login
Symmetric matrix based on (1,0,1,1,0,1,1,0,1,...), by antidiagonals.
3

%I #6 Jul 12 2012 00:39:54

%S 1,0,0,1,1,1,1,0,0,1,0,1,2,1,0,1,1,1,1,1,1,1,0,1,3,1,0,1,0,1,2,1,1,2,

%T 1,0,1,1,1,2,3,2,1,1,1,1,0,1,3,1,1,3,1,0,1,0,1,2,1,2,4,2,1,2,1,0,1,1,

%U 1,2,3,2,2,3,2,1,1,1,1,0,1,3,1,2,5,2,1,3,1,0,1,0,1,2,1,2,4,2,2

%N Symmetric matrix based on (1,0,1,1,0,1,1,0,1,...), by antidiagonals.

%C Let s be the periodic sequence (1,0,1,1,0,1,...) and let T be the infinite square matrix whose n-th row is formed by putting n-1 zeros before the terms of s. Let T' be the transpose of T. Then A203947 represents the matrix product M=T'*T. M is the self-fusion matrix of s, as defined at A193722. See A203948 for characteristic polynomials of principal submatrices of M, with interlacing zeros.

%e Northwest corner:

%e 1 0 1 1 0 1 1 0

%e 0 1 0 1 1 0 1 1

%e 1 0 2 1 1 0 1 1

%e 1 1 1 3 1 2 3 1

%e 0 1 1 1 3 1 2 3

%e 1 0 2 2 1 4 2 2

%e 1 1 1 3 2 2 5 2

%t t = {1, 0, 1};

%t t1 = Flatten[{t, t, t, t, t, t, t, t, t, t}];

%t s[k_] := t1[[k]];

%t U = NestList[Most[Prepend[#, 0]] &, #, Length[#] - 1] &[

%t Table[s[k], {k, 1, 15}]];

%t L = Transpose[U]; M = L.U; TableForm[M] (* A203947 *)

%t m[i_, j_] := M[[i]][[j]];

%t Flatten[Table[m[i, n + 1 - i], {n, 1, 12}, {i, 1, n}]]

%Y Cf. A203948, A202453.

%K nonn,tabl

%O 1,13

%A _Clark Kimberling_, Jan 08 2012