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Search: id:A138157
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| A138157 |
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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. |
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2
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| 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
(list; graph; listen)
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OFFSET
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0,3
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COMMENT
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Row n has 1+n(n+1)/2 terms.
Row sums are the Catalan numbers (A000108).
Column sums yield A095830.
Sum(k*T(n,k),k>=0)=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).
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REFERENCES
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D. E. Knuth, The Art of Computer Programming. Addison-Wesley, Reading, MA, 1997, Vol. 1, p. 405 (exercise 5) and p. 595 (solution).
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FORMULA
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G.f.=G(t,z) satisfies G(t,z)=1 + 2tzG(t,tz) + [tzG(t,tz)]^2. The row generating polynomials P[n]=P[n](t) (n=1,2,...) are given by P[n]=t^n (2P[n-1] + sum(P[i]P[n-2-i], i=0..n-2)0; P[0]=1.
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EXAMPLE
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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;
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MAPLE
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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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CROSSREFS
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Cf. A000108, A095830, A138156.
Sequence in context: A052553 A045847 A137586 this_sequence A073429 A123634 A091866
Adjacent sequences: A138154 A138155 A138156 this_sequence A138158 A138159 A138160
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KEYWORD
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nonn,tabf
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AUTHOR
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Emeric Deutsch (deutsch(AT)duke.poly.edu), Mar 20 2008
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