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Symmetric matrix based on f(i,j)=max(3i-2j, 3j-2i), by antidiagonals.
5

%I #5 Mar 30 2012 18:58:07

%S 1,4,4,7,2,7,10,5,5,10,13,8,3,8,13,16,11,6,6,11,16,19,14,9,4,9,14,19,

%T 22,17,12,7,7,12,17,22,25,20,15,10,5,10,15,20,25,28,23,18,13,8,8,13,

%U 18,23,28,31,26,21,16,11,6,11,16,21,26,31,34,29,24,19,14,9,9,14

%N Symmetric matrix based on f(i,j)=max(3i-2j, 3j-2i), by antidiagonals.

%C A204158 represents the matrix M given by f(i,j)=max(3i-2j, 3j-2i) for i>=1 and j>=1. See A204159 for characteristic polynomials of principal submatrices of M, with interlacing zeros. See A204016 for a guide to other choices of M.

%e Northwest corner:

%e 1....4....7....10...13

%e 4....2....5....8....11

%e 7....5....3....6....9

%e 10...8....6....4....7

%t f[i_, j_] := Max[3 i - 2 j, 3 j - 2 i];

%t m[n_] := Table[f[i, j], {i, 1, n}, {j, 1, n}]

%t TableForm[m[8]] (* 8x8 principal submatrix *)

%t Flatten[Table[f[i, n + 1 - i],

%t {n, 1, 15}, {i, 1, n}]] (* A204158 *)

%t p[n_] := CharacteristicPolynomial[m[n], x];

%t c[n_] := CoefficientList[p[n], x]

%t TableForm[Flatten[Table[p[n], {n, 1, 10}]]]

%t Table[c[n], {n, 1, 12}]

%t Flatten[%] (* A204159 *)

%t TableForm[Table[c[n], {n, 1, 10}]]

%Y Cf. A204159, A204016, A202453.

%K nonn,tabl

%O 1,2

%A _Clark Kimberling_, Jan 12 2012