In how many ways can a $ 2 times n $ rectangle be tiled by $ 2 times 1 $ or $ 1 times 1 $ tiles?












2












$begingroup$


This problem is from the book "Problem Solving Strategies" by Arthur Engel (Chapter 9, problem 64) and the solution given there is $ a_0 = 1, a_1 = 2, a_2 = 7$ and the recurrence relation being $ a_n = 3a_{n-1} + a_{n-2} - a_{n-3}$

I'd like to know how did we get this recurrence.






sequences-and-series combinatorics recurrence-relations tiling







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  • $begingroup$
    @RossMillikan my bad, sorry for not writing the question properly.
    $endgroup$
    – Shafin Ahmed
    Dec 13 '18 at 17:30
















2












$begingroup$


This problem is from the book "Problem Solving Strategies" by Arthur Engel (Chapter 9, problem 64) and the solution given there is $ a_0 = 1, a_1 = 2, a_2 = 7$ and the recurrence relation being $ a_n = 3a_{n-1} + a_{n-2} - a_{n-3}$

I'd like to know how did we get this recurrence.






sequences-and-series combinatorics recurrence-relations tiling







share|cite|improve this question











$endgroup$












  • $begingroup$
    @RossMillikan my bad, sorry for not writing the question properly.
    $endgroup$
    – Shafin Ahmed
    Dec 13 '18 at 17:30














2












2








2





$begingroup$


This problem is from the book "Problem Solving Strategies" by Arthur Engel (Chapter 9, problem 64) and the solution given there is $ a_0 = 1, a_1 = 2, a_2 = 7$ and the recurrence relation being $ a_n = 3a_{n-1} + a_{n-2} - a_{n-3}$

I'd like to know how did we get this recurrence.






sequences-and-series combinatorics recurrence-relations tiling







share|cite|improve this question











$endgroup$




This problem is from the book "Problem Solving Strategies" by Arthur Engel (Chapter 9, problem 64) and the solution given there is $ a_0 = 1, a_1 = 2, a_2 = 7$ and the recurrence relation being $ a_n = 3a_{n-1} + a_{n-2} - a_{n-3}$

I'd like to know how did we get this recurrence.






sequences-and-series combinatorics recurrence-relations tiling




sequences-and-series combinatorics recurrence-relations tiling






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 13 '18 at 17:28







Shafin Ahmed

















asked Dec 13 '18 at 17:12









Shafin AhmedShafin Ahmed

707




707












  • $begingroup$
    @RossMillikan my bad, sorry for not writing the question properly.
    $endgroup$
    – Shafin Ahmed
    Dec 13 '18 at 17:30


















  • $begingroup$
    @RossMillikan my bad, sorry for not writing the question properly.
    $endgroup$
    – Shafin Ahmed
    Dec 13 '18 at 17:30
















$begingroup$
@RossMillikan my bad, sorry for not writing the question properly.
$endgroup$
– Shafin Ahmed
Dec 13 '18 at 17:30




$begingroup$
@RossMillikan my bad, sorry for not writing the question properly.
$endgroup$
– Shafin Ahmed
Dec 13 '18 at 17:30










1 Answer
1






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2












$begingroup$

We can define $A(n)$ as the number of ways to tile a $2 times n$ rectangle, $B(n)$ as the number of ways to tile a $2 times n$ rectangle plus one square and $C(n)$ as the number of ways to tile a $2 times n$ rectangle plus one horizontal domino. We can set up a set of coupled recurrences by imagining we add one piece that covers the leftmost uncovered square or the top one if there are two.
$$A(n)=A(n-1)+B(n-1)+C(n-2)\B(n)=A(n)+B(n-1)+C(n-1)\C(n)=A(n)$$
Because we can get a $2 times n$ rectangle by adding a vertical domino to a $2 times n-1$ rectangle or filling in the hole on the other shapes, we can get a $B$ shape by adding $1 times 1$ to a rectangle, by adding a $1 times 2$ to a B shape or by adding a $1 times 1$ to a C shape, and we can only get a C shape by adding a horizontal $1 times 2$ to a rectangle. Then we get
$$A(n)=A(n-1)+B(n-1)+A(n-2)\
B(n)=A(n)+B(n-1)+A(n-1)\
B(n-1)-B(n-2)=A(n-1)+A(n-2)\
A(n-1)=A(n-2)+B(n-2)+A(n-3)\
A(n)-A(n-1)=A(n-1)+B(n-1)-B(n-2)-A(n-3)\
A(n)=3A(n-1)+A(n-2)-A(n-3)$$



