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A069269
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Second level generalization of Catalan triangle (0th level is Pascal's triangle A007318; first level is Catalan triangle A009766; 3rd level is A069270).
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3
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1, 1, 1, 1, 2, 3, 1, 3, 7, 12, 1, 4, 12, 30, 55, 1, 5, 18, 55, 143, 273, 1, 6, 25, 88, 273, 728, 1428, 1, 7, 33, 130, 455, 1428, 3876, 7752, 1, 8, 42, 182, 700, 2448, 7752, 21318, 43263, 1, 9, 52, 245, 1020, 3876, 13566, 43263, 120175, 246675, 1, 10, 63, 320, 1428
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OFFSET
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0,5
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COMMENTS
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For the m-th level generalization of Catalan triangle T(n,k) = C(n+mk,k)*(n-k+1)/(n+(m-1)k+1); for n >= k+m: T(n,k) = T(n-m+1,k+1) - T(n-m,k+1); and T(n,n) = T(n+m-1,n-1) = C((m+1)n,n)/(mn+1).
With offset 1 for n and k, T(n,k) is (conjecturally) the number of permutations of [n] that avoid the patterns 4-2-3-1 and 4-2-5-1-3 and for which the last ascent ends at position k (k=1 if there are no ascents). For example, T(4,1) = 1 counts 4321; T(4,2) = 3 counts 1432, 2431, 3421; T(4,3) = 7 counts 1243, 1342, 2143, 2341, 3142, 3241, 4132. - David Callan, Jul 22 2008
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LINKS
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Tad White, Quota Trees, arXiv:2401.01462 [math.CO], 2024. See p. 18.
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FORMULA
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T(n, k) = C(n+2k, k)*(n-k+1)/(n+k+1).
For n >= k+2: T(n, k) = T(n-1, k+1) - T(n-2, k+1).
T(n, n) = T(n+1, n-1) = C(3n, n)/(2n+1).
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EXAMPLE
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Rows start
1;
1, 1;
1, 2, 3;
1, 3, 7, 12;
1, 4, 12, 30, 55;
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CROSSREFS
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KEYWORD
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AUTHOR
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STATUS
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approved
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