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

Author

Paul D. Hanna, Feb 04 2006

Keywords

• nonn

Examples

			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
		

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.

Crossrefs

Cf. A033184.

Programs

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

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

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 *)

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),", "))

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

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