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A187660
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Triangle read by rows: T(n,k) = (-1)^(floor(3*k/2))*binomial(floor((n+k)/2),k), 0 <= k <= n.
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7
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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
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OFFSET
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0,8
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COMMENTS
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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
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LINKS
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FORMULA
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EXAMPLE
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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;
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MAPLE
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MATHEMATICA
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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 *)
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CROSSREFS
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Absolute values of a(n) form a reflected version of A065941, which is considered the main entry.
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KEYWORD
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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