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.
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
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).
Links
- Alois P. Heinz, Rows n = 0..50, flattened
- Marilena Barnabei, Flavio Bonetti, and Niccolò Castronuovo, Motzkin and Catalan Tunnel Polynomials, J. Int. Seq., Vol. 21 (2018), Article 18.8.8.
Programs
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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
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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.
Comments