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Number of 4 X n Fibonacci minimal checkerboards.
1

%I #22 Jan 05 2025 19:51:42

%S 1,15,50,254,1202,5774,27650,132494,634802,3041534,14572850,69822734,

%T 334540802,1602881294,7679865650,36796446974,176302369202,

%U 844715399054,4047274626050,19391657731214,92911014030002,445163412418814,2132906048064050,10219366827901454,48963928091443202

%N Number of 4 X n Fibonacci minimal checkerboards.

%C a(n) is the number of tilings of a 4 X n board by monominoes and dominoes with all dominoes placed horizontally which cannot be decomposed into two or more such tilings placed end to end. - _Andrew Howroyd_, Feb 28 2023

%H Andrew Howroyd, <a href="/A349817/b349817.txt">Table of n, a(n) for n = 1..1000</a>

%H Yifan Zhang and George Grossman, <a href="https://web.archive.org/web/2024*/https://www.fq.math.ca/Papers1/55-3/ZhangGrossman03112017.pdf">A Combinatorial Proof for the Generating Function of Powers of the Fibonacci Sequence</a>, Fib. Q., 55:3 (2017), 235-242.

%F Lemma 2.23 of Zhang-Grossman gives a formula.

%F G.f.: 1 - 1/B(x) where x*B(x) is the g.f. of A056571. - _Andrew Howroyd_, Feb 28 2023

%e a(2) = 15. Each row of a 4 X 2 board can be tiled with either a domino or two monominoes giving a total of 2^4 = 16 tilings. The tiling consisting of all monominoes is not minimal so a(2) = 16 - 1 = 15. - _Andrew Howroyd_, Feb 28 2023

%o (PARI) \\ x*F(n,4) gives g.f. of A056571.

%o F(n,k)=sum(i=0, n, fibonacci(i+1)^k*x^i, O(x*x^n))

%o Vec(1 - 1/F(25,4)) \\ _Andrew Howroyd_, Feb 28 2023

%Y Cf. A056571.

%K nonn

%O 1,2

%A _N. J. A. Sloane_, Dec 24 2021

%E Terms a(11) and beyond from _Andrew Howroyd_, Feb 28 2023