A120406 Triangle read by rows: related to series expansion of the square root of 2 linear factors.

1, 2, 2, 5, 6, 5, 14, 18, 18, 14, 42, 56, 60, 56, 42, 132, 180, 200, 200, 180, 132, 429, 594, 675, 700, 675, 594, 429, 1430, 2002, 2310, 2450, 2450, 2310, 2002, 1430, 4862, 6864, 8008, 8624, 8820, 8624, 8008, 6864, 4862

Offset: 0

Author

David Callan, Jul 03 2006

Keywords

• nonn
• tabl

Comments

The numbers T(n,k) arise in the expansion of the square root of 2 generic linear factors: 1 - sqrt((1-a*x)*(1-b*x)) = (a+b)*x/2 + (1/8)*(b-a)^2*x^2*Sum_{n>=0} (Sum_{k=0..n} T(n,k)*a^k*b^(n-k))*(x/4)^n. (The g.f. below simply reformulates this fact.) A combinatorial interpretation of T(n,k) would be very interesting.

Examples

			Table begins
  \ k..0....1....2....3....4....5....6
  n
  0 |..1
  1 |..2....2
  2 |..5....6....5
  3 |.14...18...18...14
  4 |.42...56...60...56...42
  5 |132..180..200..200..180..132
  6 |429..594..675..700..675..594..429
		

Crossrefs

Column k=0 is the Catalan numbers A000108 (offset). The middle-of-row entries form A005566. Cf. A067804.

Programs

Mathematica

Table[2 Binomial[n,k]^2 Binomial[2n+2,n]/ Binomial[2n+2,2k+1],{n,0,9},{k,0,n}]

Maxima

solve(A=x*(A^2*y^2-2*A^2*y-2*A*y+A^2-2*A+1),A); /* Vladimir Kruchinin, Oct 24 2020 */

Formula

T(n,k) = 2*binomial(n,k)^2*binomial(2n+2,n)/binomial(2n+2,2k+1). This shows that T(n,k) is positive and the rows are symmetric.

T(n,k) = (k+1)*CatalanNumber(n+1) - 2*Sum_{j=0..k-1} (k-j)*CatalanNumber(j)*CatalanNumber(n-j). This shows that T(n,k) is an integer.

G.f.: F(x,y):=Sum_{n>=0, k=0..n} T(n,k) x^n y^k is given by F(x,y) = ( 1-2x-2x*y-sqrt(1-4x)*sqrt(1-4x*y) )/( 2x^2*(1-y)^2 ). This shows that the row sums are the powers of 4 (A000302) because lim_{y->1} F(x,y) = 1/(1-4x).

1 + x*(d/dx)(log(F(x,y))) = 1 + (2 + 2*y)*x + (6 + 4*y + 6*y^2)*x^2 + ... is the o.g.f. for A067804. - Peter Bala, Jul 17 2015

G.f. A(x,y) = -G(-x,y), G(x,y) satisfies G(x,y) = x/A008459(G(x,y))^2. - Vladimir Kruchinin, Oct 24 2020