

A086615


Antidiagonal sums of triangle A086614.


9



1, 2, 4, 8, 17, 38, 89, 216, 539, 1374, 3562, 9360, 24871, 66706, 180340, 490912, 1344379, 3701158, 10237540, 28436824, 79288843, 221836402, 622599625, 1752360040, 4945087837, 13988490338, 39658308814, 112666081616
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OFFSET

0,2


COMMENTS

Partial sums of the Motzkin sequence (A001006).  Emeric Deutsch, Jul 12 2004
a(n) = number of distinct ordered trees obtained by branchreducing the ordered trees on n+1 edges.  David Callan, Oct 24 2004
a(n)= the number of consecutive horizontal steps at height 0 of all Motzkin paths from (0,0) to (n,0) starting with a horizontal step.  Charles Moore (chamoore(AT)howard.edu), Apr 15 2007
Equals row sums of triangle A136788  Gary W. Adamson, Jan 21 2008
The subsequence of prime partial sums of the Motzkin sequence begins: 2, 17, 89, no more through a(27). [From Jonathan Vos Post, Feb 11 2010]
This sequence (with offset 1 instead of 0) occurs in Section 7 of K. Grygiel, P. Lescanne (2015), see g.f. N.  N. J. A. Sloane, Nov 09 2015
Also number of plain (untyped) normal forms of lambdaterms (terms that cannot be further betareduced.) [Bendkowski et al., 2016].  N. J. A. Sloane, Nov 22 2017


LINKS

Vincenzo Librandi, Table of n, a(n) for n = 0..200
P. Barry, Continued fractions and transformations of integer sequences, JIS 12 (2009) 09.7.6
Maciej Bendkowski, K Grygiel, P Tarau, Random generation of closed simplytyped lambdaterms: a synergy between logic programming and Boltzmann samplers, arXiv preprint arXiv:1612.07682, 2016
K. Grygiel, P. Lescanne, A natural counting of lambda terms, SOFSEM 2016. Preprint 2015


FORMULA

G.f.: A(x) = 1/(1x)^2 + x^2*A(x)^2.
a(n)=sum{k=0..floor((n+1)/2), binomial(n+1, 2k+1)binomial(2k, k)/(k+1)}  Paul Barry, Nov 29 2004
a(n) = n + 1 + sum_k a(k1)a(nk1), starting from a(n)=0 for n negative.  Henry Bottomley, Feb 22 2005
a(n)=sum{k=0..n, sum{j=0..nk, C(j)C(nk, 2j)}};  Paul Barry, Aug 19 2005
G.f.: c(x^2/(1x)^2)/(1x)^2, c(x) the g.f. of A000108; a(n)=sum{k=0..floor(n/2), C(n+1,n2k)C(k)};  Paul Barry, May 31 2006
Binomial transform of doubled Catalan sequence 1,1,1,1,2,2,5,5,14,14,...  Paul Barry, Nov 17 2005
Row sums of PascalCatalan triangle A086617.  Paul Barry, Nov 17 2005
g(z)=(1zsqrt(12z3z^2))/(2z2z^2)/z  Charles Moore (chamoore(AT)howard.edu), Apr 15 2007, corrected by Vaclav Kotesovec, Feb 13 2014
Conjecture: (n+2)*a(n) +3*(n1)*a(n1) +(n+4)*a(n2) +3*(n1)*a(n3)=0.  R. J. Mathar, Nov 30 2012
a(n) ~ 3^(n+5/2) / (4 * sqrt(Pi) * n^(3/2)).  Vaclav Kotesovec, Feb 13 2014


EXAMPLE

a(0)=1, a(1)=2, a(2)=3+1=4, a(3)=4+4=8, a(4)=5+10+2=17, a(5)=6+20+12=38, are upward antidiagonal sums of triangle A086614:
{1},
{2,1},
{3,4,2},
{4,10,12,5},
{5,20,42,40,14},
{6,35,112,180,140,42}, ...
For example with n=2, the 5 ordered trees (A000108) on 3 edges are
......./\.../\.../\..
../.\................
.......................
Suppressing nonroot vertices of outdegree 1 (branchreducing) yields
......./\.../\../\..
.../.\.................
of which 4 are distinct. So a(2)=4.
a(4)=8 because we have HHHH, HHUD, HUDH, HUHD


MATHEMATICA

CoefficientList[Series[(1xSqrt[12*x3*x^2])/(2*x2*x^2)/x, {x, 0, 20}], x] (* Vaclav Kotesovec, Feb 13 2014 *)


CROSSREFS

Cf. A086614 (triangle), A086616 (row sums).
Cf. A001006.
Cf. A136788.
Sequence in context: A257300 A229202 A003007 * A081124 A090901 A101516
Adjacent sequences: A086612 A086613 A086614 * A086616 A086617 A086618


KEYWORD

nonn


AUTHOR

Paul D. Hanna, Jul 24 2003


EXTENSIONS

Edited by N. J. A. Sloane, Oct 16 2006


STATUS

approved



