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%I #5 Apr 09 2012 18:39:37
%S 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,
%T 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,
%U 160,240,320,200,240,80,80,15,10,1,64,112,160,240,160
%N Matrix square of lower triangular array A211226.
%C Analog of square of Pascal's triangle.
%F T(2*n,2*k) = T(2*n+1,2*k+1) = (n+2-k)*binomial(n,k)*2^(n-k-1);
%F 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).
%F Recurrence equations:
%F 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);
%F 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).
%F 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.
%F Row polynomials:
%F R(2*n,x) = (x^2+2*n*x+n+2)*(x^2+2)^(n-1);
%F R(2*n+1,x) = (x^3+2*x^2+(n+2)*x+4)*(x^2+2)^(n-1).
%F Column 0 = A211227. Row sums A211231.
%e Triangle begins
%e .n\k.|....0....1....2....3....4....5....6....7....8....9
%e = = = = = = = = = = = = = = = = = = = = = = = = = = = = =
%e ..0..|....1
%e ..1..|....2....1
%e ..2..|....3....2....1
%e ..3..|....4....3....2....1
%e ..4..|....8....8....6....4....1
%e ..5..|....8....8....8....6....2....1
%e ..6..|...20...24...24...24....9....6....1
%e ..7..|...16...20...24...24...12....9....2....1
%e ..8..|...48...64...80...96...48...48...12....8....1
%e ..9..|...32...48...64...80...48...48...16...12....2....1
%e ...
%Y Cf. A211226, A211231 (row sums).
%K nonn,easy,tabl
%O 0,2
%A _Peter Bala_, Apr 05 2012
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