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%I #10 Feb 23 2024 04:25:25
%S 1,-2,1,-1,-1,1,0,-1,-1,1,-1,-1,0,-1,1,1,0,-2,0,-1,1,-1,-1,0,-1,0,-1,
%T 1,0,0,0,-1,-1,0,-1,1,0,0,-1,-1,0,-1,0,-1,1,1,0,-1,1,-2,0,-1,0,-1,1,
%U -1,-1,0,-1,0,-1,0,-1,0,-1,1,0,1,1,-1,0,-1,-1,0,-1,0,-1,1,-1,-1,0,-1,0,-1,0,-1,0,-1,0,-1,1
%N Matrix inverse of A178534.
%C Except for first term row sums equal a signed version of A023022.
%e Table begins:
%e 1
%e -2 1
%e -1 -1 1
%e 0 -1 -1 1
%e -1 -1 0 -1 1
%e 1 0 -2 0 -1 1
%e -1 -1 0 -1 0 -1 1
%e 0 0 0 -1 -1 0 -1 1
%e 0 0 -1 -1 0 -1 0 -1 1
%e 1 0 -1 1 -2 0 -1 0 -1 1
%e -1 -1 0 -1 0 -1 0 -1 0 -1 1
%p A178535 := proc(n, l)
%p option remember;
%p a := 0 ;
%p if n = l then
%p a := 1 ;
%p end if;
%p for k from l to n-1 do
%p a := a-A178534(n, k)*procname(k, l) ;
%p end do:
%p a/A178534(n, n) ;
%p end proc:
%p seq(seq(A178535(n,k),k=1..n),n=1..12) ; # _R. J. Mathar_, Oct 28 2010
%t nmax = 13;
%t (* T is A178534 *)
%t T[n_, 1] := Fibonacci[n+1];
%t T[n_, k_] := T[n, k] = If[k > n, 0, Sum[T[n-i, k-1], {i, 1, k-1}] - Sum[T[n-i, k], {i, 1, k-1}]];
%t A178535 = Inverse[Array[T, {nmax, nmax}]];
%t Table[A178535[[n, k]], {n, 1, nmax}, {k, 1, n}] // Flatten (* _Jean-François Alcover_, Feb 23 2024 *)
%Y Cf. First column is A178536.
%K sign,tabl
%O 1,2
%A _Mats Granvik_, May 29 2010
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