Show that f_nāˆ’1+L_n=2f_n.So we need to find a 2 to 1 correspondence. Set 1: Tilings an n-board. Set 2: Tiling of an nāˆ’1-board or tiling of an š‘›-bracelet. So we need to decompose a tiling of an š‘›-board to a tiling of an nāˆ’1-board or a tiling of an nāˆ’1-bracelet?

omvamen71 2022-10-02 Answered
Show that f n āˆ’ 1 + L n = 2 f n .
So we need to find a 2 to 1 correspondence.
Set 1: Tilings an n-board.
Set 2: Tiling of an n āˆ’ 1-board or tiling of an š‘›-bracelet.
So we need to decompose a tiling of an š‘›-board to a tiling of an n āˆ’ 1-board or a tiling of an n āˆ’ 1-bracelet?
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Answers (2)

Marshall Horne
Answered 2022-10-03 Author has 8 answers
f n denotes number of tilings of the n Ɨ 1 board by 2 Ɨ 1 and 1 Ɨ 1 pieces.
l n is almost the same thing, but the ends of the board is joined so that they form a kind of "bracelet" (i.e., one domino can cover the first and the last position; since they are adjacent now.) Also it should be mentioned that the position where the ends are glued together is fixed.
The above gives a combinatorial interpretation of Fibonacci and Lucas numbers, since f n = F n + 1 and l n = L n .
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Domianpv
Answered 2022-10-04 Author has 2 answers
Your identity is equivalent to L n = 2 f n āˆ’ f n āˆ’ 1 = f n + f n āˆ’ 2 (using f n āˆ’ f n āˆ’ 1 = f n āˆ’ 2 ). A combinatorial proof of this identity is easy (and it is given in the book you quote as identity 32).
Basically the same proof, just formulated another way, is rewriting the identity as f n āˆ’ f n āˆ’ 1 = L n āˆ’ f n and to check that both expressions are equal to the number of tilings of ( n āˆ’ 2 )-board.
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