The sequence is given in OEIS A030186 and begins $$1, 2, 7, 22, 71, 228, 733, 2356, 7573, 24342, 78243, 251498, 808395, 2598440, 8352217, 26846696, 86293865, 277376074, 891575391, 2865808382, 9211624463, 29609106380, 95173135221, 305916887580, 983314691581, 3160687827102 $$

The fourth comment says this is the count of the tilings we want.






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    $begingroup$

    We can define $A(n)$ as the number of ways to tile a $2 times n$ rectangle, $B(n)$ as the number of ways to tile a $2 times n$ rectangle plus one square and $C(n)$ as the number of ways to tile a $2 times n$ rectangle plus one horizontal domino. We can set up a set of coupled recurrences by imagining we add one piece that covers the leftmost uncovered square or the top one if there are two.
    $$A(n)=A(n-1)+B(n-1)+C(n-2)\B(n)=A(n)+B(n-1)+C(n-1)\C(n)=A(n)$$
    Because we can get a $2 times n$ rectangle by adding a vertical domino to a $2 times n-1$ rectangle or filling in the hole on the other shapes, we can get a $B$ shape by adding $1 times 1$ to a rectangle, by adding a $1 times 2$ to a B shape or by adding a $1 times 1$ to a C shape, and we can only get a C shape by adding a horizontal $1 times 2$ to a rectangle. Then we get
    $$A(n)=A(n-1)+B(n-1)+A(n-2)\
    B(n)=A(n)+B(n-1)+A(n-1)\
    B(n-1)-B(n-2)=A(n-1)+A(n-2)\
    A(n-1)=A(n-2)+B(n-2)+A(n-3)\
    A(n)-A(n-1)=A(n-1)+B(n-1)-B(n-2)-A(n-3)\
    A(n)=3A(n-1)+A(n-2)-A(n-3)$$



    The sequence is given in OEIS A030186 and begins $$1, 2, 7, 22, 71, 228, 733, 2356, 7573, 24342, 78243, 251498, 808395, 2598440, 8352217, 26846696, 86293865, 277376074, 891575391, 2865808382, 9211624463, 29609106380, 95173135221, 305916887580, 983314691581, 3160687827102 $$

    The fourth comment says this is the count of the tilings we want.






    share|cite|improve this answer









    $endgroup$


















      2












      $begingroup$

      We can define $A(n)$ as the number of ways to tile a $2 times n$ rectangle, $B(n)$ as the number of ways to tile a $2 times n$ rectangle plus one square and $C(n)$ as the number of ways to tile a $2 times n$ rectangle plus one horizontal domino. We can set up a set of coupled recurrences by imagining we add one piece that covers the leftmost uncovered square or the top one if there are two.
      $$A(n)=A(n-1)+B(n-1)+C(n-2)\B(n)=A(n)+B(n-1)+C(n-1)\C(n)=A(n)$$
      Because we can get a $2 times n$ rectangle by adding a vertical domino to a $2 times n-1$ rectangle or filling in the hole on the other shapes, we can get a $B$ shape by adding $1 times 1$ to a rectangle, by adding a $1 times 2$ to a B shape or by adding a $1 times 1$ to a C shape, and we can only get a C shape by adding a horizontal $1 times 2$ to a rectangle. Then we get
      $$A(n)=A(n-1)+B(n-1)+A(n-2)\
      B(n)=A(n)+B(n-1)+A(n-1)\
      B(n-1)-B(n-2)=A(n-1)+A(n-2)\
      A(n-1)=A(n-2)+B(n-2)+A(n-3)\
      A(n)-A(n-1)=A(n-1)+B(n-1)-B(n-2)-A(n-3)\
      A(n)=3A(n-1)+A(n-2)-A(n-3)$$



