A211230 Matrix square of lower triangular array A211226.

1, 2, 1, 3, 2, 1, 4, 3, 2, 1, 8, 8, 6, 4, 1, 8, 8, 8, 6, 2, 1, 20, 24, 24, 24, 9, 6, 1, 16, 20, 24, 24, 12, 9, 2, 1, 48, 64, 80, 96, 48, 48, 12, 8, 1, 32, 48, 64, 80, 48, 48, 16, 12, 2, 1, 112, 160, 240, 320, 200, 240, 80, 80, 15, 10, 1, 64, 112, 160, 240, 160

Offset

0,2

Comments

Analog of square of Pascal's triangle.

Links

Table of n, a(n) for n=0..70.

Formula

T(2*n,2*k) = T(2*n+1,2*k+1) = (n+2-k)*binomial(n,k)*2^(n-k-1);

T(2*n,2*k+1) = n*binomial(n-1,k)*2^(n-k); T(2*n+1,2*k) = binomial(n,k)*2^(n-k+1).

Recurrence equations:

T(2*n,2*k) = n/k*T(2*n-1,2*k-1), T(2*n,2*k+1) = n*T(2*n-1,2*k);

T(2*n+1,2*k) = 1/k*T(2*n,2*k-1), T(2*n+1,2*k+1) = T(2*n,2*k).

O.g.f.: P(x,t)/Q(x,t), where P(x,t) = 1 + (x+2)*t - (1-x)^2*t^2 - (x^3+2*x^2+x+4)*t^3 and Q(x,t) = (1-(x^2+2)*t^2)^2.

Row polynomials:

R(2*n,x) = (x^2+2*n*x+n+2)*(x^2+2)^(n-1);

R(2*n+1,x) = (x^3+2*x^2+(n+2)*x+4)*(x^2+2)^(n-1).

Column 0 = A211227. Row sums A211231.

Example

Triangle begins
.n\k.|....0....1....2....3....4....5....6....7....8....9
= = = = = = = = = = = = = = = = = = = = = = = = = = = = =
..0..|....1
..1..|....2....1
..2..|....3....2....1
..3..|....4....3....2....1
..4..|....8....8....6....4....1
..5..|....8....8....8....6....2....1
..6..|...20...24...24...24....9....6....1
..7..|...16...20...24...24...12....9....2....1
..8..|...48...64...80...96...48...48...12....8....1
..9..|...32...48...64...80...48...48...16...12....2....1
...

Crossrefs

Cf. A211226, A211231 (row sums).

Sequence in context: A052310 A052313 A271355 * A049085 A193173 A331581

Adjacent sequences: A211227 A211228 A211229 * A211231 A211232 A211233

Keyword

nonn,easy,tabl

Author

Peter Bala, Apr 05 2012

Status

approved