A247352 Rectangular array read by upwards by columns: T = T(n,k) = number of paths from (0,1) to (n,k), where 0 <= k <= 3, consisting of segments given by the vectors (1,1), (1,2), (1,-1).

0, 1, 0, 0, 1, 0, 1, 1, 0, 2, 2, 1, 2, 2, 3, 4, 2, 5, 8, 5, 5, 10, 12, 13, 10, 17, 28, 22, 17, 38, 49, 45, 38, 66, 100, 87, 66, 138, 191, 166, 138, 257, 370, 329, 257, 508, 724, 627, 508, 981, 1392, 1232, 981, 1900, 2721, 2373, 1900, 3702, 5254, 4621, 3702

Offset: 0

Author

Clark Kimberling, Sep 15 2014

Keywords

• nonn
• tabl
• easy

Comments

Also, T(n,k) = number of strings s(0)..s(n) of integers such that s(0) = 1, s(n) = k, and for i > 0, s(i) is in {0,1,2,3} and s(i) - s(i-1) is in {-1,1,2} for 1 <= i <= n.

Examples

			First 10 columns:
0 .. 1 .. 1 .. 4 .. 5 .. 13 .. 22 .. 45 .. 87 ... 166
0 .. 1 .. 2 .. 3 .. 8 .. 12 .. 28 .. 49 .. 100 .. 191
1 .. 0 .. 2 .. 2 .. 5 .. 10 .. 17 .. 38 .. 66 ... 138
0 .. 1 .. 0 .. 2 .. 2 .. 5 ... 10 .. 17 .. 38 ... 66
T(5,0) counts these 5 paths, given as vector sums applied to (0,1):
(1,1) + (1,1) + (1,-1) + (1,-1) + (1 -1)
(1,1) + (1,-1) + (1,1) + (1,-1) + (1,-1)
(1,-1) + (1,1) + (1,1) + (1,-1) + (1,-1)
(1,1) + (1,-1) + (1,-1) + (1,1) + (1,-1)
(1,-1) + (1,1) + (1,-1) + (1,1) + (1,-1)
Partial sums of second components in each vector sum give the 3 integer strings described in comments:
(1,2,3,2,1,0),
(1,2,1,2,1,0),
(1,0,1,2,1,0),
(1,2,1,0,1,0),
(1,0,1,0,1,0).
		

Links

• Clark Kimberling, Table of n, a(n) for n = 0..1000

Crossrefs

Cf. A247049, A247321, A247353, A247354, A247355.

Programs

Mathematica

z = 50; t[0, 0] = 0; t[0, 1] = 1; t[0, 2] = 0; t[0, 3] = 0;
t[1, 3] = 1; t[n_, 0] := t[n, 0] = t[n - 1, 1];
t[n_, 1] := t[n, 1] = t[n - 1, 0] + t[n - 1, 2]
t[n_, 2] := t[n, 2] = t[n - 1, 0] + t[n - 1, 1] + t[n - 1, 3]
t[n_, 3] := t[n, 3] = t[n - 1, 1] + t[n - 1, 2]
u = Flatten[Table[t[n, k], {n, 0, z}, {k, 0, 3}]] (* A247352 *)
TableForm[Reverse[Transpose[Table[t[n, k], {n, 0, 12}, {k, 0, 3}]]]] (* array  *)
v = Map[Total, u1]  (* A247353 *)
Table[t[n, 0], {n, 0, z}]   (* row 0, A247354*)
Table[t[n, 1], {n, 0, z}]   (* row 1, cf. row 0 *)
Table[t[n, 2], {n, 0, z}]   (* row 2, A247355 *)
Table[t[n, 3], {n, 0, z}]   (* row 3, A247325 *)

Formula

The four rows and column sums all empirically satisfy the linear recurrence r(n) = 3*r(n-2) + 2*r(n-3) - r(n-4), with g.f. of the form p(x)/q(x), where q(x) = 1 - 3 x^2 - 2 x^3 + x^4. Initial terms and p(x) follow:

(row 0, the bottom row): 0,1,0,2; x - x^3

(row 1): 1,0,0,2; 1 - x^2

(row 2): 0,1,2,3; x +2*x^2

(row 3): 0,1,1,4; x + x^2 + x^3

(n-th column sum) = 1,3,5,11; 1 + 3*x + 2*x^2.