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A007889 Number of intransitive (or alternating) trees: vertices are [0,n] and for no i<j<k are both (i,j) and (j,k) edges. +0
5
1, 1, 2, 7, 36, 246, 2104, 21652, 260720, 3598120, 56010096, 971055240, 18558391936, 387665694976, 8787898861568, 214868401724416, 5636819806209792, 157935254554567296, 4707152127520549120, 148704074888134683520 (list; graph; listen)
OFFSET

0,3

COMMENT

Number of local binary search trees (i.e. labeled binary trees such that every left child has a smaller label than its parent, and every right child has a larger label than its parent) on n vertices. Example: a(3)=7 because we have 3L2L1, 2L1R3, 3L1R2, 1R2R3, 1R3L2, 2R3L1 (Li means left child labeled i, RI means right child labeled i) and root 2 with left child 1 and right child 3. - Emeric Deutsch (deutsch(AT)duke.poly.edu), Nov 24 2004

REFERENCES

C. Chauve, S. Dulucq and A. Rechnitzer, Enumerating alternating trees, J. Combin. Theory Ser. A 94 (2001), 142-151.

I. M. Gelfand, M. I. Graev and A. Postnikov, Combinatorics of hypergeometric functions associated with positive roots, in Arnold-Gelfand Mathematical Seminars: Geometry and Singularity Theory, Birkhauser, 1997.

A. Postnikov, Intransitive Trees, J. Combin. Theory Ser. A 79 (1997), 360-366.

R. P. Stanley, Enumerative Combinatorics, Cambridge, Vol. 2, 1999; see Problem 5.41(a).

LINKS

A. Postnikov, home page

Index entries for sequences related to trees

FORMULA

E.g.f. satisfies: A(x) = exp( x*(1 + A(x))/2 ). E.g.f. A(x) equals the inverse function of 2*log(x)/(1+x). - Paul D. Hanna (pauldhanna(AT)juno.com), Mar 29 2008

E.g.f.: -2/x*LambertW(-1/2*x*exp(1/2*x)). - Vladeta Jovovic (vladeta(AT)Eunet.yu), Mar 29 2008

Comments from Vladeta Jovovic and Paul D. Hanna, Apr 03 2008 (Start): Powers of e.g.f.: If A(x)^p = Sum_{n>=0} a(n,p)*x^n/n! then a(n,p) = (1/2^n)* Sum_{k=0..n} binomial(n,k)*p*(k+p)^(n-1).

Let A(x) = e.g.f. of A007889, B(x) = e.g.f. of A138860 where B(x) = exp( x*[B(x) + B(x)^2]/2 ); then B(x) = A(x*B(x)) = (1/x)*Series_Reversion(x/A(x)) and A(x) = B(x/A(x)) = x/Series_Reversion(x*B(x)). (End)

MAPLE

f := (n)->1/(2^n*(n+1))*sum(binomial(n+1, k)*k^n, 'k'=1..(n+1));

PROGRAM

(PARI) {a(n)=local(A=1+x); for(i=0, n, A=exp(x*(1+A)/2 +x*O(x^n))); n!*polcoeff(A, n)} - Paul D. Hanna (pauldhanna(AT)juno.com), Mar 29 2008

(PARI) /* Coefficients of A(x)^p are given by: */ {a(n, p=1)=(1/2^n)*sum(k=0, n, binomial(n, k)*p*(k+p)^(n-1))} - Vladeta Jovovic and Paul D. Hanna, Apr 03 2008

CROSSREFS

Cf. A038049.

Cf. A138860.

Adjacent sequences: A007886 A007887 A007888 this_sequence A007890 A007891 A007892

Sequence in context: A088313 A095793 A029768 this_sequence A125033 A034430 A112293

KEYWORD

nonn,easy,nice

AUTHOR

Alexander Postnikov [ apost(AT)math.mit.edu ]

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Last modified May 16 01:24 EDT 2008. Contains 139630 sequences.


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