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Symmetric matrix based on f(i,j) = max{3i+j-3,i+3j-3}, by antidiagonals.
7

%I #24 Jun 24 2017 00:58:59

%S 1,4,4,7,5,7,10,8,8,10,13,11,9,11,13,16,14,12,12,14,16,19,17,15,13,15,

%T 17,19,22,20,18,16,16,18,20,22,25,23,21,19,17,19,21,23,25,28,26,24,22,

%U 20,20,22,24,26,28,31,29,27,25,23,21,23,25,27,29,31,34,32,30

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

%C A204008 represents the matrix M given by f(i,j)=max{3i+j-3,i+3j-3}for i>=1 and j>=1. See A204011 for characteristic polynomials of principal submatrices of M, with interlacing zeros.

%C General case A206772. Let m be natural number. Table T(n,k)=max{m*n+k-m,n+m*k-m} read by antidiagonals.

%C For m=1 the result is A002024,

%C for m=2 the result is A204004,

%C for m=3 the result is A204008,

%C for m=4 the result is A206772. - _Boris Putievskiy_, Jan 24 2013

%H Boris Putievskiy, <a href="http://arxiv.org/abs/1212.2732">Transformations [Of] Integer Sequences And Pairing Functions</a>, arXiv preprint arXiv:1212.2732, 2012.

%F From _Boris Putievskiy_, Jan 24 2013: (Start)

%F For the general case, a(n) = m*A002024(n) + (m-1)*max{-A002260(n),-A004736(n)}.

%F a(n) = m*(t+1) + (m-1)*max{t*(t+1)/2-n,n-(t*t+3*t+4)/2}, where t=floor((-1+sqrt(8*n-7))/2).

%F For m=3, a(n) = 3*(t+1) + 2*max{t*(t+1)/2-n,n-(t*t+3*t+4)/2}, where t=floor((-1+sqrt(8*n-7))/2). (End)

%e Northwest corner:

%e 1, 4, 7, 10

%e 4, 5, 8, 11

%e 7, 8, 9, 12

%e 10, 11, 12, 13

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

%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}]] (* A204008 *)

%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[%] (* A204011 *)

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

%Y Cf. A204011, A202453, A002024, A204004, A002260, A004736, A206772.

%K nonn,tabl

%O 1,2

%A _Clark Kimberling_, Jan 09 2012