cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

A247050 Rectangular array read upwards by columns: T = T(n,k) = number of paths from (0,1) to (n,k), where 0 >= k <= 2, consisting of segments given by the vectors (,1,1), (1,2), (1,-1).

Original entry on oeis.org

0, 1, 0, 1, 0, 1, 0, 2, 1, 2, 1, 2, 1, 4, 3, 4, 4, 5, 4, 9, 8, 9, 12, 13, 12, 22, 21, 22, 33, 34, 33, 56, 55, 56, 88, 89, 88, 145, 144, 145, 232, 233, 232, 378, 377, 378, 609, 610, 609, 988, 987, 988, 1596, 1597, 1596, 2585, 2584, 2585, 4180, 4181, 4180
Offset: 0

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Author

Clark Kimberling, Sep 11 2014

Keywords

Comments

Also, T(n,k) = number of strings s(0)..s(n) of integers such that s(0) = 1, s(n) = k, and if i > 0, then s(i) is in {0,1,2} and s(i) - s(i-1) is in {1,2,-1}. The column sums form the Fibonacci sequence (A000045).
This is a 3-rowed array read upwards by columns. - N. J. A. Sloane, Sep 14 2014

Examples

			First 10 columns:
0 .. 1 .. 1 .. 2 .. 3 .. 5 .. 8 .. 13 .. 21 .. 34
1 .. 0 .. 2 .. 1 .. 4 .. 4 .. 9 .. 12 .. 22 .. 33
0 .. 1 .. 0 .. 2 .. 1 .. 4 .. 4 .. 9 ... 12 .. 22
T(4,1) counts these 4 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)
Partial sums of second components in each vector sum give the 4 integer strings described in Comments: (1,2,1,2,1), (1,0,1,2,1), (1,2,1,0,1), (1,0,1,0,1).
		

Crossrefs

Programs

  • Mathematica
    t[0, 0] = 0; t[0, 1] = 1; t[0, 2] = 0; 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]; TableForm[ Reverse[Transpose[Table[t[n, k], {n, 0, 12}, {k, 0, 2}]]]] (* array *)
    u = Flatten[Table[t[n, k], {n, 0, 20}, {k, 0, 2}]] (* sequence *)

Formula

Let F = A000045, the Fibonacci numbers. Then (row 0, the bottom row) = F(n-1) - (-1)^n for n >= 0; (row 1, the middle row) = F(n) + (-1)^n for n >=0; (row 2, the top row) = F.
Conjectures from Chai Wah Wu, Apr 16 2025: (Start)
a(n) = 2*a(n-6) + a(n-9) for n > 8.
G.f.: (-x^8 - x^5 - x^3 - x)/(x^9 + 2*x^6 - 1). (End)