A191579 -id:A191579 - cp's OEIS frontend

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

Emeric Deutsch, Sep 11 2004

Column 0 is A004148 (RNA secondary structure numbers).
This triangle appears identical to A191579 (apart from offsets). - Philippe Deléham, Jan 26 2014
Conjecture: the row reverse triangle is the triangle of connection constants for expressing the polynomial u(n,x+1) as a linear combination of the polynomials u(k,x), 0 <= k <= n, where u(n,x) = U(n,x/2) with U(n,x) the n-th Chebyshev polynomial of the second kind. An example is given below.  Cf. A205810. - Peter Bala, Jun 26 2025

			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

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

Cf. A004148, A191579, A091964 (row sums), A205810.

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

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 *)

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

A192062 Square Array T(ij) read by antidiagonals (from NE to SW) with columns 2j being the denominators of continued fraction convergents to square root of (j^2 + 2j).

Original entry on oeis.org

0, 0, 1, 0, 1, 1, 0, 1, 1, 1, 0, 1, 1, 2, 2, 0, 1, 1, 3, 3, 1, 0, 1, 1, 4, 4, 5, 3, 0, 1, 1, 5, 5, 11, 8, 1, 0, 1, 1, 6, 6, 19, 15, 13, 4, 0, 1, 1, 7, 7, 29, 24, 41, 21, 1, 0, 1, 1, 8, 8, 41, 35, 91, 56, 34, 5, 0, 1, 1, 9, 9, 55, 48, 169, 115, 153, 55, 1, 0, 1, 1, 10, 10, 71, 63, 281, 204, 436, 209, 89, 6

Offset: 0

Kenneth J Ramsey, Jun 21 2011

Column j=1 is the Fibonacci sequence A000045.  Column 2 is A002530; column 4 is A041011; column 6 is A041023; column 8 is A041039, column 10 is A041059, column 12 is A041083, column 14 is A041111 corresponding the denominators of continued fraction convergents to square root of 3,8,15,24,35,48 and 63.
  T(2*i-1,j)*T(2*i,j)^2*T(2*i+1,j)*j/2 appears to be always a triangular number, T(j*T(2*i,j)^2).
  T(2*i,j)*T(2*i+1,j)^2*T(2*i+2)*j/2 appears to always equal a triangular number, T(j*T(2*i,j)*T(2*i+2,j)).
Conjecture re relation of A192062 to the sequence of primes: T(2*n,j) = A(n,j)*T(n,j) where A(n,j) is from the square array A191971. There, A(3*n,j) = A(n,j)*B(n,j) where B(n,j) are integers. It appears further that B(5*n,j)=B(n,j)*C(n,j); C(7*n,j)= C(n,j)*D(n,j); D(11*n,j) = D(n,j)*E(n,j); E(13*n,j) = E(n,j)*F(n,j) and F(17*n,j) = F(n,j)*G(n,j) where C(n,j), D(n,j) etc. are all integers. My conjecture is that this property continues indefinitely and follows the sequence of primes.

			Array as meant by the definition
First column has index j=0
0  0  0   0   0   0   0 ...
1  1  1   1   1   1   1 ...
1  1  1   1   1   1   1 ...
1  2  3   4   5   6   7 ...
2  3  4   5   6   7   8 ...
1  5 11  19  29  41  55 ...
3  8 15  24  35  48  63 ...
1 13 41  91 169 281 433 ...
4 21 56 115 204 329 496 ...
.
.
.

Kenneth J Ramsey,Triangular and Fibonacci Numbers

Cf. A191579, A191971.

Each column j is a recursive sequence defined by T(0,j)=0, T(1,j) = 1, T(2i,j)= T(2i-2,j)+T(2i-1,j) and T(2i+1,j) = T(2i-1,j)+j*T(2i,j). Also, T(n+2,j) = (j+2)*T(n,j)-T(n-2,j).
T(2n,j) = Sum(k=1 to n) C(k)*T(2*k,j-1) where the C(k) are the n-th row of the triangle A191579.
T(2*i,j) = T(i,j)*A(i,j) where A(i,j) is from the table A(i,j) of A191971.
T(4*i,j) = (T(2*i+1)^2 - T(2*i-1)^2)/j
T(4*i+2,j) = T(2*i+2,j)^2 - T(2*i,j)^2

Corrected and edited by Olivier Gérard, Jul 05 2011

cp's OEIS Frontend

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.

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