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A204008
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Symmetric matrix based on f(i,j) = max{3i+j-3,i+3j-3}, by antidiagonals.
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7
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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, 17, 19, 22, 20, 18, 16, 16, 18, 20, 22, 25, 23, 21, 19, 17, 19, 21, 23, 25, 28, 26, 24, 22, 20, 20, 22, 24, 26, 28, 31, 29, 27, 25, 23, 21, 23, 25, 27, 29, 31, 34, 32, 30
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OFFSET
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1,2
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COMMENTS
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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.
General case A206772. Let m be natural number. Table T(n,k)=max{m*n+k-m,n+m*k-m} read by antidiagonals.
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LINKS
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FORMULA
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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).
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)
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EXAMPLE
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Northwest corner:
1, 4, 7, 10
4, 5, 8, 11
7, 8, 9, 12
10, 11, 12, 13
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MATHEMATICA
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f[i_, j_] := Max[3 i + j - 3, 3 j + i - 3];
m[n_] := Table[f[i, j], {i, 1, n}, {j, 1, n}]
TableForm[m[6]] (* 6x6 principal submatrix *)
Flatten[Table[f[i, n + 1 - i],
{n, 1, 12}, {i, 1, n}]] (* A204008 *)
p[n_] := CharacteristicPolynomial[m[n], x];
c[n_] := CoefficientList[p[n], x]
TableForm[Flatten[Table[p[n], {n, 1, 10}]]]
Table[c[n], {n, 1, 12}]
TableForm[Table[c[n], {n, 1, 10}]]
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CROSSREFS
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KEYWORD
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AUTHOR
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STATUS
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approved
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