cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

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.

Original entry on oeis.org

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

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Author

Emeric Deutsch, Sep 11 2004

Keywords

Comments

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

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
		

Crossrefs

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

Programs

  • 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
  • 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