l, c-r}.
a(3) = 34 because we have every W_2 concatenated with {l,r,c,c-} except for those with immediate violations (lr etc.) and those two which go in a triangle {c^3, c^-3}; hence a(3) = 3*a(2) - 2 = 3*12 - 2 = 36 - 2 = 34.
-
w = [[[(0, 0)]]]
for n in range(1, 15):
nw = []
for walk in w[-1]:
(x, t) = walk[-1]
nss = [(x-1, t), (x+1, t), (x, (t+1)%3), (x, (t-1)%3)]
for ns in nss:
if ns not in walk:
nw.append(walk[:] + [ns])
w.append(nw)
print([len(x) for x in w])
# Andrey Zabolotskiy, Sep 19 2019
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