A187660 Triangle read by rows: T(n,k) = (-1)^(floor(3*k/2))*binomial(floor((n+k)/2),k), 0 <= k <= n.

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

Offset: 0

Author

L. Edson Jeffery, Mar 12 2011

Keywords

• sign
• easy
• tabl

Comments

Conjecture: (i) Let n > 1 and N=2*n+1. Row n of T gives the coefficients of the characteristic polynomial p_N(x)=Sum_{k=0..n} T(n,k)*x^(n-k) of the n X n Danzer matrix D_{N,n-1} = {{0,...,0,1}, {0,...,0,1,1}, ..., {0,1,...,1}, {1,...,1}}. (ii) Let S_0(t)=1, S_1(t)=t and S_r(t)=t*S_(r-1)(t)-S_(r-2)(t), r > 1 (cf. A049310). Then p_N(x)=0 has solutions w_{N,j}=S_(n-1)(phi_{N,j}), where phi_{N,j}=2*(-1)^(j+1)*cos(j*Pi/N), j = 1..n. - L. Edson Jeffery, Dec 18 2011

Examples

			Triangle begins:
  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;
		

Links

• L. E. Jeffery, Danzer matrices
• Guoce Xin and Yueming Zhong, Proving some conjectures on Kekulé numbers for certain benzenoids by using Chebyshev polynomials, arXiv:2201.02376 [math.CO], 2022.

Crossrefs

Signed version of A046854.

Absolute values of a(n) form a reflected version of A065941, which is considered the main entry.

Cf. A046854, A066170, A130777, A267482.

Programs

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

Mathematica

t[n_, k_] := (-1)^Floor[3 k/2] Binomial[Floor[(n + k)/2], k]; Table[t[n, k], {n, 0, 11}, {k, 0, n}] (* L. Edson Jeffery, Oct 20 2017 *)

Formula

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)).

Extensions

Edited and corrected by L. Edson Jeffery, Oct 20 2017