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A204004
Symmetric matrix based on f(i,j) = max{2i+j-2,i+2j-2}, by antidiagonals.
5
1, 3, 3, 5, 4, 5, 7, 6, 6, 7, 9, 8, 7, 8, 9, 11, 10, 9, 9, 10, 11, 13, 12, 11, 10, 11, 12, 13, 15, 14, 13, 12, 12, 13, 14, 15, 17, 16, 15, 14, 13, 14, 15, 16, 17, 19, 18, 17, 16, 15, 15, 16, 17, 18, 19, 21, 20, 19, 18, 17, 16, 17, 18, 19, 20, 21, 23, 22, 21, 20, 19, 18
OFFSET
1,2
COMMENTS
A204004 represents the matrix M given by f(i,j)=max{2i+j,i+2j}for i>=1 and j>=1. See A204005 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.
For m=1 the result is A002024,
for m=2 the result is A204004,
for m=3 the result is A204008,
for m=4 the result is A206772. - Boris Putievskiy, Jan 24 2013
LINKS
Boris Putievskiy, Transformations [Of] Integer Sequences And Pairing Functions, arXiv preprint arXiv:1212.2732 [math.CO], 2012.
FORMULA
From Boris Putievskiy, Jan 24 2013: (Start)
For the general case, a(n) = m*A002024(n) + (m-1)*max{-A002260(n),-A004736(n)}.
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=2, a(n) = 2*(t+1) + max{t*(t+1)/2-n,n-(t*t+3*t+4)/2}, where t=floor((-1+sqrt(8*n-7))/2). (End)
EXAMPLE
Northwest corner:
1, 3, 5, 7, 9
3, 4, 6, 8, 10
5, 6, 7, 9, 11
7, 8, 9, 10, 12
MATHEMATICA
f[i_, j_] := Max[2 i + j - 2, 2 j + i - 2];
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}]] (* A204004 *)
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}]
Flatten[%] (* A204005 *)
TableForm[Table[c[n], {n, 1, 10}]]
KEYWORD
nonn,tabl
AUTHOR
Clark Kimberling, Jan 09 2012
STATUS
approved