%I
%S 2,5,31,173,1521,1056,16709,184183,1370009,474809,13478513,150399317,
%T 1034714947,2897704261
%N Probability P(n) of the occurrence of a 2D selftrapping walk of length n: Numerator.
%C A comparison of the exact probabilities with simulation results obtained for 1.2*10^10 random walks is given under "Results of simulation, comparison with exact probabilities" in the first link. The behavior of P(n) for larger values of n is illustrated in "Probability density for the number of steps before trapping occurs" at the same location. P(n) has a maximum for n=31 (P(31)~=1/85.01) and drops exponentially for large n (P(800)~=1/10^9). The average walk length determined by the numerical simulation is sum n=7..infinity (n*P(n))=70.7598+0.001
%D See under A001411
%D Alexander Renner: Self avoiding walks and lattice polymers. Diplomarbeit University of Vienna, December 1994
%D More references are given in the sci.math NG posting in the second link
%H Hugo Pfoertner, <a href="http://www.randomwalk.de/stw2d.html">Results for the 2D SelfTrapping Random Walk</a>
%H Hugo Pfoertner, <a href="https://groups.google.com/group/sci.math/msg/205ce31e504d1a0a">Selftrapping random walks on square lattice in 2D (cubic in 3D).Posting in NG sci.math dated March 4, 2002</a>
%F P(n) = a077483(n) / ( 3^(n1) * 2^a077484(n) )
%e A077483(10)=173 and A077484(10)=1 because there are 4 different probabilities for the 50 (=2*A077482(10)) walks: 4 walks with probability p1=1/6561, 14 walks with p2=1/8748, 22 walks with p3=1/13122, 10 walks with p4=1/19683. The sum of all probabilities is P(10) = 4*p1+14*p2+22*p3+10*p4 = (12*4+9*14+6*22+4*10)/78732 = 346/78732 = 173 / (3^9 * 2^1)
%o FORTRAN program provided at first link
%Y Cf. A077484, A077482, A001411.
%K frac,more,nonn,walk
%O 7,1
%A _Hugo Pfoertner_, Nov 08 2002
