A116363 a(n) = dot product of row n in Catalan triangle A033184 with row n in Pascal's triangle.

1, 2, 7, 30, 141, 698, 3571, 18686, 99385, 535122, 2908863, 15932766, 87809541, 486421770, 2706138987, 15110359038, 84637982961, 475381503266, 2676447372535, 15100548901790, 85357620588541, 483304834607322

Offset

0,2

Links

G. C. Greubel, Table of n, a(n) for n = 0..1000

D. Drake, Bijections from Weighted Dyck Paths to Schröder Paths, J. Int. Seq. 13 (2010) # 10.9.2.

Formula

a(n) = Sum_{k=0..n} C(n,k)*C(2*n-k+1,n-k)*(k+1)/(2*n-k+1).

G.f. A(x) satisfies: d/dx[log(1 - 4*x*A(x))] = -4*(1-5*x)/(1-13*x+43*x^2-7*x^3).

O.g.f.: 2*(R+x)/(R*(R+x+1)), where R = sqrt(x^2+6*x+1). [Dan Drake, May 19 2010]

Conjecture: +(2*n+5)*(n+1)*a(n) +4*(-3*n^2-9*n+5)*a(n-1) +(2*n+7)*(n-1)*a(n-2)=0. - R. J. Mathar, Jun 22 2016

Example

The dot product of Catalan row 4 and Pascal row 4 equals
a(4) = [14,14,9,4,1]*[1,4,6,4,1] = 141
which is equivalent to obtaining the final term
in these repeated partial sums of Pascal row 4:
1,4, 6, 4, 1
.5,11,15,16
..16,31,47
...47,94
....141

Mathematica

Table[Sum[Binomial[n, j]*Binomial[2*n-j+1, n-j]*(j+1)/(2*n-j+1), {j, 0, n} ], {n, 0, 30}] (* G. C. Greubel, May 12 2019 *)

Prog

(PARI) a(n)=sum(k=0, n, binomial(n, k)*binomial(2*n-k+1, n-k)*(k+1)/(2*n-k+1))
for(n=0, 30, print1(a(n), ", "))
(Magma) [(&+[Binomial(n, j)*Binomial(2*n-j+1, n-j)*(j+1)/(2*n-j+1): j in [0..n]]): n in [0..30]]; // G. C. Greubel, May 12 2019
(Sage) [sum(binomial(n, j)*binomial(2*n-j+1, n-j)*(j+1)/(2*n-j+1) for j in (0..n)) for n in (0..30)] # G. C. Greubel, May 12 2019
(GAP) List([0..30], n-> Sum([0..n], j-> Binomial(n, j)*Binomial(2*n-j+1, n-j)*(j+1)/(2*n-j+1))) # G. C. Greubel, May 12 2019

Crossrefs

Cf. A033184.

Sequence in context: A027136 A353288 A299296 * A186858 A360102 A369441

Adjacent sequences: A116360 A116361 A116362 * A116364 A116365 A116366

Keyword

nonn

Author

Paul D. Hanna, Feb 04 2006

Status

approved