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.

A271592 Array read by antidiagonals: T(n,m) = number of directed Hamiltonian walks from NW to SW corners on a grid with n rows and m columns.

Original entry on oeis.org

1, 0, 1, 0, 1, 1, 0, 1, 0, 1, 0, 1, 2, 1, 1, 0, 1, 0, 4, 0, 1, 0, 1, 4, 8, 8, 1, 1, 0, 1, 0, 23, 0, 16, 0, 1, 0, 1, 8, 55, 86, 47, 32, 1, 1, 0, 1, 0, 144, 0, 397, 0, 64, 0, 1, 0, 1, 16, 360, 948, 1770, 1584, 264, 128, 1, 1, 0, 1, 0, 921, 0, 11658, 0, 6820, 0, 256, 0, 1
Offset: 1

Views

Author

Andrew Howroyd, Apr 10 2016

Keywords

Examples

			The start of the sequence as table:
* 1 0   0   0     0     0       0       0          0 ...
* 1 1   1   1     1     1       1       1          1 ...
* 1 0   2   0     4     0       8       0         16 ...
* 1 1   4   8    23    55     144     360        921 ...
* 1 0   8   0    86     0     948       0      10444 ...
* 1 1  16  47   397  1770   11658   59946     359962 ...
* 1 0  32   0  1584     0   88418       0    4999752 ...
* 1 1  64 264  6820 52387  909009 8934966  130373192 ...
* 1 0 128   0 28002     0 7503654       0 2087813834 ...
* ...
		

Crossrefs

Column 4 is aerated A014524, column 5 is A014585.
Rows include A181688, A181689.
Main diagonal is A000532.
Cf. A333580.

Programs

  • Python
    # Using graphillion
    from graphillion import GraphSet
    import graphillion.tutorial as tl
    def A271592(n, k):
        if k == 1: return 1
        universe = tl.grid(k - 1, n - 1)
        GraphSet.set_universe(universe)
        start, goal = 1, n
        paths = GraphSet.paths(start, goal, is_hamilton=True)
        return paths.len()
    print([A271592(j + 1, i - j + 1) for i in range(12) for j in range(i + 1)])  # Seiichi Manyama, Mar 28 2020

Formula

T(n,m)=0 for n odd and m even, T(1,n)=0 for n>1.
T(2,n)=T(n,1)=T(2*n,2)=1, T(3,2*n+1)=T(n+1,3)=2^n.