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%I #22 May 17 2024 03:27:02
%S 1,2,4,1,8,6,16,24,1,32,80,10,64,240,60,1,128,672,280,14,256,1792,
%T 1120,112,1,512,4608,4032,672,18,1024,11520,13440,3360,180,1,2048,
%U 28160,42240,14784,1320,22,4096,67584,126720,59136,7920,264,1,8192,159744,366080,219648,41184,2288,26
%N 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).
%C 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.
%H Jean-Luc Baril, Pamela E. Harris, and José L. Ramírez, <a href="https://arxiv.org/abs/2405.05357">Flattened Catalan Words</a>, arXiv:2405.05357 [math.CO], 2024. See pp. 8-9.
%F G.f.: (1-2*x)*x*y/(1 - 4*x + 4*x^2 - x^2*y).
%F T(n,k) = 2^(n-2*k+1)*binomial(n-1, 2*k-2).
%F T(n,1) = A000079(n-1).
%F T(n,2) = A001788(n-2).
%F T(n,3) = A003472(n-1).
%F T(n,4) = A002409(n-7).
%F T(n,5) = A140325(n-9).
%F T(n,6) = A172242(n-1).
%F Sum_{k>=0} T(n,k) = A007051(n-1).
%e The irregular triangle begins:
%e 1;
%e 2;
%e 4, 1;
%e 8, 6;
%e 16, 24, 1;
%e 32, 80, 10;
%e 64, 240, 60, 1;
%e 128, 672, 280, 14;
%e 256, 1792, 1120, 112, 1;
%e ...
%e 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.
%t 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 T[n_,k_]:=2^(n-2k+1)Binomial[n-1,2k-2]; Table[T[n,k],{n,14},{k,Ceiling[n/2]}]
%Y Cf. A000079, A001788, A002409, A003472, A007051 (row sums), A110654 (row lengths), A140325, A172242.
%Y Cf. A372852, A372972.
%K nonn,tabf
%O 1,2
%A _Stefano Spezia_, May 15 2024
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