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Remark. Writing out (3.11) and (3.12) explicitly, one gets the following pair of identities, the first of which occurs in [6] and [7], the second as (V82) in Benjamin and Quinn [2, p. 145):

Figure 7 Remark. Writing out (3.11) and (3.12) explicitly, one gets the following pair of identities, the first of which occurs in [6] and [7], the second as (V82) in Benjamin and Quinn [2, p. 145):

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Throughout this section, we consider the important special case of |1 : n| in which the a; and b; are both periodic sequences. For m > 1, we say that a board has period m if dj+m = aj and bjim = 0; for all 7. It will be convenient to think of such tilings within a rectangular grid of width m, where a domino’s left half may end one row whenever its right half starts the next. Periodic Boards
Throughout this section, we consider the important special case of |1 : n| in which the a; and b; are both periodic sequences. For m > 1, we say that a board has period m if dj+m = aj and bjim = 0; for all 7. It will be convenient to think of such tilings within a rectangular grid of width m, where a domino’s left half may end one row whenever its right half starts the next. Periodic Boards
Observe that tailswapping is its own inverse since it does not change the location of the last fault. It also preserves weight by m-periodicity, since all tiles stay in their original
Observe that tailswapping is its own inverse since it does not change the location of the last fault. It also preserves weight by m-periodicity, since all tiles stay in their original
Figure 3.2: Tiling after tailswap. Figure 3.1: Tiling before tailswap, m = 7.
Figure 3.2: Tiling after tailswap. Figure 3.1: Tiling before tailswap, m = 7.
Proof. Tf X € Fam, then arrange A in two parallel rows of length m, breaking a domino covering cells m and m-+ 1 if necessary. Suppose that 7 is the largest 7, 0 <i < m, such that neither boundary 7 nor boundary i + m is covered by a domino (in which case, we say that a fault occurs at i; we'll assume for now that A has at least one fault). Exchange the portion of the first row to the right of boundary 7 with the portion of the second row to the right of boundary 7 + m (i.e., perform a tailswap). This yields a pair of tilings: row 1 is an ordinary m-tiling, but row 2 is a Lucas m-tiling because it can begin and end with a half-domino. Specifically, this happens if row 1 of the original tiling ends in a half-domino, because the tailswap in this case moves the half-domino to row 2 where the other half is ocated. as illustrated below.
Proof. Tf X € Fam, then arrange A in two parallel rows of length m, breaking a domino covering cells m and m-+ 1 if necessary. Suppose that 7 is the largest 7, 0 <i < m, such that neither boundary 7 nor boundary i + m is covered by a domino (in which case, we say that a fault occurs at i; we'll assume for now that A has at least one fault). Exchange the portion of the first row to the right of boundary 7 with the portion of the second row to the right of boundary 7 + m (i.e., perform a tailswap). This yields a pair of tilings: row 1 is an ordinary m-tiling, but row 2 is a Lucas m-tiling because it can begin and end with a half-domino. Specifically, this happens if row 1 of the original tiling ends in a half-domino, because the tailswap in this case moves the half-domino to row 2 where the other half is ocated. as illustrated below.
Figure 3.3: The fault-free configuration, m = 7. column. There is exactly one fault-free configuration (accounted for by the @ term) and it must consist of all dominos in staggered formation (since any square will necessarily have a fault line either to its left or to its right). A quick exercise in visualization shows that this configuration has weight [[j_, b, and belongs to Fo, if m is odd and to Fy, x Lm if m is even, as shown below.
Figure 3.3: The fault-free configuration, m = 7. column. There is exactly one fault-free configuration (accounted for by the @ term) and it must consist of all dominos in staggered formation (since any square will necessarily have a fault line either to its left or to its right). A quick exercise in visualization shows that this configuration has weight [[j_, b, and belongs to Fo, if m is odd and to Fy, x Lm if m is even, as shown below.
Figure 3.4: The fault-free configuration, m = 6.
Figure 3.4: The fault-free configuration, m = 6.
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