|
%I #9 Jun 05 2018 22:33:11
%S 1,-1,1,-1,-1,1,0,-1,-1,1,0,0,-1,-1,1,0,0,0,-1,-1,1,0,0,0,0,-1,-1,1,0,
%T 0,0,0,0,-1,-1,1,0,0,0,0,0,0,-1,-1,1,0,0,0,0,0,0,0,-1,-1,1,0,0,0,0,0,
%U 0,0,0,-1,-1,1
%N Triangle of coefficients of the inverse of A058071.
%C The row sums are {1, 0, -1, -1, -1, -1, -1, -1, -1, -1, -1, ...}.
%C The inverse is a tridiagonal lower triangular matrix.
%H G. C. Greubel, <a href="/A141679/b141679.txt">Rows n=1..100 of triangle, flattened</a>
%F A058071(n,m) = if(m <= n, Fibonacci(n - m + 1)*Fibonacci(m + 1), 0), t(n,m) = Fibonacci(n)*Inverse(A058071(n,m)).
%e {1},
%e {-1, 1},
%e {-1, -1, 1},
%e {0, -1, -1, 1},
%e {0, 0, -1, -1, 1},
%e {0, 0,0, -1, -1, 1},
%e {0, 0, 0, 0, -1, -1, 1},
%e {0, 0, 0, 0, 0, -1, -1, 1},
%e {0, 0, 0, 0, 0, 0, -1, -1, 1},
%e {0, 0, 0, 0, 0, 0, 0, -1, -1, 1},
%e {0, 0, 0, 0, 0, 0, 0, 0, -1, -1, 1}
%t Clear[t, n, m, M] (*A058071*) t[n_, m_] = If[m <= n, Fibonacci[n - m + 1]*Fibonacci[m + 1], 0]; Table[Table[t[n, m], {m, 0, n}], {n, 0, 10}]; Flatten[%]; M = Inverse[Table[Table[t[n, m], {m, 0, 10}], {n, 0, 10}]]; Table[Table[Fibonacci[n]*M[[n, m]], {m, 1, n}], {n, 1, 11}]; Flatten[%]
%Y Cf. A058071.
%Y As a sequence, quite similar to A136705. - _N. J. A. Sloane_, Dec 14 2014
%K tabl,sign
%O 1,1
%A _Roger L. Bagula_ and _Gary W. Adamson_, Sep 07 2008
%E Edited by _N. J. A. Sloane_, Jan 05 2009
|
