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.

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

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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).