A367995 - OEIS

%I #8 Dec 21 2023 15:37:06

%S 1,1,3,3,21,21,7,21,21,1001,77,77,77,1001,77,77,1001,1001,77,91,77,

%T 89089,785603,143,143,24297,143,25924899,97097,785603,97097,25924899,

%U 143,89089,143,97097,97097,143,25924899,97097,97097,143,143,97097,143,97097,143,143,97097,97097,97097,143,97097,291291,291291,143

%N a(n) is the denominator of the probability that the free polyomino with binary code A246521(n+1) appears as the image of a simple random walk on the square lattice.

%C In a simple random walk on the square lattice, draw a unit square around each visited point. A367994(n)/a(n) is the probability that, when the appropriate number of distinct points have been visited, the drawn squares form the free polyomino with binary code A246521(n+1).

%C Can be read as an irregular triangle, whose n-th row contains A000105(n) terms, n >= 1.

%H Pontus von Brömssen, <a href="/A367995/b367995.txt">Table of n, a(n) for n = 1..6473</a> (rows 1..10).

%H <a href="/index/Pol#polyominoes">Index entries for sequences related to polyominoes</a>.

%F A367994(n)/a(n) = (A368000(n)/A368001(n))*A335573(n+1).

%e As an irregular triangle:

%e      1;

%e      1;

%e      3,  3;

%e     21, 21,  7, 21,   21;

%e   1001, 77, 77, 77, 1001, 77, 77, 1001, 1001, 77, 91, 77;

%e   ...

%e There are only one monomino and one free domino, so both of these appear with probability 1, and a(1) = a(2) = 1.

%e For three squares, the probability for an L (or right) tromino (whose binary code is 7 = A246521(4)) is 2/3, so a(3) = 3. The probability for the straight tromino (whose binary code is 11 = A246521(5)) is 1/3, so a(4) = 3.

%Y Cf. A000105, A246521, A335573, A367672, A367761, A367994 (numerators), A367997, A367999, A368000, A368001.

%K nonn,frac,tabf

%O 1,3

%A _Pontus von Brömssen_, Dec 08 2023