A283240 Triangle read by rows: T(n,k) = number of directed self-avoiding walks (SAWs) of length k in an n-ladder graph that use all rows of the graph.
2, 2, 0, 4, 8, 8, 0, 0, 4, 12, 20, 16, 0, 0, 0, 4, 16, 32, 40, 28, 0, 0, 0, 0, 4, 20, 48, 72, 72, 44, 0, 0, 0, 0, 0, 4, 24, 68, 120, 144, 120, 64, 0, 0, 0, 0, 0, 0, 4, 28, 92, 188, 264, 264, 188, 88, 0, 0, 0, 0, 0, 0, 0, 4, 32, 120, 280, 452, 528, 452, 280, 116
Offset: 1
Examples
Triangle T(n,k) begins: n/k 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 1 2 2 2 0 4 8 8 3 0 0 4 12 20 16 4 0 0 0 4 16 32 40 28 5 0 0 0 0 4 20 48 72 72 44 6 0 0 0 0 0 4 24 68 120 144 120 64 7 0 0 0 0 0 0 4 28 92 188 264 264 188 88 8 0 0 0 0 0 0 0 4 32 120 280 452 528 452 280 116 9 0 0 0 0 0 0 0 0 4 36 152 400 732 980 980 732 400 148 10 0 0 0 0 0 0 0 0 0 4 40 188 552 1132 1712 1960 1712 1132 552 184 e.g., there are T(3,3) = 12 directed SAWs of length 3 in L_3 that use at least one node from each row. Six shapes walked in both directions. > _ _ > | | | | |_ _| > | |_ | _| | |
Links
- OEIS, Self-avoiding walks.
- Wikipedia, Ladder graph.
- Wolfram MathWorld, Ladder Graph.
Crossrefs
Programs
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Python
maxN=20 Tnk=[[0 for column in range(0, row*2)] for row in range(0,maxN+1)] Tnk[1][0]=2 # initial values for the special case of 1-ladder. Two single nodes. Tnk[1][1]=2 # SAW of length 1 on a L_1, either left or right for row in range(2,maxN+1): for column in range(0, row*2): if(column+1 < row): # path is smaller than ladder - no possible SAW using all rows Tnk[row][column] = 0 elif(column+1 == row): # vertical SAW, 2 possible in 2 directions Tnk[row][column] = 4 elif(column == row*2 -1): # n-ladder Hamiltonian A137882 Tnk[row][column] = 2*(row*row - row + 2) elif(column == 2*(row-1)): # Grow SAW including Hamiltonians from previous row, 4 extra SAWs from Hamiltonians Tnk[row][column] = Tnk[row-1][column-1] + Tnk[row-1][column-2] + 4 else: # Grow SAW from previous SAWs. Either adding one or two edges Tnk[row][column] = Tnk[row-1][column-1] + Tnk[row-1][column-2] print(Tnk)
Formula
T(n,k) = 0 when k+1 < n
T(n,k) = 4 when k+1 = n
T(n,k) = 2(n^2-n+2) when k = 2n-1
T(n,k) = T(n-1,k-1) + T(n-1,k-2) + 4 when k = 2n-2
T(n,k) = T(n-1,k-1) + T(n-1,k-2) otherwise
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