OFFSET
1,3
COMMENTS
a(n) is also the number of ways to tile an unbreakable 3 X 2n bracelet with dominoes and with upside-down T-shaped tetrominoes which looks like this:
_
_| |_
|_____|
The g.f. and order-6 recurrence by Colin Barker are correct. Proof: a length-m mino comes in colors(m) = floor((m-1)/2) - 1 colors (m >= 5), with piece g.f. P(x) = Sum_{m>=5} colors(m)*x^m = x^5/((1 - x)^2*(1 + x)). On the labeled cycle Z/(2n), conditioning on the unique mino covering cell 0 (its length m, color, and m rotational offsets, the rest a linear run) gives the cycle-lemma g.f. C(x) = x*P'(x)/(1 - P(x)), where 1/(1 - P(x)) is the g.f. of linear tilings. Taking the even part of C(x) and setting y = x^2 yields, in exact arithmetic, Sum_{n>=0} a(n)*y^n = (6*y^3 - 8*y^4 + 7*y^5 - 2*y^6)/((1 - y)*(1 - 3*y + 3*y^2 - 3*y^3 + 2*y^4 - y^5)), which is identically Barker's g.f.; clearing its denominator gives his recurrence for all n > 6 (the y^6 term of the numerator forces the range). - Yiming Beckmann, Jun 06 2026
FORMULA
From Colin Barker, Sep 06 2020: (Start)
G.f.: x^3*(6 - 8*x + 7*x^2 - 2*x^3) / ((1 - x)*(1 - 3*x + 3*x^2 - 3*x^3 + 2*x^4 - x^5)).
a(n) = 4*a(n-1) - 6*a(n-2) + 6*a(n-3) - 5*a(n-4) + 3*a(n-5) - a(n-6) for n>6.
(End)
EXAMPLE
For n=5 the a(5) = 35 tilings are as follows: we can use 3 colors of 10-minoes, each of which can be rotated to 10 different positions or "phases", giving us 30, and we can use two (single-color) 5-minoes in five different "phases", giving us another 5 tilings, with a grand total of 30 + 5 = 35.
MATHEMATICA
B[1] = 0; B[2] = 0; B[3] = 0; B[4] = 0; B[5] = 5;
B[n_?IntegerQ] :=
B[n] = Floor[(n - 3)/2]*n +
Sum[Floor[(i + 1)/2]*B[n - 4 - i], {i, 1, n - 5}];
Table[B[2 n], {n, 1, 30}]
CROSSREFS
KEYWORD
nonn,changed
AUTHOR
Tianshu Ouyang and Greg Dresden, Sep 05 2020
STATUS
approved