|
%I #32 Mar 07 2014 10:13:58
%S 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,
%T 978,2238,3319,1,14,102,505,1874,5406,11941,17498,1,16,133,754,3235,
%U 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
%N 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).
%H Alois P. Heinz, <a href="/A223092/b223092.txt">Rows n = 0..140, flattened</a>
%H M. Dziemianczuk, <a href="http://dx.doi.org/10.1007/s00373-013-1357-1">Counting Lattice Paths With Four Types of Steps</a>, Graphs and Combinatorics, September 2013
%H N. J. A. Sloane, <a href="/A223092/a223092.pdf">Illustration of initial terms of A223092 and A064641</a>
%F 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
%e Triangle begins:
%e [1]
%e [1, 2]
%e [1, 4, 7]
%e [1, 6, 18, 29]
%e [1, 8, 33, 86, 133]
%e [1, 10, 52, 179, 431, 650]
%e [1, 12, 75, 316, 978, 2238, 3319]
%e ...
%e The T(n,k) array begins:
%e 4: 0 0 0 0 1 10 ...
%e 3: 0 0 0 1 8 52 ...
%e 2: 0 0 1 6 33 179 ...
%e 1: 0 1 4 18 86 431 ...
%e 0: 1 2 7 29 133 650 ...
%e -------------------------
%e k/n:0 1 2 3 4 5 ...
%e 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
%e This is also Dziemianczuk's array N(-i,i+j) read by antidiagonals:
%e 1,2,7,29,133,650,3319,17498, ...
%e 1,4,18,86,431,2238,11941,65086, ...
%e 1,6,33,179,978,5406,30241,171045, ...
%e 1,8,52,316,1874,11020,64698,380400, ...
%e 1,10,75,505,3235,20202,124455,761160, ...
%e ... - _N. J. A. Sloane_, Dec 05 2013
%p T:= proc(n, k) option remember; `if`(n=0 and k=0, 1,
%p `if`(n<0 or k<0 or k>n, 0, add(T(n-l[1], k-l[2]),
%p l=[[1, 1], [1, 0], [1, -1], [0, -1]]) ))
%p end:
%p seq(seq(T(n, n-j), j=0..n), n=0..10); # _Alois P. Heinz_, Apr 08 2013
%t 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_ *)
%Y Cf. A064641 (T(n,0)), A071943, A052709.
%K nonn,tabl
%O 0,3
%A _N. J. A. Sloane_, Mar 29 2013
|
