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
Offset: 0
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, Birkhäuser, 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.
- Richard P. Stanley, Hyperplane arrangements, interval orders, and trees, Proc. Natl. Acad. Sci. USA 93 (1996), 2620-2625.
- Index entries for sequences related to trees
Programs
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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);
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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 *) -
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
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]
Comments