0,8

a(n,k)=0 unless k and n have the same parity and 0<=k<=n.

Contribution from Emeric Deutsch (deutsch(AT)duke.poly.edu), Oct 05 2008: (Start)

Sum(k*a(n,k),k=0..n)=A014300(n).

For the case of even-length ascents see A143950. (End)

a(n, k)=binom[(n+k)/2, (n-k)/2]binom[(3n-k)/2+1, (n+k)/2]/((3n-k)/2+1)

Equivalently, a(2n+k, k) = binom[3n+k, k]*T[n] where T[n] := bi[3n, n]/(2n+1) is A001764. Proof. Given a Dyck (2n+k)-path with k ascents of odd length, delete the peaks (UD) that terminate odd-length ascents. This is a mapping to Dyck (2n)-paths all of whose ascents have even length; there are T[n] such paths. The mapping is clearly onto and is binom[3n+k, k]-to-1 as follows. A Dyck (2n)-path all of whose ascents have even length has exactly 3n+1 vertices that are (i) not incident with an upstep, or (ii) incident with an upstep and at even distance (possibly 0) from the start of the ascent they lie in. The k deleted UDs can be inserted arbitrarily at these vertices, repetition allowed, to get the preimages---binom[3n+k, k] choices.

G.f.: G(z, t) + H(z, t) where G satisfies G^3*(t^2 - 1)*z^2 - G^2*t*z*(2 + t*z) + G*(1 + 2*t*z) - 1 = 0 and H satisfies H^3*(t^2 - 1)*z^2 + H^2*t*z*(2 + t*z) - H*t^2*(1 - t*z) + t^3*z = 0. Here z marks size (n) and t marks number of odd-length ascents (k). G is gf for paths that start with an even-length ascent and H is gf for paths that start with an odd-length ascent. - David Callan (callan(AT)stat.wisc.edu), Sep 03 2005

Contribution from Emeric Deutsch (deutsch(AT)duke.poly.edu), Oct 05 2008: (Start)

G.f. G=G(t,z) satisfies G = 1 + zG(t + zG)/(1 - z^2*G^2).

The trivariate g.f. H=H(t,s,z), where t (s) marks odd-length (even-length) ascents satisfies H = 1 + zH(t+szH)/(1-z^2*H^2). (End)

Table begins

\ k 0, 1, 2, ...

n

0 | 1

1 | 0, 1

2 | 1, 0, 1

3 | 0, 4, 0, 1

4 | 3, 0, 10, 0, 1

5 | 0, 21, 0, 20, 0, 1

6 | 12, 0, 84, 0, 35, 0, 1

7 | 0, 120, 0, 252, 0, 56, 0, 1

8 | 55, 0, 660, 0, 630, 0, 84, 0, 1

a(4,0)=3 because the Dyck 4-paths containing no odd-length ascents are UUUUDDDD,UUDUUDDD,UUDDUUDD.

bi[n_, k_] := If[IntegerQ[k], Binomial[n, k], 0]; TableForm[Table[bi[(n+k)/2, (n-k)/2]bi[(3n-k)/2+1, (n+k)/2]/((3n-k)/2+1), {n, 0, 10}, {k, 0, n}]]

The nonzero entries in column k=0 give A001764, in k=1 give A045721, in k=2 give A090763. The row sums are the Catalan numbers A000108.

A143950 [From Emeric Deutsch (deutsch(AT)duke.poly.edu), Oct 05 2008]

Sequence in context: A126222 A071637 A141277 this_sequence A155998 A127538 A096008

Adjacent sequences: A096790 A096791 A096792 this_sequence A096794 A096795 A096796

nonn,tabl

David Callan (callan(AT)stat.wisc.edu), Aug 17 2004