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An invertible partition matrix.
1

%I #3 Mar 30 2012 18:59:12

%S 1,1,1,0,2,1,0,1,2,1,0,0,2,2,1,0,0,1,2,2,1,0,0,0,2,2,2,1,0,0,0,1,2,2,

%T 2,1,0,0,0,0,2,2,2,2,1,0,0,0,0,1,2,2,2,2,1,0,0,0,0,0,2,2,2,2,2,1,0,0,

%U 0,0,0,1,2,2,2,2,2,1,0,0,0,0,0,0,2,2,2,2,2,2,1

%N An invertible partition matrix.

%C Row sums are n+1, A000027. Diagonal sums are 1,1,1,2,2,2,3,3,3,.... or A008620. Inverse is A114115. Product with first difference matrix (1-x,x) is A114117.

%F Number triangle T(n, k)=sum{j=0..n, C(floor((n+j)/2), k)C(k, floor((n+j)/2))}.

%e Triangle begins

%e 1.................=1

%e 1,1...............=2

%e 0,2,1.............=3

%e 0,1,2,1...........=4

%e 0,0,2,2,1.........=5

%e 0,0,1,2,2,1.......=6

%e 0,0,0,2,2,2,1.....=7

%e 0,0,0,1,2,2,2,1...=8

%K easy,nonn,tabl

%O 0,5

%A _Paul Barry_, Nov 13 2005