A103469 Number of polyominoes consisting of 3 regular unit n-gons.
1, 2, 2, 3, 2, 3, 3, 4, 4, 5, 4, 5, 5, 6, 6, 7, 6, 7, 7, 8, 8, 9, 8, 9, 9, 10, 10, 11, 10, 11, 11, 12, 12, 13, 12, 13, 13, 14, 14, 15, 14, 15, 15, 16, 16, 17, 16, 17, 17, 18, 18, 19, 18, 19, 19, 20, 20, 21, 20, 21, 21, 22, 22, 23, 22, 23, 23, 24, 24, 25, 24, 25, 25, 26, 26, 27, 26, 27
Offset: 3
Examples
a(3)=1 because there is only one polyiamond consisting of 3 triangles and a(4)=2 because there are 2 polyominoes consisting of 3 squares.
Links
- Colin Barker, Table of n, a(n) for n = 3..1000
- M. Koch and S. Kurz, Enumeration of generalized polyominoes, arXiv:math/0605144 [math.CO], 2006.
- S. Kurz, k-polyominoes.
- Lorenzo Sauras Altuzarra, Some arithmetical problems that are obtained by analyzing proofs and infinite graphs, arXiv:2002.03075 [math.NT], 2020.
- Index entries for linear recurrences with constant coefficients, signature (1,0,0,0,0,1,-1).
Programs
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Mathematica
LinearRecurrence[{1,0,0,0,0,1,-1},{1,2,2,3,2,3,3},80] (* Harvey P. Dale, Sep 18 2016 *)
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PARI
Vec(-x^3*(x^6-x^5+x^4-x^3-x-1)/((x-1)^2*(x+1)*(x^2-x+1)*(x^2+x+1)) + O(x^100)) \\ Colin Barker, Jan 19 2015
Formula
a(n) = floor((n-2)/2) - floor((n-1)/6) + 1.
G.f.: -x^3*(x^6-x^5+x^4-x^3-x-1) / ((x-1)^2*(x+1)*(x^2-x+1)*(x^2+x+1)). - Colin Barker, Jan 19 2015
Comments