Date: Sun, 17 Feb 2002 18:08:38 -0800 From: Dean Hickerson .___. a(1)=1: |___| .___. ._______. |___| a(2)=2: |___|___| |___| ._____. ._____. .___________. |___| | | | | | a(3)=3: |___|___|___| |___|_| |_|_|_| ._______. ._______. ._______. ._______________. |___| | | | |___| | | | | | | a(4)=4: |___|___|___|___| |___|_|_| |_|___|_| |_|_|_|_| We abbreviate the last three arrangement to 211, 121, 112, using the convention that 1 means vertical mat, 2 means 2 horizontal mats. For n=5, there's 1 way to cover a 1x10 room and 4 ways to cover a 2x5 (212, 2111, 1211, 11111). So a(5) = 1+4 = 5. For n=6, there's 1 way to cover 1x12, 6 ways to cover 2x6 (2112, 2121, 21111, 12111, 11211, 111111), and 2 ways to cover 3x4: ._______. ._______. | | |___| | |___| | |_|_| | | |_|___|_| |___|_|_| |___|___| So a(6) = 1+6+2 = 9. For n=7, there's 1 way to cover 1x14 and 8 ways to cover 2x7 (21112, 21121, 21211, 12121, 211111, 121111, 112111, 1111111), so a(7) = 1+8 = 9.