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%I #22 Oct 13 2022 04:55:41
%S 1,-1,1,0,-1,1,1,0,-2,1,-1,2,0,-2,1,-1,-3,6,0,-3,1,4,-3,-7,8,0,-3,1,
%T -1,18,-18,-13,17,0,-4,1,-19,-4,56,-28,-22,20,0,-4,1,31,-127,60,136,
%U -98,-34,36,0,-5,1,120,163,-511,80,288,-126,-50,40,0,-5,1
%N Matrix inverse of Losanitsch's triangle A034851.
%C Also the adjoint matrix of Losanitsch's triangle, since the determinants of n X n subarrays, rooted at (0,0), of Losanitsch's triangle are 1 and since for a matrix A, A^(-1) = 1/det(A) * (adjoint of A). - _Gerald McGarvey_, Oct 30 2007
%H Michel Marcus, <a href="/A055138/b055138.txt">Rows n=0..50 of triangle, flattened</a>
%e Triangle begins:
%e 1;
%e -1, 1;
%e 0,-1, 1;
%e 1, 0,-2, 1;
%e -1, 2, 0,-2, 1;
%e ...
%p b:= proc(n, k) (binomial(irem(n, 2, 'i'), irem(k, 2, 'j'))*
%p binomial(i, j)+binomial(n, k))/2 end:
%p T:= n-> (M-> seq(M[n+1, j], j=1..n+1))(Matrix(n+1,
%p (i, j)-> b(i-1, j-1))^(-1)):
%p seq(T(n), n=0..10); # _Alois P. Heinz_, Mar 01 2022
%t nmax = 10;
%t b[n_?EvenQ, k_?OddQ] := Binomial[n, k]/2;
%t b[n_, k_] := (Binomial[n, k] + Binomial[Quotient[n, 2], Quotient[k, 2]])/2;
%t M = Table[b[n, k], {n, 0, nmax}, {k, 0, nmax}] // Inverse;
%t T[n_, k_] := M[[n+1, k+1]];
%t Table[T[n, k], {n, 0, nmax}, {k, 0, n}] // Flatten (* _Jean-François Alcover_, Oct 13 2022 *)
%o (PARI) T(n, k) = (1/2)*(binomial(n, k) + binomial(n%2, k%2) * binomial(n\2, k\2)); \\ A034851
%o row(n) = my(m=matrix(n+1, n+1, i, j, i--; j--; T(i, j))); vector(n+1, i, (1/m)[n+1,i]); \\ _Michel Marcus_, Mar 01 2022
%Y Cf. A034851.
%K sign,tabl
%O 0,9
%A _Christian G. Bower_, May 09 2000
%E Typo in definition corrected by _Georg Fischer_, Mar 01 2022
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