%I #76 Mar 18 2021 23:59:29
%S 0,0,2,3,1,3,1,2,1,2,1,2,1,1,1,2,1,2,1,2,1,1,1,2,1,1,1,2,1,2,1,2,1,1,
%T 1,2,1,1,1,2,1,2,1,1,1,1,1,2,1,1,1,1,1,2,1,2,1,1,1,2,1,1,1,2,1,2,1,1,
%U 1,1,1,2,1,1,1,1,1,2,1,2,1,1,1,2,1,1,1,2,1,2
%N Number of parts in the symmetric representation of antisigma(n).
%C In order to construct this sequence and the diagram of the symmetric representation of antisigma(n) = A024816(n) we use the following rules:
%C At stage 1 in the first quadrant of the square grid we draw the symmetric representation of sigma(n) using the two Dyck paths described in the rows n and n-1 of A237593. The area of the region that is below the symmetric representation of sigma(n) equals A024916(n-1).
%C At stage 2 we draw a pair of orthogonal line segments (if it's necessary) such that in the drawing appears totally formed a square n X n. The area of the region that is above the symmetric representation of sigma(n) equals A004125(n). Then we draw a zig-zag path with line segments of length 1 from (0,n-1) to (n-1,0) such that appears a staircase with n-1 steps. The area of the region (or regions) that is below the symmetric representation of sigma(n) and above the staircase equals A244048(n) = A153485(n-1). The area of the region that is below the staircase equals A000217(n-1).
%C At stage 3 we turn OFF the cells of the symmetric representation of sigma(n) and also the cells that are below the staircase. Then we turn ON the rest of the cells that are in the square n X n. The result is that the ON cell form the diagram of the symmetric representation of antisigma(n) = A024816(n). See the Example section.
%C For n >= 7; if A237271(n) = 1 or n is a term of A262259 then a(n) = 2 otherwise a(n) = 1.
%e Illustration of the symmetric representation of antisigma(n) = AS(n) = A024816(n), for n = 1..6:
%e . y| _ _
%e . y| _ _ | _ _ |_ |
%e . y| _ | _ _| | | |_ | |_|
%e . y| _ | _ |_| | |_ _| | |_|_ _
%e . y| | _|_| | |_|_ | |_ | | |_ |
%e . y| | | |_| | |_| | |_| | |_|
%e . |_ _ |_ _ _ |_ _ _ _ |_ _ _ _ _ |_ _ _ _ _ _ |_ _ _ _ _ _ _
%e . x x x x x x
%e .
%e n: 1 2 3 4 5 6
%e a(n): 0 0 2 3 1 3
%e AS(n): 0 0 2 3 9 9
%e .
%e Illustration of the symmetric representation of antisigma(n) = AS(n) = A024816(n), for n = 7..9:
%e . y| _ _ _ _
%e . y| _ _ _ | _ _ _ _| |
%e . y| _ _ _ | _ _ _ | | | |_ _ _ |
%e . | _ _ _| | | |_ | |_ | | |_ |_ | |
%e . | |_ | | |_ |_ |_ _| | |_ |_| _|
%e . | |_ _| | |_ |_ _ | |_ |
%e . | |_ | | |_ | | |_ |
%e . | |_ | | |_ | | |_ |
%e . | |_| | |_| | |_|
%e . |_ _ _ _ _ _ _ _ |_ _ _ _ _ _ _ _ _ |_ _ _ _ _ _ _ _ _ _
%e . x x x
%e .
%e n: 7 8 9
%e a(n): 1 2 1
%e AS(n): 20 21 32
%e .
%e For n = 9 the figures 1, 2 and 3 below show respectively the three stages described in the Comments section as follows:
%e .
%e . y|_ _ _ _ _ 5 y|_ _ _ _ _ _ _ _ _ y| _ _ _ _
%e . |_ _ _ _ _| |_ _ _ _ _| | | _ _ _ _| |
%e . | |_ _ 3 | |_ |_ _ R | | |_ _ _ |
%e . | |_ | | |_ |_ | | | |_ |_ | |
%e . | |_|_ _ 5 | |_ T |_|_ _| | |_ |_| _|
%e . | | | | |_ | | | |_ |
%e . | Q | | | |_ | | | |_ |
%e . | | | | W |_ | | | |_ |
%e . | | | | |_| | | |_|
%e . |_ _ _ _ _ _ _ _|_|_ |_ _ _ _ _ _ _ _|_|_ |_ _ _ _ _ _ _ _ _ _
%e . x x x
%e . Figure 1. Figure 2. Figure 3.
%e . Symmetric Symmetric Symmetric
%e . representation representation representation
%e . of sigma(9) of sigma(9) of antisigma(9)
%e . A000203(9) = 13 A000203(9) = 13 A024816(9) = 32
%e . and of and of
%e . Q = A024916(8) = 56 R = A004125(9) = 12
%e . T = A244048(9) = 20
%e . T = A153485(8) = 20
%e . W = A000217(8) = 36
%e .
%e Note that the symmetric representation of antisigma(9) contains a hole formed by three cells because these three cells were the central part of the symmetric representation of sigma(9).
%Y Cf. A000203, A000217, A000290, A004125, A024816, A024916, A153485, A174973, A236104, A237270, A237271, A237593, A238443, A239660, A239931, A239932, A239933, A239934, A244048, A262259.
%K nonn
%O 1,3
%A _Omar E. Pol_, Mar 08 2021