A096793 Triangle read by rows: a(n,k) is the number of Dyck n-paths containing k odd-length ascents.

1, 0, 1, 1, 0, 1, 0, 4, 0, 1, 3, 0, 10, 0, 1, 0, 21, 0, 20, 0, 1, 12, 0, 84, 0, 35, 0, 1, 0, 120, 0, 252, 0, 56, 0, 1, 55, 0, 660, 0, 630, 0, 84, 0, 1, 0, 715, 0, 2640, 0, 1386, 0, 120, 0, 1, 273, 0, 5005, 0, 8580, 0, 2772, 0, 165, 0, 1, 0, 4368, 0, 25025, 0, 24024, 0, 5148, 0, 220, 0, 1

Offset: 0

Author

David Callan, Aug 17 2004

Keywords

• nonn
• tabl

Comments

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

From Emeric Deutsch, Oct 05 2008: (Start)

Sum_{k=0..n} k*a(n,k) = A014300(n).

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

Examples

			Table begins
.
n |k = 0    1    2    3    4    5    6    7    8
--+---------------------------------------------
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.
		

Crossrefs

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.

Cf. A143950. - Emeric Deutsch, Oct 05 2008

Programs

Mathematica

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}]]

Formula

a(n, k) = binomial((n+k)/2, (n-k)/2)*binomial((3n-k)/2+1, (n+k)/2)/((3n-k)/2+1).

Equivalently, a(2n+k, k) = binomial(3n+k, k)*T(n) where T(n) = binomial(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 binomial(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 -- binomial(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, Sep 03 2005

From Emeric Deutsch, 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)