

A000992


"HalfCatalan numbers": a(n) = Sum_{k=1 ... floor(n/2)} a(k)a(nk) with a(1) = 1.
(Formerly M0793 N0300)


17



1, 1, 1, 2, 3, 6, 11, 24, 47, 103, 214, 481, 1030, 2337, 5131, 11813, 26329, 60958, 137821, 321690, 734428, 1721998, 3966556, 9352353, 21683445, 51296030, 119663812, 284198136, 666132304, 1586230523, 3734594241, 8919845275
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OFFSET

1,4


COMMENTS

From David Callan, Nov 02 2006: (Start)
a(n) = number of (unlabeled, rooted) ordered trees on n1 vertices in which all outdegrees are <=2 and, for each vertex of outdegree 2, the sizes of its two subtrees are weakly increasing left to right (n>=2). The number b(n) of such trees on n vertices satisfies the recurrence b[1]=1; b[n_]/;n>=2 := b[n] = b[n1] + Sum[b[i]b[n1i],{i,Floor[(n1)/2]}], the first term counting trees whose root has outdegree 1 and the sum counting trees whose root has outdegree 2 by size of the left subtree. This recurrence generates b(n)=a(n+1), n>=1. For example, the a(5)=3 such trees are:
........../\..
..../.\.......
............... (End)
From R. J. Mathar, Mar 27 2009: (Start)
The connection with the Rayleigh polynomials Phi(2n,x) of A158616 is that Phi(2n,x)= sum_{i=1..a(n)} 2^(n_i) product_{j=2..n1} (x+j)^(n_ij), as described by Kishore.
So a(n) counts the terms in the representation of the polynomial Phi(2n,x) as a sum over these "base" polynomials.
For example Phi(12,x) = 2^4*(x+2)^2*(x+3) + 2^2*(x+2)*(x+3)^2 + 2^3*(x+2)*(x+3)*(x+4) + 2^3*(x+2)*(x+3)*(x+5) + 2^2*(x+2)*(x+4)*(x+5) + 2*(x+3)^2*(x+5) has a(6)=6 terms. (End)
From Wolfdieter Lang, Jan 06 2012: (Start)
The o.g.f. G(x):=sum(a(n)*x^n,n=0..infty), with a(0)=0, satisfies the relation (G(x))^2  2*G(x) + G2(x^2) + 2*x = 0, with the o.g.f. G2(x):=sum(a(n)^2*x^n,n=0..infty) of the squares. This can be proved from the connection to the halfconvolution of the sequence with itself (for this notion see a comment on A201204, where also the rule for the o.g.f. is given). (End)
From Stanislav Sykora, Oct 13 2014: (Start)
For n>1, a(n) is the number of rooted binary trees with n leaves or, equivalently, the number of rooted binary trees with n1 internal vertices.
Also, a symmetric binary operation f(x,y) = f(y,x) can be used to form at most a(n) nonequivalent expressions over n distinct arguments (when f(x,y) has additional properties such as f(x,f(y,z))=f(y,f(x,z)), the count can be smaller).
(End)
Limit n>infinity (a(n))^(1/n) = 2.49086422... .  Vaclav Kotesovec, Oct 15 2014


REFERENCES

N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).


LINKS

T. D. Noe and Vaclav Kotesovec, Table of n, a(n) for n = 1..2500 (first 200 terms from T. D. Noe)
E. Bohl, P. Lancaster, Implementation of a Markov model for phylogenetic trees, J. Theor. Biol. 239 (3) (2006) 324333
T. Feil, K.Hutson, R. J. Kretchmar, Tree traversals and permutations, Congr. Numer. 175 (2005) 201221 (mentions the sequence only in the references, not in the text)
N. Kishore, A structure of the Rayleigh polynomial, Duke Math. J., 31 (1964), 513518.


EXAMPLE

From Stanislav Sykora, Oct 13 2014: (Start)
The a(5)=3 rooted binary trees with 5 leaves are:
(1,((1,1),(1,1))); (1,(1,(1,(1,1)))); (1,((1,1),(1,(1,1)));
The corresponding expressions on 5 arguments x1,x2,x3,x4,x5 based on a binary operation f(x,y)=f(y,x) are:
f(x1,f(f(x2,x3),f(x4,x5)));
f(x1,f(x2,f(x3,f(x4,x5))));
f(f(x1,x2),f(x3,f(x4,x5))));
These are equivalent for f(x,y)=x+y or f(x,y)=xy, but nonequivalent in more general cases, such as f(x,y)=f(y,x)=x^2+y^2.
(End)
G.f. = x + x^2 + x^3 + 2*x^4 + 3*x^5 + 6*x^6 + 11*x^7 + 24*x^8 + 47*x^9 + ...


MAPLE

al := 1/2; M1 := 30; a[ 0 ] := 1; for n from 0 to M1 do n0 := floor(al*n);
a[ n+1 ] := sum( a[ i ]*a[ ni ], i=0..n0); i := 'i'; od: [ seq(a[ j ], j=0..M1) ];


MATHEMATICA

a[1]=1; a[n_]:=a[n]=Sum[a[k] a[nk], {k, 1, Floor[n/2]}]; Table[a[n], {n, 1, 32}] (* JeanFrançois Alcover, Mar 21 2011 *)


PROG

(PARI) A000992_list(n)={for(i=4, #n=vector(n, i, 1), n[i]=sum(j=1, i\2, n[j]*n[ij])); n} \\ M. F. Hasler, Dec 20 2011
(Haskell)
a000992 n = a000992_list !! (n1)
a000992_list = 1 : f 1 0 [1] where
f x y zs = z : f (x + y) (1  y) (z:zs) where
z = sum $ take x $ zipWith (*) zs $ reverse zs
 Reinhard Zumkeller, Dec 21 2011


CROSSREFS

Compare recurrence for A000108 (the Catalan numbers).
A093637 counts above trees without the restriction that all outdegrees are <=2.
Cf. A001190, A124973, A248748.
Sequence in context: A217312 A006787 A176425 * A036648 A047750 A072187
Adjacent sequences: A000989 A000990 A000991 * A000993 A000994 A000995


KEYWORD

nonn,easy,nice


AUTHOR

N. J. A. Sloane


STATUS

approved



