%I A086615
%S A086615 1,2,4,8,17,38,89,216,539,1374,3562,9360,24871,66706,180340,490912,
%T A086615 1344379,3701158,10237540,28436824,79288843,221836402,622599625,
%U A086615 1752360040,4945087837,13988490338,39658308814,112666081616
%N A086615 Antidiagonal sums of triangle A086614.
%C A086615 Partial sums of the Motzkin sequence (A001006). - Emeric Deutsch (deutsch(AT)duke.poly.edu), 
               Jul 12 2004
%C A086615 a(n) = number of distinct ordered trees obtained by branch-reducing the 
               ordered trees on n+1 edges. - David Callan (callan(AT)stat.wisc.edu), 
               Oct 24 2004
%C A086615 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
%C A086615 Equals row sums of triangle A136788 - Gary W. Adamson (qntmpkt(AT)yahoo.com), 
               Jan 21 2008
%F A086615 G.f.: A(x) = 1/(1-x)^2 + x^2*A(x)^2.
%F A086615 a(n)=sum{k=0..floor((n+1)/2), binomial(n+1, 2k+1)binomial(2k, k)/(k+1)} 
               - Paul Barry (pbarry(AT)wit.ie), Nov 29 2004
%F A086615 a(n) = n + 1 + sum_k a(k-1)a(n-k-1), starting from a(n)=0 for n negative. 
               - Henry Bottomley (se16(AT)btinternet.com), Feb 22 2005
%F A086615 a(n)=sum{k=0..n, sum{j=0..n-k, C(j)C(n-k, 2j)}}; - Paul Barry (pbarry(AT)wit.ie), 
               Aug 19 2005
%F A086615 G.f.: c(x^2/(1-x)^2)/(1-x)^2, c(x) the g.f. of A000108; a(n)=sum{k=0..floor(n/
               2), C(n+1,n-2k)C(k)}; - Paul Barry (pbarry(AT)wit.ie), May 31 2006
%F A086615 Binomial transform of doubled Catalan sequence 1,1,1,1,2,2,5,5,14,14,
               ... - Paul Barry (pbarry(AT)wit.ie), Nov 17 2005
%F A086615 Row sums of Pascal-Catalan triangle A086617. - Paul Barry (pbarry(AT)wit.ie), 
               Nov 17 2005
%F A086615 g(z)=(1-z-sqrt(1-2z-3z^2))/2z-2z^2 - Charles Moore (chamoore(AT)howard.edu), 
               Apr 15 2007
%e A086615 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:
%e A086615 {1},
%e A086615 {2,1},
%e A086615 {3,4,2},
%e A086615 {4,10,12,5},
%e A086615 {5,20,42,40,14},
%e A086615 {6,35,112,180,140,42}, ...
%e A086615 For example with n=2, the 5 ordered trees (A000108) on 3 edges are
%e A086615 |...|..../\.../\.../|\..
%e A086615 |../.\..|......|........
%e A086615 |.......................
%e A086615 Suppressing nonroot vertices of outdegree 1 (branch-reducing) yields
%e A086615 |...|..../\.../\../|\..
%e A086615 .../.\.................
%e A086615 of which 4 are distinct. So a(2)=4.
%e A086615 a(4)=8 because we have HHHH, HHUD, HUDH, HUHD
%Y A086615 Cf. A086614 (triangle), A086616 (row sums).
%Y A086615 Cf. A001006.
%Y A086615 Cf. A136788.
%Y A086615 Sequence in context: A025276 A006461 A003007 this_sequence A081124 A090901 
               A101516
%Y A086615 Adjacent sequences: A086612 A086613 A086614 this_sequence A086616 A086617 
               A086618
%K A086615 nonn
%O A086615 0,2
%A A086615 Paul D. Hanna (pauldhanna(AT)juno.com), Jul 24 2003
%E A086615 Edited by N. J. A. Sloane (njas(AT)research.att.com), Oct 16 2006