A223092 Triangle read by rows: let T(n,k) (for n >= 0, 0 <= k <= n) be the number of walks from (0,0) to (n,k) using steps (1,1), (1,0), (1,-1) and (0,-1); n-th row of triangle gives T(n,n), T(n,n-1), ..., T(n,0).
1, 1, 2, 1, 4, 7, 1, 6, 18, 29, 1, 8, 33, 86, 133, 1, 10, 52, 179, 431, 650, 1, 12, 75, 316, 978, 2238, 3319, 1, 14, 102, 505, 1874, 5406, 11941, 17498, 1, 16, 133, 754, 3235, 11020, 30241, 65086, 94525, 1, 18, 168, 1071, 5193, 20202, 64698, 171045, 360897, 520508, 1, 20, 207, 1464, 7896, 34362, 124455, 380400, 977040, 2029490, 2910895
Offset: 0
Examples
Triangle begins: [1] [1, 2] [1, 4, 7] [1, 6, 18, 29] [1, 8, 33, 86, 133] [1, 10, 52, 179, 431, 650] [1, 12, 75, 316, 978, 2238, 3319] ... The T(n,k) array begins: 4: 0 0 0 0 1 10 ... 3: 0 0 0 1 8 52 ... 2: 0 0 1 6 33 179 ... 1: 0 1 4 18 86 431 ... 0: 1 2 7 29 133 650 ... ------------------------- k/n:0 1 2 3 4 5 ... T(5,2) = T(5,3) + T(4,3) + T(4,2) + T(4,1) = 52 + 8 + 33 + 86 = 179.- _Philippe Deléham_, Mar 29 2013 This is also Dziemianczuk's array N(-i,i+j) read by antidiagonals: 1,2,7,29,133,650,3319,17498, ... 1,4,18,86,431,2238,11941,65086, ... 1,6,33,179,978,5406,30241,171045, ... 1,8,52,316,1874,11020,64698,380400, ... 1,10,75,505,3235,20202,124455,761160, ... ... - _N. J. A. Sloane_, Dec 05 2013
Links
- Alois P. Heinz, Rows n = 0..140, flattened
- M. Dziemianczuk, Counting Lattice Paths With Four Types of Steps, Graphs and Combinatorics, September 2013
- N. J. A. Sloane, Illustration of initial terms of A223092 and A064641
Programs
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Maple
T:= proc(n, k) option remember; `if`(n=0 and k=0, 1, `if`(n<0 or k<0 or k>n, 0, add(T(n-l[1], k-l[2]), l=[[1, 1], [1, 0], [1, -1], [0, -1]]) )) end: seq(seq(T(n, n-j), j=0..n), n=0..10); # Alois P. Heinz, Apr 08 2013 -
Mathematica
max = 10; T[0, 0] = 1; T[n_ /; n >= 0, k_ /; 0 <= k <= max] := T[n, k] = T[n, k+1]+T[n-1, k+1]+T[n-1, k]+T[n-1, k-1]; T[n_, k_] = 0; Table[Table[T[n, k], {k, n, 0, -1}], {n, 0, max}] // Flatten (* Jean-François Alcover, Mar 07 2014, after Philippe Deléham *)
Formula
T(n,k) = T(n,k+1) + T(n-1,k+1) + T(n-1,k) + T(n-1,k-1). - Philippe Deléham, Mar 29 2013