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 A187660 Triangle in which T(n,k) = (-1)^(floor[3*k/2])*binomial[floor[(n+k)/2],k] is entry k of row n, where 0<=k<=n. 7
 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, -1, 1, -4, -6, 10, 5, -6, -1, 1, 1, -4, -10, 10, 15, -6, -7, 1, 1, 1, -5, -10, 20, 15, -21, -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 (list; table; graph; refs; listen; history; text; internal format)
 OFFSET 0,8 COMMENTS (Start) Let T(n,k) denote entry k (0<=k<=n) in row n of the triangle which begins {1} {1,-1} {1,-1,-1} {1,-2,-1,1} {1,-2,-3,1,1},... Let n>1 and N=2*n+1. Row n gives the coefficients of the characteristic polynomial p_N(x)=Sum[k=0..n, T(n,k)*x^{(n-1)*(n-k)}] of the n X n Danzer matrix (see [Jeffery]) D_{n-1} = [0,...,0,1; 0,...,0,1,1; ...; 0,1,...,1; 1,...,1]. Let {S_r(t)} be a sequence of polynomials with recurrence relation S_0(t)=1, S_1(t)=t, S_r(t)=t*S_(r-1)(t)-S_(r-2)(t) (r>1). (cf. A049310.) Then p_N(x)=0 has solutions w_j=S_(n-1)(phi_j), where phi_j=2*cos((2*j-1)*Pi/N), j = 1,2,...,n; that is, the w_j are the eigenvalues of D_{n-1}. (See the example below.) (End) Also taking the above exponents in reverse order gives the characteristic polynomial Sum[k=0..n, T(n,k)*x^{(n-1)*k}] for the n X n Danzer matrix D_{1} (again see [Jeffery]). - L. Edson Jeffery, Dec 18 2011 LINKS L. E. Jeffery, Danzer matrices FORMULA T(n,k) = (-1)^(floor(3*k/2))*binomial(floor((n+k)/2),k) T(n,k) = (-1)^n*A066170(n,k) abs(T(n,k)) = A046854(n,k) = abs(A066170(n,k)) = abs(A130777(n,k)) abs(T(n,k)) = A065941(n,n-k) = abs(A108299(n,n-k)) EXAMPLE For n=3 (N=7), k = 0,1,2,3, {T(3,k)} = {1,-2,-1,1}, giving the coefficients of the characteristic function p_7(x)=x^6-2*x^4-x^2+1=0 for the 3 X 3 Danzer matrix D_2=[0,0,1;0,1,1;1,1,1] with eigenvalues w_j=S_2(phi_j)=[2*cos((2*j-1)*Pi/7)]^2-1, j=1,2,3. Hence entries in row n>0 of the triangle are also given, up to a sign, by the elementary symmetric polynomials e_k in the w_j: defining e_0=1, for all n, then, again for n=3 (N=7), we have e_0=T(3,0)=1, e_1=-T(3,1)=w_1+w_2+w_3=2, e_2=T(3,2)=w_1*w_2+w_1*w_3+w_2*w_3=-1, e_3=-T(3,3)=w_1*w_2*w_3=-1, so also p_7(x)=Sum[k=0..3, (-1)^k*e_k*x^{2*(3-k)}]. MAPLE A187660 := proc(n, k): (-1)^(floor(3*k/2))*binomial(floor((n+k)/2), k) end: seq(seq(A187660(n, k), k=0..n), n=0..11); [Johannes W. Meijer, Aug 08 2011] CROSSREFS Signed version of A046854. Absolute values of a(n) form a reflected version of A065941, which is considered the main entry. Cf. A066170, A130777. Sequence in context: A267482 A130777 A046854 * A066170 A184957 A228349 Adjacent sequences:  A187657 A187658 A187659 * A187661 A187662 A187663 KEYWORD sign,easy,tabl AUTHOR L. Edson Jeffery, Mar 12 2011 EXTENSIONS Edited by L. Edson Jeffery, Jan 04 2013 STATUS approved

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