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A372868
Irregular triangle read by rows: T(n,k) is the number of flattened Catalan words of length n with exactly k runs of weak ascents, with 1 <= k <= ceiling(n/2).
2
1, 2, 4, 1, 8, 6, 16, 24, 1, 32, 80, 10, 64, 240, 60, 1, 128, 672, 280, 14, 256, 1792, 1120, 112, 1, 512, 4608, 4032, 672, 18, 1024, 11520, 13440, 3360, 180, 1, 2048, 28160, 42240, 14784, 1320, 22, 4096, 67584, 126720, 59136, 7920, 264, 1, 8192, 159744, 366080, 219648, 41184, 2288, 26
OFFSET
1,2
COMMENTS
With offset 0 for the variable k, T(n,k) is the number of flattened Catalan words of length n with exactly k peaks. In such case, T(4,1) = 6 corresponds to 6 flattened Catalan words of length 4 with 1 peak: 0010, 0100, 0110, 0101, 0120, and 0121. See Baril et al. at page 20.
LINKS
Jean-Luc Baril, Pamela E. Harris, and José L. Ramírez, Flattened Catalan Words, arXiv:2405.05357 [math.CO], 2024. See pp. 8-9.
FORMULA
G.f.: (1-2*x)*x*y/(1 - 4*x + 4*x^2 - x^2*y).
T(n,k) = 2^(n-2*k+1)*binomial(n-1, 2*k-2).
T(n,1) = A000079(n-1).
T(n,2) = A001788(n-2).
T(n,3) = A003472(n-1).
T(n,4) = A002409(n-7).
T(n,5) = A140325(n-9).
T(n,6) = A172242(n-1).
Sum_{k>=0} T(n,k) = A007051(n-1).
EXAMPLE
The irregular triangle begins:
1;
2;
4, 1;
8, 6;
16, 24, 1;
32, 80, 10;
64, 240, 60, 1;
128, 672, 280, 14;
256, 1792, 1120, 112, 1;
...
T(4,2) = 6 since there are 6 flattened Catalan words of length 4 with 2 runs of weak ascents: 0010, 0100, 0101, 0110, 0120, and 0121.
MATHEMATICA
T[n_, k_]:=SeriesCoefficient[(1-2x)*x*y/(1-4*x+4*x^2-x^2*y), {x, 0, n}, {y, 0, k}]; Table[T[n, k], {n, 14}, {k, Ceiling[n/2]}] //Flatten (* or *)
T[n_, k_]:=2^(n-2k+1)Binomial[n-1, 2k-2]; Table[T[n, k], {n, 14}, {k, Ceiling[n/2]}]
CROSSREFS
Cf. A000079, A001788, A002409, A003472, A007051 (row sums), A110654 (row lengths), A140325, A172242.
Sequence in context: A065260 A257794 A345881 * A091894 A127151 A193034
KEYWORD
nonn,tabf
AUTHOR
Stefano Spezia, May 15 2024
STATUS
approved