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.

A138157 Triangle read by rows: T(n,k) is the number of binary trees with n edges and path length k; 0<=k<=n*(n+1)/2.

Original entry on oeis.org

1, 0, 2, 0, 0, 1, 4, 0, 0, 0, 0, 4, 2, 8, 0, 0, 0, 0, 0, 0, 6, 8, 8, 4, 16, 0, 0, 0, 0, 0, 0, 0, 0, 4, 24, 4, 28, 16, 16, 8, 32, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 24, 36, 48, 24, 56, 40, 56, 32, 32, 16, 64, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 8, 60, 72, 144, 26, 168, 104, 128, 64, 176, 80, 112
Offset: 0

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Author

Emeric Deutsch, Mar 20 2008

Keywords

Comments

Row n has 1+n*(n+1)/2 terms.
Row sums are the Catalan numbers (A000108).
Column sums yield A095830.
Sum_{k>=0} k*T(n,k) = A138156(n).
The g.f. B(w,z) in the Knuth reference is related to the above G(t,z) through B(t,z)=1+zG(t,z).

Examples

			T(2,3) = 4 because we have the path trees LL, LR, RL and RR, where L (R) denotes a left (right) edge; each of these four trees has path length 3.
Triangle starts:
  1;
  0,2;
  0,0,1,4;
  0,0,0,0,4,2,8;
  0,0,0,0,0,0,6,8,8,4,16;
  0,0,0,0,0,0,0,0,4,24,4,28,16,16,8,32;
		

References

  • D. E. Knuth, The Art of Computer Programming. Addison-Wesley, Reading, MA, 1997, Vol. 1, p. 405 (exercise 5) and p. 595 (solution).

Crossrefs

Programs

  • Maple
    P[0]:=1: for n to 7 do P[n]:=sort(expand(t^n*(2*P[n-1]+add(P[i]*P[n-2-i],i= 0..n-2)))) end do: for n from 0 to 7 do seq(coeff(P[n],t,j),j=0..(1/2)*n*(n+1)) end do; # yields sequence in triangular form
  • Mathematica
    p[n_] := p[n] = t^n*(2p[n-1] + Sum[p[i]*p[n-2-i], {i, 0, n-2}]); p[0] = 1; Flatten[ Table[ CoefficientList[ p[n], t], {n, 0, 7}]] (* Jean-François Alcover, Jul 22 2011, after Maple prog. *)

Formula

G.f.: G(t,z) satisfies G(t,z) = 1 + 2*t*z*G(t,t*z) + (t*z*G(t,t*z))^2. The row generating polynomials P[n] = P[n](t) (n=1,2,...) are given by P[n] = t^n*(2*P[n-1] + Sum_{i=0..n-2} P[i]*P[n-2-i]); P[0] = 1.