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Search: id:A126181
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| A126181 |
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Triangle read by rows: T(n,k) is the number of hex trees with n edges and k nodes having median children (i.e. k vertical edges; 0<=k<=n). A hex tree is a rooted tree where each vertex has 0, 1, or 2 children and, when only one child is present, it is either a left child, or a median child, or a right child (name due to an obvious bijection with certain tree-like polyhexes; see the Harary-Read paper). |
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+0
2
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| 1, 2, 1, 5, 4, 1, 14, 15, 6, 1, 42, 56, 30, 8, 1, 132, 210, 140, 50, 10, 1, 429, 792, 630, 280, 75, 12, 1, 1430, 3003, 2772, 1470, 490, 105, 14, 1, 4862, 11440, 12012, 7392, 2940, 784, 140, 16, 1, 16796, 43758, 51480, 36036, 16632, 5292, 1176, 180, 18, 1, 58786
(list; table; graph; listen)
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OFFSET
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0,2
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COMMENT
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Sum of terms in row n = A002212(n+1). T(n,0)=A000108(n+1) (the Catalan numbers). Sum(kT(n,k),k=0..n)=A026376(n).
Also, with offset 1, triangle read by rows: T(n,k) is the number of skew Dyck paths of semilength n and having k left steps (n>=1; 0<=k<=n-1). A skew Dyck path is a path in the first quadrant which begins at the origin, ends on the x-axis, consists of steps U=(1,1)(up), D=(1,-1)(down), and L=(-1,-1)(left) so that up and left steps do not overlap. The length of a path is defined to be the number of its steps. For example, T(4,2)=6 because we have UDUUUDLL, UUUUDLLD, UUDUUDLL, UUUUDLDL, UUUDUDLL, and UUUUDDLL.
Also, with offset 1, number of skew Dyck paths of semilength and having k UDU's. Example: T(3,1)=4 because we have (UDU)UDD, (UDU)UDL, U(UDU)DD, and U(UDU)DL (the UDU's are shown between parentheses). Row sums yield A002212. T(n,0)=binom(2n,n)/(n+1)=A000108(n) (the Catalan numbers). Sum(k*T(n,k),k=0..n-1)=A026376(n-1). Mirror image of A108198.
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REFERENCES
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F. Harary and R. C. Read, The enumeration of tree-like polyhexes, Proc. Edinburgh Math. Soc. (2) 17 (1970), 1-13.
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FORMULA
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T(n,k)=binomial(n,k)*c(n-k+1), where c(m)=binom(2m,m)/(m+1) is a Catalan number (A000108). Proof: There are c(n-k+1) binary trees with n-k edges. We can insert k vertical edges at the n-k+1 vertices (repetitions possible) in binom(n-k+1+k-1,k)=binom(n,k) ways. G.f.=G=G(t,z) satisfies G=1+(2+t)zG+z^2*G^2.
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EXAMPLE
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Triangle starts:
1;
2,1;
5,4,1;
14,15,6,1;
42,56,30,8,1;
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MAPLE
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c:=n->binomial(2*n, n)/(n+1): T:=proc(n, k) if k<=n then binomial(n, k)*c(n-k+1) else 0 fi end: for n from 0 to 10 do seq(T(n, k), k=1..n) od; # yields sequence in triangular form
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CROSSREFS
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Cf. A002212, A000108, A026376, A108198.
Sequence in context: A104710 A039598 A128738 this_sequence A104259 A137650 A110271
Adjacent sequences: A126178 A126179 A126180 this_sequence A126182 A126183 A126184
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KEYWORD
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nonn,tabl
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AUTHOR
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Emeric Deutsch (deutsch(AT)duke.poly.edu), Dec 19 2006, Mar 30 2007
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EXTENSIONS
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Edited by njas at the suggestion of Andrew Plewe, Jun 13 2007
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