OFFSET
0,2
COMMENTS
M has 2's on the main diagonal, -1's on the two adjacent diagonals, except that the entry M[floor(n/2)+1, floor(n/2)] is -2.
For n=4, M is the Cartan matrix of type F_4.
Row sums are 1, 1, -1, -2, -2, 0, 1, 1, -1, -2, -2, ...
FORMULA
T(n, m, d)= If[ n == m, 2, If[n == Floor[d/2] + 1 && m == Floor[d/2], -2, If[(n == m - 1 || n == m + 1), -1, 0]]]
EXAMPLE
Triangle begins:
{1},
{2, -1},
{2, -4, 1},
{2, -9, 6, -1},
{1, -16, 20, -8,1},
{0, -24, 50, -35, 10, -1},
{-2, -32, 104, -112, 54, -12, 1},
{-4, -38, 190, -293, 210, -77, 14, -1},
{-7, -40, 314, -664, 659, -352, 104, -16, 1},
{-10, -35, 478, -1349, 1772, -1286, 546, -135, 18, -1},
{-14, -20,677, -2512, 4234, -3992, 2274, -800, 170, -20, 1}
...
For n=4, the matrix M is
[2,-1,0,0],
[-1,2,-1,0],
[0,-2,2,-1],
[0,0,-1,2],
which has charpoly x^4-8*x^3+20*x^2-16*x+1. The coefficients (starting with the constant term) give row 4 of the triangle.
MATHEMATICA
T[n_, m_, d_] := If[ n == m, 2, If[n == Floor[d/2] + 1 && m == Floor[d/2], -2, If[(n == m - 1 || n == m + 1), -1, 0]]]; M[d_] := Table[T[n, m, d], {n, 1, d}, {m, 1, d}]; a0 = Table[M[d], {d, 1, 10}]; Table[Det[M[d]], {d, 1, 10}]; g = Table[Det[M[d] - x*IdentityMatrix[d]], {d, 1, 10}]; a = Join[{{1}}, Table[CoefficientList[Det[M[d] - x*IdentityMatrix[d]], x], {d, 1, 10}]]; Flatten[a] MatrixForm[a];
CROSSREFS
KEYWORD
tabl,sign
AUTHOR
Roger L. Bagula, Apr 05 2008
EXTENSIONS
Edited by N. J. A. Sloane, Jan 27 2014
STATUS
approved