A322419 Number of n-step self-avoiding walks on L-lattice.
1, 2, 4, 8, 12, 20, 32, 52, 84, 136, 220, 356, 564, 904, 1448, 2320, 3684, 5872, 9376, 14960, 23688, 37652, 59912, 95316, 150744, 239080, 379528, 602424, 951788, 1507136, 2388252, 3784344, 5973988, 9447880, 14950796, 23658540, 37321752, 58965260, 93206864, 147333080, 232286272
Offset: 0
Examples
a(1) = 2 because there are only two possible directions at each intersection; for the same reason a(2) = 2*2 and a(3) = 2*4 ; but a(4) = 12 (not 16) because four paths return to the starting point and are not self-avoiding. See the 12 paths under "links".
Links
- Sean A. Irvine, Table of n, a(n) for n = 0..50
- Robert FERREOL, The a(4)=12 walks in L-lattice
- Keh-Ying Lin and Yee-Mou Kao, Universal amplitude combinations for self-avoiding walks and polygons on directed lattices, J. Phys. A: Math. Gen. 32 (1999), page 6929.
- A. Malakis, Self-avoiding walks on oriented square lattices, Journal of Physics A: Mathematical and General, Volume 8, Number 12 (1975), page 1890.
- Wikipedia, Connective constant
Crossrefs
Programs
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Maple
walks:=proc(n) option remember; local i,father,End,X,walkN,dir,u,x,y; if n=1 then [[[0,0]]] else father:=walks(n-1): walkN:=NULL: for i to nops(father) do u:=father[i]:End:=u[n-1]:if n mod 2 = 0 then dir:=[[1,0], [-1, 0]] else dir := [[0,1], [0, -1]] fi: for X in dir do if not(member(End+X,u)) then walkN:=walkN,[op(u),End+X] fi; od od: [walkN] fi end: n:=5:L:=walks(n):N:=nops(L); # This program explicitly gives the a(n) walks. -
Mathematica
mo = {{1, 0}, {-1, 0}}; moo = {{0, 1}, {0, -1}}; a[0] = 1; a[tg_, p_: {{0, 0}}] := Module[{e, mv}, If[Mod[tg, 2] == 0, mv = Complement[Last[p] + # & /@ mo, p], mv = Complement[Last[p] + # & /@ moo, p]]; If[tg == 1, Length@mv, Sum[a[tg - 1, Append[p, e]], {e, mv}]]]; a /@ Range[0, 20] (* after the program from Giovanni Resta at A001411 *) -
Python
def add(L, x): M = [y for y in L] M.append(x) return M plus = lambda L, M: [x + y for x, y in zip(L, M)] mo = [[1, 0], [-1, 0]] moo = [[0, 1], [0, -1]] def a(n, P=[[0, 0]]): if n == 0: return 1 if n % 2 == 0: mv1 = [plus(P[-1], x) for x in mo] else: mv1 = [plus(P[-1], x) for x in moo] mv2 = [x for x in mv1 if x not in P] if n == 1: return len(mv2) else: return sum(a(n - 1, add(P, x)) for x in mv2) [a(n) for n in range(21)]
Formula
a(n) = 4*A189722(n) for n >= 2.
It is proved that a(n)^(1/n) has a limit mu called the "connective constant" of the L-lattice; approximate value of mu: 1.5657. It is only conjectured that a(n + 1) ~ mu * a(n).
Comments