A262048 Irregular triangle read by rows: T(n, k) = A262045(n, k) * A237593(n, k), n >= 1, 1 <= k <= 2 * A003056(n).
1, 1, 2, 2, 2, 0, 0, 2, 3, 1, 1, 3, 3, 0, 0, 3, 4, 1, 2, 2, 1, 4, 4, 0, 0, 0, 0, 4, 5, 2, 1, 1, 2, 5, 5, 0, 2, 2, 0, 5, 6, 2, 1, 0, 0, 1, 2, 6, 6, 0, 0, 0, 0, 0, 0, 6, 7, 2, 4, 2, 2, 4, 2, 7, 7, 0, 0, 0, 0, 0, 0, 7, 8, 3, 1, 0, 0, 1, 3, 8, 8, 0, 2, 1, 2, 2, 1, 2, 0, 8, 9, 3, 2, 1, 1, 1, 1, 2, 3, 9
Offset: 1
Examples
For the 9th row: [5 2 2 2 2 5] in A237593 and [1 0 1 1 0 1] in A262045 result in [5 0 2 2 0 5] in this sequence which defines three regions between the Dyck paths for 8 and 9. The areas of the three regions are 5, 3 and 5; note that the width at the center of the middle region is 1 and is counted twice when adding up contiguous subsequences of nonzero numbers. The sum of the three areas is sigma(9) = 13. For the 19th row: [10 4 2 2 1 1 2 2 4 10] in A237593 and [1 0 0 0 0 0 0 0 0 1] in A262045 result in [10 0 0 0 0 0 0 0 0 10] in this sequence which defines two disjoint regions of area 10 each between the Dyck paths for 18 and 19; their sum is sigma(19) = 20. The first 20 rows of the triangle: ------------------------------------- n\k 1 2 3 4 5 6 7 8 9 10 ------------------------------------- 1: 1, 1; 2: 2, 2; 3: 2, 0, 0, 2; 4: 3, 1, 1, 3; 5: 3, 0, 0, 3; 6: 4, 1, 2, 2, 1, 4; 7: 4, 0, 0, 0, 0, 4; 8: 5, 2, 1, 1, 2, 5; 9: 5, 0, 2, 2, 0, 5; 10: 6, 2, 1, 0, 0, 1, 2, 6; 11: 6, 0, 0, 0, 0, 0, 0, 6; 12: 7, 2, 4, 2, 2, 4, 2, 7; 13: 7, 0, 0, 0, 0, 0, 0, 7; 14: 8, 3, 1, 0, 0, 1, 3, 8; 15: 8, 0, 2, 1, 2, 2, 1, 2, 0, 8; 16: 9, 3, 2, 1, 1, 1, 1, 2, 3, 9; 17: 9, 0, 0, 0, 0, 0, 0, 0, 0, 9; 18: 10, 3, 4, 2, 1, 1, 2, 4, 3, 10; 19: 10, 0, 0, 0, 0, 0, 0, 0, 0, 10; 20: 11, 4, 2, 1, 4, 4, 1, 2, 4, 11; ... A diagram of the first six rows shows the regions with an "x" marking the cells on the diagonal that are double counted. . _ _ _ _ . |_|_|_|_|_ . |_|x|_ . |x|_|_| . [4 1 2 2 1 4] & sigma(6)=12 |_| . |_| . _ _ _ |_| . |_|_|_| . |_ _ . [3 0 0 3] & sigma(5)=6 |_| . |_| . _ _ _ |_| . |_|_|_|_ . |x|_| . |_| [3 1 1 3] & sigma(4)=7 . _ _ |_| . |_|_|_ . |_| [2 0 0 2] & sigma(3)=4 . |_| . _ _ . |_ x| [2 2] & sigma(2)=3 . |_| . _ . |x| [1 1] & sigma(1)=1 . From _Omar E. Pol_, Sep 30 2015: (Start) Illustration of the 12th row of the triangle: . . 7 . _ _ _ _ _ _ _ 2 . / / / / / / /| 4 . /|_ _ 2 . / / /| . / / / 2 . / _ 4 . / /| 2 . / / /|_ _ . / / / /| . /| . /| . /| 7 . /| . /| . /| . The original diagram of the symmetric representation of sigma(12) in the first quadrant has been partitioned into two octants. Note that now there are two main diagonals in the diagram. The sums of the widths associated to the successive line segments of the partitioned zig-zag path give [7, 2, 4, 2, 2, 4, 2, 7], the same as the 12th row of the triangle. (End)
Crossrefs
Programs
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Mathematica
(* auxiliary functions are defined in A237048, A237593 and A262045 respectively *) s[n_,k_]:=Ceiling[(n+1)/k-(k+1)/2] - Ceiling[(n+1)/(k+1)-(k+2)/2] t[n_,k_]:=If[k<=row[n], s[n, k], s[n, 2*row[n]+1-k]] a262048[n_]:=Map[t[n, #]&, Range[2*row[n]]] a262045[n] Flatten[Map[a262048, Range[16]]] (* data *)
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