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%I #33 Mar 08 2019 03:29:23
%S 1,-1,1,1,-1,-1,-1,2,1,-1,1,-2,-3,1,1,-1,3,3,-4,-1,1,1,-3,-6,4,5,-1,
%T -1,-1,4,6,-10,-5,6,1,-1,1,-4,-10,10,15,-6,-7,1,1,-1,5,10,-20,-15,21,
%U 7,-8,-1,1,1,-5,-15,20,35,-21,-28,8,9,-1,-1,-1,6,15,-35,-35,56,28,-36,-9,10,1,-1,1,-6,-21,35,70,-56,-84,36,45,-10,-11
%N Triangle read by rows: T(n,k) = (-1)^n*(-1)^(floor(3*k/2))*binomial(floor((n+k)/2),k), 0 <= k <= n, n >= 0.
%C 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
%D Thomas Koshy, "Fibonacci and Lucas Numbers with Applications", John Wiley and Sons, 2001 (Chapter 14)
%H Indranil Ghosh, <a href="/A066170/b066170.txt">Rows 0..125, flattened</a>
%H Henry W. Gould, <a href="http://www.fq.math.ca/Scanned/3-4/gould.pdf"> A Variant of Pascal's Triangle </a>, The Fibonacci Quarterly, Vol. 3, Nr. 4, Dec. 1965, p. 257-271, with <a href="http://www.fq.math.ca/Scanned/4-1/corrections2.pdf">corrections</a>.
%H P. Steinbach, <a href="http://www.jstor.org/stable/2691048">Golden fields: a case for the heptagon</a>, Math. Mag. 70 (1997), no. 1, 22-31.
%F From _L. Edson Jeffery_, Mar 23 2011: (Start)
%F T(n,k) = (-1)^n*(-1)^(floor(3*k/2))*binomial(floor((n+k)/2),k);
%F T(n,k) = (-1)^n*A187660(n,k). (End)
%F From _Johannes W. Meijer_, Aug 08 2011: (Start)
%F abs(T(n,k)) = A046854(n,k) = abs(A108299(n,n-k))
%F abs(T(n,n-k)) = A065941(n,k). (End)
%e The table begins {1}; {-1, 1}; {1, -1, -1}; {-1, 2, 1, -1}; ...
%e The characteristic function of
%e ( 1 1 1 )
%e ( 1 1 0 )
%e ( 1 0 0 )
%e 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}.
%p 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
%t 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 *)
%Y Cf. A007700, A059455, A065941. For another version see A030111.
%K sign,easy,tabl
%O 0,8
%A _Floor van Lamoen_, Dec 14 2001
%E More terms from _Vladeta Jovovic_, Jan 02 2002
%E Corrected and edited by _Johannes W. Meijer_, Aug 08 2011
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