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Sum_[k, 0<=k<=n}T(n,k)*x^(n-k) = F(n+1,-x) where F(n,x)is the n-th Fibonacci polynomial in x defined in A011973. - From DELEHAM Philippe, Feb 22 2013
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Convolution of A001792 with itself. - DELEHAM Philippe, Feb 21 2013
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allocated for DELEHAM PhilippeA convolution triangle of numbers obtained from A146559.
1, 0, 1, 0, 1, 1, 0, 0, 2, 1, 0, -2, 1, 3, 1, 0, -4, -4, 3, 4, 1, 0, -4, -12, -5, 6, 5, 1, 0, 0, -16, -24, -4, 10, 6, 1, 0, 8, -4, -42, -39, 0, 15, 7, 1, 0, 16, 32, -24, -88, -55, 8, 21, 8, 1, 0, 16, 80, 72, -80
0,9
Triangle T(n,k) given by (0, 1, -1, 2, 0, 0, 0, 0, 0, 0, 0, ...) DELTA (1, 0, 0, 0, 0, 0, 0, 0, ...) where DELTA is the operator defined in A084938.
G.f. for the k-th column: ((x-x^2)/(1-2*x+2*x^2))^k.
G.f.: (1-2*x+2*x^2)/(1-2*x+2*x^2-x*y+x^2*y).
T(n,k) = 2*T(n-1,k) + T(n-1,k-1) - 2*T(n-2,k) - T(n-2,k-1), T(0,0)=1, T(1,0) = T(1,1) = 1, T(n,k) = 0 if k<0 or if k>n.
T(n,k) = (-1)^(n-k)*A181472(n-1,k-1) for n>0 and k>0.
T(n,1) = A146559(n-1).
T(n+1,n) = n = A001477(n).
T(n+2,n) = (n^2-n)/2 = A161680(n).
Sum_{k, 0<=k<=n} T(n,k) = A057682(n) for n>0.
Triangle begins:
1
0, 1
0, 1, 1
0, 0, 2, 1
0, -2, 1, 3, 1
0, -4, -4, 3, 4, 1
0, -4, -12, -5, 6, 5, 1
0, 0, -16, -24, -4, 10, 6, 1
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DELEHAM Philippe, Feb 20 2013
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