      The sequence is given in OEIS A030186 and begins $$1, 2, 7, 22, 71, 228, 733, 2356, 7573, 24342, 78243, 251498, 808395, 2598440, 8352217, 26846696, 86293865, 277376074, 891575391, 2865808382, 9211624463, 29609106380, 95173135221, 305916887580, 983314691581, 3160687827102 $$

      The fourth comment says this is the count of the tilings we want.






      share|cite|improve this answer









      $endgroup$
















        2












        2








        2





        $begingroup$

        We can define $A(n)$ as the number of ways to tile a $2 times n$ rectangle, $B(n)$ as the number of ways to tile a $2 times n$ rectangle plus one square and $C(n)$ as the number of ways to tile a $2 times n$ rectangle plus one horizontal domino. We can set up a set of coupled recurrences by imagining we add one piece that covers the leftmost uncovered square or the top one if there are two.
        $$A(n)=A(n-1)+B(n-1)+C(n-2)\B(n)=A(n)+B(n-1)+C(n-1)\C(n)=A(n)$$
        Because we can get a $2 times n$ rectangle by adding a vertical domino to a $2 times n-1$ rectangle or filling in the hole on the other shapes, we can get a $B$ shape by adding $1 times 1$ to a rectangle, by adding a $1 times 2$ to a B shape or by adding a $1 times 1$ to a C shape, and we can only get a C shape by adding a horizontal $1 times 2$ to a rectangle. Then we get
        $$A(n)=A(n-1)+B(n-1)+A(n-2)\
        B(n)=A(n)+B(n-1)+A(n-1)\
        B(n-1)-B(n-2)=A(n-1)+A(n-2)\
        A(n-1)=A(n-2)+B(n-2)+A(n-3)\
        A(n)-A(n-1)=A(n-1)+B(n-1)-B(n-2)-A(n-3)\
        A(n)=3A(n-1)+A(n-2)-A(n-3)$$



        The sequence is given in OEIS A030186 and begins $$1, 2, 7, 22, 71, 228, 733, 2356, 7573, 24342, 78243, 251498, 808395, 2598440, 8352217, 26846696, 86293865, 277376074, 891575391, 2865808382, 9211624463, 29609106380, 95173135221, 305916887580, 983314691581, 3160687827102 $$

        The fourth comment says this is the count of the tilings we want.






        share|cite|improve this answer









        $endgroup$



        We can define $A(n)$ as the number of ways to tile a $2 times n$ rectangle, $B(n)$ as the number of ways to tile a $2 times n$ rectangle plus one square and $C(n)$ as the number of ways to tile a $2 times n$ rectangle plus one horizontal domino. We can set up a set of coupled recurrences by imagining we add one piece that covers the leftmost uncovered square or the top one if there are two.
        $$A(n)=A(n-1)+B(n-1)+C(n-2)\B(n)=A(n)+B(n-1)+C(n-1)\C(n)=A(n)$$
        Because we can get a $2 times n$ rectangle by adding a vertical domino to a $2 times n-1$ rectangle or filling in the hole on the other shapes, we can get a $B$ shape by adding $1 times 1$ to a rectangle, by adding a $1 times 2$ to a B shape or by adding a $1 times 1$ to a C shape, and we can only get a C shape by adding a horizontal $1 times 2$ to a rectangle. Then we get
        $$A(n)=A(n-1)+B(n-1)+A(n-2)\
        B(n)=A(n)+B(n-1)+A(n-1)\
        B(n-1)-B(n-2)=A(n-1)+A(n-2)\
        A(n-1)=A(n-2)+B(n-2)+A(n-3)\
        A(n)-A(n-1)=A(n-1)+B(n-1)-B(n-2)-A(n-3)\
        A(n)=3A(n-1)+A(n-2)-A(n-3)$$



        The sequence is given in OEIS A030186 and begins $$1, 2, 7, 22, 71, 228, 733, 2356, 7573, 24342, 78243, 251498, 808395, 2598440, 8352217, 26846696, 86293865, 277376074, 891575391, 2865808382, 9211624463, 29609106380, 95173135221, 305916887580, 983314691581, 3160687827102 $$

        The fourth comment says this is the count of the tilings we want.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 13 '18 at 17:49









        Ross MillikanRoss Millikan

        298k23198371




        298k23198371






























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