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Symmetric matrix based on f(i,j)=max(ceiling(i/j),ceiling(j/i)), by antidiagonals.
3

%I #6 Mar 30 2012 18:58:07

%S 1,2,2,3,1,3,4,2,2,4,5,2,1,2,5,6,3,2,2,3,6,7,3,2,1,2,3,7,8,4,2,2,2,2,

%T 4,8,9,4,3,2,1,2,3,4,9,10,5,3,2,2,2,2,3,5,10,11,5,3,2,2,1,2,2,3,5,11,

%U 12,6,4,3,2,2,2,2,3,4,6,12,13,6,4,3,2,2,1,2,2,3,4,6,13,14,7,4,3

%N Symmetric matrix based on f(i,j)=max(ceiling(i/j),ceiling(j/i)), by antidiagonals.

%C A204143 represents the matrix M given by f(i,j)=max(ceiling(i/j),ceiling(j/i)) for i>=1 and j>=1. See A204144 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 2 3 4 5 6

%e 2 1 2 2 3 3

%e 3 2 1 2 2 2

%e 4 2 2 1 2 2

%e 5 3 2 2 1 2

%e 6 3 2 2 2 1

%t f[i_, j_] := Max[Ceiling[i/j], Ceiling[j/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}]] (* A204143 *)

%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[%] (* A204144 *)

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

%Y Cf. A204144, A204016, A202453.

%K nonn,tabl

%O 1,2

%A _Clark Kimberling_, Jan 11 2012