OFFSET
0,8
COMMENTS
The original name of this sequence was: Triangle giving coefficients of characteristic function of n X n matrix in which the left upper half and the antidiagonal are filled with 1's and the right lower half is filled with 0's. As was pointed out by L. Edson Jeffery this is only correct if we multiply each triangle row by (-1)^n. For the straightforward version of the coefficients of the characteristic polynomials see A187660. - Johannes W. Meijer, Aug 08 2011
REFERENCES
Thomas Koshy, "Fibonacci and Lucas Numbers with Applications", John Wiley and Sons, 2001 (Chapter 14)
LINKS
Indranil Ghosh, Rows 0..125, flattened
Henry W. Gould, A Variant of Pascal's Triangle , The Fibonacci Quarterly, Vol. 3, Nr. 4, Dec. 1965, p. 257-271, with corrections.
P. Steinbach, Golden fields: a case for the heptagon, Math. Mag. 70 (1997), no. 1, 22-31.
FORMULA
From L. Edson Jeffery, Mar 23 2011: (Start)
T(n,k) = (-1)^n*(-1)^(floor(3*k/2))*binomial(floor((n+k)/2),k);
T(n,k) = (-1)^n*A187660(n,k). (End)
From Johannes W. Meijer, Aug 08 2011: (Start)
abs(T(n,n-k)) = A065941(n,k). (End)
EXAMPLE
The table begins {1}; {-1, 1}; {1, -1, -1}; {-1, 2, 1, -1}; ...
The characteristic function of
( 1 1 1 )
( 1 1 0 )
( 1 0 0 )
is f(x) = x^3 - 2x^2 - x + 1, so the 3rd row is (-1)^3 times the f(x) coefficients, i.e., {-1; 2; 1; -1}.
MAPLE
A066170 := proc(n, k): (-1)^n*(-1)^(floor(3*k/2))*binomial(floor((n+k)/2), k) end: seq(seq(A066170(n, k), k=0..n), n=0..11); // Johannes W. Meijer, Aug 08 2011
MATHEMATICA
Flatten[Table[(-1)^n*(-1)^Floor[3*k/2]*Binomial[Floor[(n+k)/2], k], {n, 0, 12}, {k, 0, n}]] (* Indranil Ghosh, Feb 19 2017 *)
CROSSREFS
KEYWORD
AUTHOR
Floor van Lamoen, Dec 14 2001
EXTENSIONS
More terms from Vladeta Jovovic, Jan 02 2002
Corrected and edited by Johannes W. Meijer, Aug 08 2011
STATUS
approved