A097724 Triangle read by rows: T(n,k) is the number of left factors of Motzkin paths without peaks, having length n and endpoint height k.
1, 1, 1, 1, 2, 1, 2, 3, 3, 1, 4, 6, 6, 4, 1, 8, 13, 13, 10, 5, 1, 17, 28, 30, 24, 15, 6, 1, 37, 62, 69, 59, 40, 21, 7, 1, 82, 140, 160, 144, 105, 62, 28, 8, 1, 185, 320, 375, 350, 271, 174, 91, 36, 9, 1, 423, 740, 885, 852, 690, 474, 273, 128, 45, 10, 1, 978, 1728, 2102, 2077
Offset: 0
Examples
Triangle starts: 1; 1, 1; 1, 2, 1; 2, 3, 3, 1; 4, 6, 6, 4, 1; Row n has n+1 terms. T(3,2)=3 because we have HUU, UHU and UUH, where U=(1,1) and H=(1,0). Row 7: let u(n,x) = U(n,x/2). Then u(7,x+1) = u(7,x) + 7*u(6,x) + 21*u(5,x) + 40*u(4,x) + 59*u(3,x) + 69*u(2,x) + 62*u(1,x) + 37. - _Peter Bala_, Jun 26 2025
Links
- Alois P. Heinz, Rows n = 0..140, flattened
- Naiomi T. Cameron and Asamoah Nkwanta, On Some (Pseudo) Involutions in the Riordan Group, Journal of Integer Sequences, Vol. 8 (2005), Article 05.3.7.
- Naiomi Cameron and Everett Sullivan, Peakless Motzkin paths with marked level steps at fixed height, Discrete Mathematics 344.1 (2021): 112154.
- Tian-Xiao He, A-sequences, Z-sequence, and B-sequences of Riordan matrices, Discrete Mathematics 343.3 (2020): 111718.
- A. Nkwanta and A. Tefera, Curious Relations and Identities Involving the Catalan Generating Function and Numbers, Journal of Integer Sequences, 16 (2013), #13.9.5.
- A. Panayotopoulos and P. Vlamos, Cutting Degree of Meanders, Artificial Intelligence Applications and Innovations, IFIP Advances in Information and Communication Technology, Volume 382, 2012, pp 480-489; DOI 10.1007/978-3-642-33412-2_49. - From _N. J. A. Sloane_, Dec 29 2012
Programs
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Maple
T:=proc(n,k) if k=n then 1 else (k+1)*sum(binomial(j,n-k-j)*binomial(j+k,n+1-j)/j,j=ceil((n-k+1)/2)..n-k) fi end: seq(seq(T(n,k),k=0..n),n=0..12); T:=proc(n,k) if k=n then 1 else (k+1)*sum(binomial(j,n-k-j)*binomial(j+k,n+1-j)/j,j=ceil((n-k+1)/2)..n-k) fi end: TT:=(n,k)->T(n-1,k-1): matrix(10,10,TT); # gives the sequence as a matrix
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Mathematica
T[n_, k_] := T[n, k] = If[k==n, 1, (k+1)*Sum[Binomial[j, n-k-j]*Binomial[j +k, n+1-j]/j, {j, Ceiling[(n-k+1)/2], n-k}]]; Table[T[n, k], {n, 0, 12}, {k, 0, n}] // Flatten (* Jean-François Alcover, Feb 22 2017, translated from Maple *)
Formula
T(n,k) = (k+1)*Sum_{j=ceiling((n-k+1)/2)..n-k} (C(j,n-k-j)*C(j+k,n+1-j)/j) for 0 <= k < n; T(n,n)=1.
G.f.: G/(1-tzG), where G = (1 - z + z^2 - sqrt(1 - 2z - z^2 - 2z^3 + z^4))/(2z^2) is the g.f. for the sequence A004148.
T(n,k) = T(n-1,k-1) + Sum_{j>=0} T(n-1-j,k+j), T(0,0) = 1, T(n,k) = 0 if k < 0 or if k > n. - Philippe Deléham, Jan 26 2014
Sum_{j=0..n-1} cos(2*Pi*k/3 + Pi/6)*T(n,k) = cos(Pi*n/2)*sqrt(3)/2 - cos(2*Pi*n/3 + Pi/6). - Leonid Bedratyuk, Dec 06 2017
Comments