OFFSET
0,2
COMMENTS
Analog of square of Pascal's triangle.
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).
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
KEYWORD
AUTHOR
Peter Bala, Apr 05 2012
STATUS
approved