A287548 Triangle read by rows: T(n,k), where each row begins with the Catalan number for n nonintersecting arches and transitions through k generations of eliminating and reducing arch configurations to an end row entry equal to number of semi-meander solutions for n arches.

1, 2, 1, 5, 3, 2, 14, 9, 7, 4, 42, 28, 23, 16, 10, 132, 90, 76, 57, 42, 24, 429, 297, 255, 199, 156, 108, 66, 1430, 1001, 869, 695, 563, 420, 304, 174, 4862, 3432, 3003, 2442, 2019, 1568, 1210, 836, 504

Offset: 1

Author

Roger Ford, May 26 2017

Keywords

• nonn
• tabl
• more

Examples

			Triangle begins:
n\k    1    2    3    4    5    6    7    8
1:     1
2:     2    1
3:     5    3    2
4:     14   9    7    4
5:     42   28   23   16   10
6:     132  90   76   57   42   24
7:     429  297  255  199  156  108  66
8:     1430 1001 869  695  563  420  304  174
...
Capital letters (U,D) represent beginning and end of first and last arch. Only 1 UD ends arch sequence in next generation.
Reduction of arches:            Elimination of arches:
(middle D U = new arch U D in the next arch generation)
            /\
     /\    //\\                      /\/\/\/\  = UDududUD
    //\\/\///\\\  = UudDudUuuddD        /\
        /\  /\                         /  \
     /\//\\//\\   =  UDuuddUudD       //\/\\   =  UududD
                                        end
For n=3 C(n)=5  nonintersecting arch configurations:
   UuuddD   UududD   UudDUD   UDUudD   UDudUD   T(3,1)=5
    end      end      UDUD     UDUD     UudD    T(3,2)=3
                       UD       UD       end    T(3,3)=2
		

Crossrefs

Cf. A000108, A000245, A067324, A000682.

Formula

T(n,1) = Catalan Numbers C(n)= A000108(n).

Conjectured:

T(n,2) = C(n) - C(n-1) = A000245(n-1).

T(n,3) = C(n) - C(n-1) - C(n-2) = A067324(n-3).

T(n,4) = C(n) - C(n-1) - 2*C(n-2) - C(n-3).

T(n,n) = semi-meander solutions = A000682(n-1).