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.

Showing 1-3 of 3 results.

A102892 Triangle read by rows: T(n,k) is the number of noncrossing trees with n edges in which the number of edges from the root to the first branch point is k.

Original entry on oeis.org

1, 0, 1, 1, 0, 2, 5, 3, 0, 4, 25, 16, 6, 0, 8, 130, 83, 32, 12, 0, 16, 700, 442, 166, 64, 24, 0, 32, 3876, 2420, 884, 332, 128, 48, 0, 64, 21945, 13566, 4840, 1768, 664, 256, 96, 0, 128, 126500, 77539, 27132, 9680, 3536, 1328, 512, 192, 0, 256
Offset: 0

Views

Author

Emeric Deutsch, Jan 16 2005

Keywords

Comments

The statistic "number of edges from the root to the first branchpoint" is equal to 0 if root is a branchpoint and it is equal to the total number of edges if there is no branchpoint.
Row n contains n+1 terms.
Row sums yield the ternary numbers (A001764).
Column 0 yields A102893.
Column 1 yields A030983.

Examples

			T(2,0)=1 because we have /\ and T(2,2)=2 because we have /_ and _\.
Triangle starts:
    1;
    0,  1;
    1,  0,  2;
    5,  3,  0,  4;
   25, 16,  6,  0, 8;
  130, 83, 32, 12, 0, 16;
  ...
		

Crossrefs

Programs

  • Maple
    T:=proc(n,k) if n=0 and k=0 then 1 elif k=0 then 5*binomial(3*n-1,n-2)/(3*n-1) elif k
    				
  • Mathematica
    T[n_, k_] := Which[n == 0 && k == 0, 1, k == 0, 5*Binomial[3n - 1, n - 2]/(3n - 1), kJean-François Alcover, Jul 06 2018, from Maple *)
  • PARI
    T(n, k) = {if(k==0, if(n==0, 1, 5*binomial(3*n-1, n-2)/(3*n-1)), if(n<=k, if(n==k, 2^(n-1), 0), 2^(k-1)*binomial(3*n-3*k+1, n-k)/(n-k+1) - 2^k*binomial(3*n-3*k-2, n-k-1)/(n-k)))}
    for(n=0, 10, for(k=0, n, print1(T(n, k), ", ")); print); \\ Andrew Howroyd, Nov 06 2017

Formula

T(n, 0) = 5*binomial(3n-1, n-2)/(3n-1) for n > 0.
T(n, k) = [2^(k-1)/(n-k+1)]binomial(3n-3k+1, n-k)-[2^k/(n-k)]binomial(3n-3k-2, n-k-1) for 0 < k < n.
T(n, n) = 2^(n-1) (n > 0).
G.f.: (1/2)*g(2-g) + g^2*(1-2*z)/(2*(1-2*t*z)), where g = 1 + z*g^3 is the g.f. of the ternary numbers (A001764).

A334643 a(n) is the total number of down steps between the second and third up steps in all 2_1-Dyck paths of length 3*n. A 2_1-Dyck path is a lattice path with steps (1, 2), (1, -1) that starts and ends at y = 0 and stays above the line y = -1.

Original entry on oeis.org

0, 0, 16, 53, 209, 963, 4816, 25367, 138531, 777041, 4449511, 25901655, 152818458, 911755012, 5491420104, 33343242196, 203881825163, 1254342228285, 7759025239189, 48227078649155, 301056318504165, 1886647802277315, 11864793375611820, 74854437302309175
Offset: 0

Views

Author

Benjamin Hackl, May 12 2020

Keywords

Comments

For n = 2, there is no 3rd up step, a(2) = 16 enumerates the total number of down steps between the 2nd up step and the end of the path.

Examples

			For n = 2, the 2_1-Dyck paths are UUDDDD, UDUDDD, UDDUDD, UDDDUD, DUDDUD, DUDUDD, DUUDDD. In total, there are a(2) = 4 + 3 + 2 + 1 + 1 + 2 + 3 = 16 down steps between the 2nd up step and the end of the path.
		

Crossrefs

Programs

  • SageMath
    [binomial(3*n + 1, n)/(3*n + 1) + 4*sum([binomial(3*j + 2, j)*binomial(3*(n - j), n - j)/(3*j + 2)/(n - j + 1) for j in srange(1, 3)]) - 7*(n==2) if n >= 2 else 0 for n in srange(30)] # Benjamin Hackl, May 12 2020

Formula

a(0) = a(1) = 0 and a(n) = binomial(3*n+1, n)/(3*n+1) + 4*Sum_{j=1..2}binomial(3*j+2, j)*binomial(3*(n-j), n-j)/((3*j+2)*(n-j+1)) - 7*[n=2] for n > 1, where [ ] is the Iverson bracket.

A334644 a(n) is the total number of down steps between the third and fourth up steps in all 2_1-Dyck paths of length 3*n. A 2_1-Dyck path is a lattice path with steps (1, 2), (1, -1) that starts and ends at y = 0 and stays above the line y = -1.

Original entry on oeis.org

0, 0, 0, 83, 299, 1263, 6076, 31307, 168561, 936161, 5321611, 30804795, 180939408, 1075636912, 6459103704, 39120216196, 238692219923, 1465783144605, 9052278085129, 56185368932615, 350293215459915, 2192731008315015, 13775745283576920, 86831135890324875
Offset: 0

Views

Author

Benjamin Hackl, May 12 2020

Keywords

Comments

For n = 3, there is no 4th up step, a(3) = 83 enumerates the total number of down steps between the 3rd up step and the end of the path.

Crossrefs

Programs

  • SageMath
    [binomial(3*n + 1, n)/(3*n + 1) + 4*sum([binomial(3*j + 2,  j) * binomial(3*(n - j), n - j)/(3*j + 2)/(n - j + 1) for j in srange(1, 4)]) - 30*(n==3) if n >= 3 else 0 for n in srange(30)] # Benjamin Hackl, May 12 2020

Formula

a(0) = a(1) = a(2) = 0 and a(n) = binomial(3*n+1, n)/(3*n+1) + 4*Sum_{j=1..3}binomial(3*j+2, j)*binomial(3*(n-j), n-j)/((3*j+2)*(n-j+1)) - 30*[n=3] for n > 2, where [ ] is the Iverson bracket.
Showing 1-3 of 3 results.