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Identifier

St001380: Integer partitions ⟶ ℤ

Values

[1] => 1

[2] => 2

[1,1] => 2

[3] => 3

[2,1] => 3

[1,1,1] => 3

[4] => 5

[3,1] => 5

[2,2] => 7

[2,1,1] => 5

[1,1,1,1] => 5

[5] => 8

[4,1] => 8

[3,2] => 10

[3,1,1] => 8

[2,2,1] => 10

[2,1,1,1] => 8

[1,1,1,1,1] => 8

[6] => 13

[5,1] => 13

[4,2] => 17

[4,1,1] => 13

[3,3] => 22

[3,2,1] => 14

[3,1,1,1] => 13

[2,2,2] => 22

[2,2,1,1] => 17

[2,1,1,1,1] => 13

[1,1,1,1,1,1] => 13

[7] => 21

[6,1] => 21

[5,2] => 27

[5,1,1] => 21

[4,3] => 32

[4,2,1] => 24

[4,1,1,1] => 21

[3,3,1] => 32

[3,2,2] => 32

[3,2,1,1] => 24

[3,1,1,1,1] => 21

[2,2,2,1] => 32

[2,2,1,1,1] => 27

[2,1,1,1,1,1] => 21

[1,1,1,1,1,1,1] => 21

[8] => 34

[7,1] => 34

[6,2] => 44

[6,1,1] => 34

[5,3] => 54

[5,2,1] => 38

[5,1,1,1] => 34

[4,4] => 71

[4,3,1] => 47

[4,2,2] => 54

[4,2,1,1] => 41

[4,1,1,1,1] => 34

[3,3,2] => 67

[3,3,1,1] => 54

[3,2,2,1] => 47

[3,2,1,1,1] => 38

[3,1,1,1,1,1] => 34

[2,2,2,2] => 71

[2,2,2,1,1] => 54

[2,2,1,1,1,1] => 44

[2,1,1,1,1,1,1] => 34

[1,1,1,1,1,1,1,1] => 34

[9] => 55

[8,1] => 55

[7,2] => 71

[7,1,1] => 55

[6,3] => 86

[6,2,1] => 62

[6,1,1,1] => 55

[5,4] => 103

[5,3,1] => 79

[5,2,2] => 86

[5,2,1,1] => 65

[5,1,1,1,1] => 55

[4,4,1] => 103

[4,3,2] => 97

[4,3,1,1] => 79

[4,2,2,1] => 79

[4,2,1,1,1] => 65

[4,1,1,1,1,1] => 55

[3,3,3] => 131

[3,3,2,1] => 97

[3,3,1,1,1] => 86

[3,2,2,2] => 103

[3,2,2,1,1] => 79

[3,2,1,1,1,1] => 62

[3,1,1,1,1,1,1] => 55

[2,2,2,2,1] => 103

[2,2,2,1,1,1] => 86

[2,2,1,1,1,1,1] => 71

[2,1,1,1,1,1,1,1] => 55

[1,1,1,1,1,1,1,1,1] => 55

[10] => 89

[9,1] => 89

[8,2] => 115

[8,1,1] => 89

[7,3] => 140

>>> Load all 1200 entries. <<<

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Description

The number of monomer-dimer tilings of a Ferrers diagram.
For a hook of length $n$, this is the $n$-th Fibonacci number.

References

[1] Triangle read by rows: T(n,k) (1 <= k <= n) = number of monomer-dimer tilings of an n X k board. OEIS:A210662

Code

from sage.combinat.tiling import TilingSolver, Polyomino

def statistic(la):
    shape = Polyomino(Partition(la).cells())
    p = Polyomino([(0, 0)])
    q = Polyomino([(0, 0), (0, 1)])
    T = TilingSolver([p, q], box=shape, reusable=True)
    return T.number_of_solutions()

Created

Apr 06, 2019 at 20:28 by Martin Rubey

Updated

Apr 06, 2019 at 20:28 by Martin Rubey