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Triangle, read by rows, which shifts one column left under matrix inverse.
8

%I #10 Jan 17 2018 03:23:41

%S 1,1,1,1,-1,1,1,-2,1,1,1,-1,1,-1,1,1,-1,1,-2,1,1,1,-1,1,-1,1,-1,1,1,

%T -1,1,-1,1,-2,1,1,1,-1,1,-1,1,-1,1,-1,1,1,-1,1,-1,1,-1,1,-2,1,1,1,-1,

%U 1,-1,1,-1,1,-1,1,-1,1,1,-1,1,-1,1,-1,1,-1,1,-2,1,1,1,-1,1,-1,1,-1,1,-1,1,-1,1,-1,1,1,-1,1,-1,1,-1,1

%N Triangle, read by rows, which shifts one column left under matrix inverse.

%C The unsigned columns of the matrix logarithm of this triangle are all equal to A110504.

%F T(n, k) = +1 when k == 0 (mod 2), T(n, k)=-1 when k == 1 (mod 2), except for T(k+2, k) = -2 when k == 1 (mod 2) and T(n, n) = 1.

%F G.f. for column k of matrix power A110503^m (ignoring leading zeros): cos(m*arccos(1-x^2/2)) + (-1)^k*sin(m*arccos(1-x^2/2))*(1-x/2)/sqrt(1-x^2/4)*(1+x)/(1-x).

%e Triangle begins:

%e 1;

%e 1, 1;

%e 1, -1, 1;

%e 1, -2, 1, 1;

%e 1, -1, 1, -1, 1;

%e 1, -1, 1, -2, 1, 1;

%e 1, -1, 1, -1, 1, -1, 1;

%e 1, -1, 1, -1, 1, -2, 1, 1;

%e 1, -1, 1, -1, 1, -1, 1, -1, 1;

%e 1, -1, 1, -1, 1, -1, 1, -2, 1, 1; ...

%e The matrix inverse drops the first column:

%e 1;

%e -1, 1;

%e -2, 1, 1;

%e -1, 1, -1, 1;

%e -1, 1, -2, 1, 1;

%e -1, 1, -1, 1, -1, 1; ...

%e The matrix logarithm equals:

%e 0;

%e 1/1!, 0;

%e 3/2!, -1/1!, 0;

%e 7/3!, -3/2!, 1/1!, 0;

%e 30/4!, -7/3!, 3/2!, -1/1!, 0;

%e 144/5!, -30/4!, 7/3!, -3/2!, 1/1!, 0;

%e 876/6!, -144/5!, 30/4!, -7/3!, 3/2!, -1/1!, 0; ...

%e unsigned columns of which all equal A110505.

%o (PARI) T(n,k)=matrix(n+1,n+1,r,c,if(r>=c, if(r==c || c%2==1,1,if(r%2==0 && r==c+2,-2,-1))))[n+1,k+1]

%Y Cf. A110504 (matrix log), A110505 (column 0 of log).

%Y Cf. A111940 (variant).

%K sign,tabl

%O 0,8

%A _Paul D. Hanna_, Jul 23 2005