|
Search: id:A086615
|
|
| |
|
| 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
(list; graph; listen)
|
|
|
OFFSET
|
0,2
|
|
|
COMMENT
|
Partial sums of the Motzkin sequence (A001006). - Emeric Deutsch (deutsch(AT)duke.poly.edu), Jul 12 2004
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
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 (qntmpkt(AT)yahoo.com), Jan 21 2008
|
|
FORMULA
|
G.f.: A(x) = 1/(1-x)^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 (pbarry(AT)wit.ie), Nov 29 2004
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
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
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
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
Row sums of Pascal-Catalan triangle A086617. - Paul Barry (pbarry(AT)wit.ie), Nov 17 2005
g(z)=(1-z-sqrt(1-2z-3z^2))/2z-2z^2 - Charles Moore (chamoore(AT)howard.edu), Apr 15 2007
|
|
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 (branch-reducing) yields
|...|..../\.../\../|\..
.../.\.................
of which 4 are distinct. So a(2)=4.
a(4)=8 because we have HHHH, HHUD, HUDH, HUHD
|
|
CROSSREFS
|
Cf. A086614 (triangle), A086616 (row sums).
Cf. A001006.
Cf. A136788.
Sequence in context: A025276 A006461 A003007 this_sequence A081124 A090901 A101516
Adjacent sequences: A086612 A086613 A086614 this_sequence A086616 A086617 A086618
|
|
KEYWORD
|
nonn
|
|
AUTHOR
|
Paul D. Hanna (pauldhanna(AT)juno.com), Jul 24 2003
|
|
EXTENSIONS
|
Edited by N. J. A. Sloane (njas(AT)research.att.com), Oct 16 2006
|
|
|
Search completed in 0.002 seconds
|