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Two sequence determinant triangle sequence: a(n)=A000045(n); b(n)=b(n-1)+b(n-2)+b(n-3) :2 start;A141036(n); t(n,m)=t(n,m)=a(m)*b(n)-b(m)*a(n).
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%I #3 Oct 12 2012 14:54:55

%S -2,-2,0,-4,2,2,-6,3,3,0,-10,6,6,2,3,-16,13,13,10,15,17,-26,25,25,24,

%T 36,47,31,-42,49,49,56,84,119,119,112,-68,95,95,122,183,271,318,385,

%U 329,-110,182,182,254,381,580,741,991,1127,963

%N Two sequence determinant triangle sequence: a(n)=A000045(n); b(n)=b(n-1)+b(n-2)+b(n-3) :2 start;A141036(n); t(n,m)=t(n,m)=a(m)*b(n)-b(m)*a(n).

%C Row sums are:{-2, -2, 0, 0, 7, 52, 162, 546, 1730, 5291}.

%C Reasoning behind the sequence is:

%C Suppose we have n affine transforms that form a group:

%C g={ a(1)*x+b(1),a(2)*x+b(2),...,a(n)*x+b(n)}

%C on the sequences a(n) and b(n).

%C We form rational projections as Moebius / bilinear transforms:

%C g(projection)={( a(1)*x+b(1))/(a(n)*x+b(n)),( a(2)*x+b(2))/(a(n)*x+b(n)),...,( a(n-1)*x+b(n-1))/(a(n)*x+b(n))

%C With determinants:

%C g_det={a(1)*b(n)-b(1)*a(n),a(2)*b(n)-b(2)*a(n),...,a(n-1)*b(n)-b(n-1)*a(n)}

%C So that we have the triangular sequences:

%C t(n,m)=a(m)*b(n)-b(m)*a(n)

%F a(n)=A000045(n); b(n)=b(n-1)+b(n-2)+b(n-3) :2 start;A141036; t(n,m)=t(n,m)=a(m)*b(n)-b(m)*a(n).

%e {-2},

%e {-2, 0},

%e {-4, 2, 2},

%e {-6, 3, 3, 0},

%e {-10, 6, 6, 2, 3},

%e {-16, 13, 13, 10, 15, 17},

%e {-26, 25, 25, 24, 36, 47, 31},

%e {-42, 49, 49, 56, 84, 119, 119, 112},

%e {-68, 95, 95, 122, 183, 271, 318, 385, 329},

%e {-110, 182, 182, 254, 381, 580, 741, 991, 1127, 963}

%t Clear[a, b, t, n, m] a[n_] := Fibonacci[n]; b[0] = 2; b[1] = 1; b[2] = 1; b[n_] := b[n] = b[n - 1] + b[n - 2] + b[n - 3]; t[n_, m_] := a[m]*b[n] - b[m]*a[n]; Table[Table[t[n, m], {m, 0, n - 1}], {n, 1, 10}]; Flatten[%]

%Y Cf. A141036, A000045.

%K uned,sign

%O 1,1

%A _Roger L. Bagula_ and _Gary W. Adamson_, Oct 07 2008