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Symmetric matrix based on f(i,j)=max{i(j+1)-1,j(i+1)-1}, by antidiagonals.
3

%I #7 Mar 30 2012 18:58:07

%S 1,3,3,5,5,5,7,8,8,7,9,11,11,11,9,11,14,15,15,14,11,13,17,19,19,19,17,

%T 13,15,20,23,24,24,23,20,15,17,23,27,29,29,29,27,23,17,19,26,31,34,35,

%U 35,34,31,26,19,21,29,35,39,41,41,41,39,35,29,21,23,32,39,44

%N Symmetric matrix based on f(i,j)=max{i(j+1)-1,j(i+1)-1}, by antidiagonals.

%C A203998 represents the matrix M given by f(i,j)=max{i(j+1)-1,j(i+1)-1}for i>=1 and j>=1. See A203999 for characteristic polynomials of principal submatrices of M, with interlacing zeros.

%e Northwest corner:

%e 1...3....5....7....9

%e 3...5....8....11...14

%e 5...8....11...15...19

%e 7...11...15...19...24

%t f[i_, j_] := Max[i (j + 1) - 1, j (i + 1) - 1];

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

%t TableForm[m[6]] (* 6x6 principal submatrix *)

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

%t {n, 1, 12}, {i, 1, n}]] (* A203998 *)

%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[%] (* A203999 *)

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

%Y Cf. A203999, A202453.

%K nonn,tabl

%O 1,2

%A _Clark Kimberling_, Jan 09 2012