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Correlation triangle for the sequence 2-0^n.
0

%I #14 Jan 15 2016 02:44:48

%S 1,2,2,2,5,2,2,6,6,2,2,6,9,6,2,2,6,10,10,6,2,2,6,10,13,10,6,2,2,6,10,

%T 14,14,10,6,2,2,6,10,14,17,14,10,6,2,2,6,10,14,18,18,14,10,6,2,2,6,10,

%U 14,18,21,18,14,10,6,2

%N Correlation triangle for the sequence 2-0^n.

%C Row sums are (n+1)^2 (A000290(n+1)). Diagonal sums are the Molien series A007980. T(2n,n) is 4n+1 (A016813), the partial sums of (2-0^n)^2. T(2,n)-T(2n,n+1) is 3-2*0^n.

%C From _Mats Granvik_, Jul 06 2010: (Start)

%C If seen as a square array:

%C 1, 2, 2, 2

%C 2, 5, 6, 6

%C 2, 6, 9, 10

%C 2, 6, 10, 13

%C then the matrix inverse contains the same values, only signed and in reversed order:

%C 13, -10, 6, -2

%C -10, 9, -6, 2

%C 6, -6, 5, -2

%C -2, 2, -2, 1

%C (End)

%F G.f.: (1+x)(1+x*y)/((1-x)(1-x*y)(1-x^2*y)).

%F T(n, k) = sum{j=0..n, [j<=k]*(2-0^(k-j))*[j<=n-k]*(2-0^(n-k-j))}.

%e Triangle begins

%e 1;

%e 2,2;

%e 2,5,2;

%e 2,6,6,2;

%e 2,6,9,6,2;

%e 2,6,10,10,6,2;

%t Flatten[Table[Table[If[n - k + 1 == k, 4*(n - k + 1 - 1) + 1, If[n - k + 1 > k, 4*(k - 1) + 2, 4*(n - k + 1 - 1) + 2]], {k, 1, n}], {n, 1, 11}]] (* _Mats Granvik_, Jan 06 2016 *)

%K easy,nonn,tabl

%O 0,2

%A _Paul Barry_, Jan 19 2006

%E a(65)-a(66) from _Mats Granvik_, Jan 06 2016