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 A007889 Number of intransitive (or alternating, or Stanley) trees: vertices are [0,n] and for no i
 1, 1, 2, 7, 36, 246, 2104, 21652, 260720, 3598120, 56010096, 971055240, 18558391936, 387665694976, 8787898861568, 214868401724416, 5636819806209792, 157935254554567296, 4707152127520549120, 148704074888134683520, 4963548160096887021056, 174553183413968718996736 (list; graph; refs; listen; history; text; internal format)
 OFFSET 0,3 COMMENTS 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, Nov 24 2004 REFERENCES 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. R. P. Stanley, Enumerative Combinatorics, Cambridge, Vol. 2, 1999; see Problem 5.41(a). LINKS Vincenzo Librandi, Table of n, a(n) for n = 0..200 C. Chauve, S. Dulucq and A. Rechnitzer, Enumerating alternating trees, J. Combin. Theory Ser. A 94 (2001), 142-151. Sergey Fomin and Grigory Mikhalkin, Labeled floor diagrams for plane curves, arXiv:0906.3828, 2009-2010.  [N. J. A. Sloane, Sep 27 2010] G. Hetyei, Efron's coins and the Linial arrangement, arXiv preprint arXiv:1511.04482 [math.CO], 2015. D. E. Knuth, Letter to Daniel Ullman and others, Apr 29 1997 [Annotated scanned copy, with permission] A. Postnikov, Intransitive Trees, J. Combin. Theory Ser. A 79 (1997), 360-366. FORMULA a(n) = (1/((n+1)*2^n))*Sum_{k=1..n+1} C(n+1,k)*k^n. E.g.f. A(x) 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, Mar 29 2008 E.g.f.: -2/x*LambertW(-1/2*x*exp(1/2*x)). - Vladeta Jovovic, Mar 29 2008 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) For n>=2, a(n)=Sum_{1,...,floor(n/2)}binomial(n-1, 2k-1)*k^(n-2). [Vladimir Shevelev, Mar 21 2010] For n>0, a(n) = A088789(n+1)*2/(n+1). [Vaclav Kotesovec, Dec 26 2011] MAPLE f:= n->1/(2^n*(n+1))*add(binomial(n+1, k)*k^n, k=1..(n+1)): seq(f(n), n=0..19); MATHEMATICA With[{nn=20}, CoefficientList[Series[-2/x LambertW[-1/2x Exp[x/2]], {x, 0, nn}], x]Range[0, nn]!] (* Harvey P. Dale, Aug 12 2011 *) Table[1/((n+1)2^n) Sum[Binomial[n+1, k]k^n, {k, n+1}], {n, 0, 20}] (* Harvey P. Dale, Apr 21 2012 *) PROG (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, 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 (Sage) def A007889(n) : return add(binomial(n, k)*(k+1)^(n-1) for k in (0..n))/2^n for n in (0..19) : print(A007889(n)) # Peter Luschny, Feb 29 2012 CROSSREFS Cf. A038049, A138860, A323841. Row sums of A029847. Sequence in context: A029768 A180271 A167199 * A125033 A034430 A143805 Adjacent sequences:  A007886 A007887 A007888 * A007890 A007891 A007892 KEYWORD nonn,easy,nice AUTHOR Alexander Postnikov [ apost(AT)math.mit.edu ] STATUS approved

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Last modified May 5 23:20 EDT 2021. Contains 343579 sequences. (Running on oeis4.)