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%I #17 Feb 12 2024 08:17:17
%S 1,0,3,0,-3,7,0,3,-7,15,0,0,0,-15,31,0,-3,7,0,-31,63,0,0,0,0,0,-63,
%T 127,0,3,-7,15,0,0,-127,255,0,0,0,0,0,0,0,-255,511,0,0,0,-15,31,0,0,0,
%U -511,1023,0,0,0,0,0,0,0,0,0,-1023,2047
%N Inverse of number triangle A(n,k) = 1/(2*2^n-1) if k <= n <= 2k, 0 otherwise.
%C Row sums are A127804.
%e Triangle begins
%e 1;
%e 0, 3;
%e 0, -3, 7;
%e 0, 3, -7, 15;
%e 0, 0, 0, -15, 31;
%e 0, -3, 7, 0, -31, 63;
%e 0, 0, 0, 0, 0, -63, 127;
%e 0, 3, -7, 15, 0, 0, -127, 255;
%e 0, 0, 0, 0, 0, 0, 0, -255, 511;
%e 0, 0, 0, -15, 31, 0, 0, 0, -511, 1023;
%e 0, 0, 0, 0, 0, 0, 0, 0, 0, -1023, 2047;
%e ...
%e Inverse of
%e 1;
%e 0, 1/3;
%e 0, 1/7, 1/7;
%e 0, 0, 1/15, 1/15;
%e 0, 0, 1/31, 1/31, 1/31;
%e 0, 0, 0, 1/63, 1/63, 1/63;
%e 0, 0, 0, 1/127, 1/127, 1/127, 1/127;
%e 0, 0, 0, 0, 1/255, 1/255, 1/255, 1/255;
%e 0, 0, 0, 0, 1/511, 1/511, 1/511, 1/511, 1/511;
%e 0, 0, 0, 0, 0, 1/1023, 1/1023, 1/1023, 1/1023, 1/1023;
%e 0, 0, 0, 0, 0, 1/2047, 1/2047, 1/2047, 1/2047, 1/2047, 1/2047;
%e ...
%p A127803 := proc(n,k)
%p A := Matrix(n+1,n+1) ;
%p for r from 0 to n do
%p for c from 0 to n do
%p if c <= r and r <= 2*c then
%p A[r+1,c+1] := 1/(2*2^r-1) ;
%p else
%p A[r+1,c+1] := 0 ;
%p end if;
%p end do:
%p end do:
%p Ainv := LinearAlgebra[MatrixInverse](A) ;
%p Ainv[n+1,k+1] ;
%p end proc:
%p seq(seq( A127803(n,k),k=0..n),n=0..12) ; # _R. J. Mathar_, Feb 12 2024
%t rows = 11;
%t A[n_, k_] := If[k <= n, If[n <= 2 k, 1/(2*2^n - 1), 0], 0];
%t T = Table[A[n, k], {n, 0, rows-1}, {k, 0, rows-1}] // Inverse;
%t Table[T[[n, k]], {n, 1, rows}, {k, 1, n}] // Flatten (* _Jean-François Alcover_, Jul 03 2018 *)
%o (PARI) B(n,k) = if(k<=n,if(n<=2*k,1/(2*2^n-1),0),0);
%o lista(nn) = {my(m = matrix(nn, nn, n, k, B(n-1,k-1))^(-1)); for (n=1, nn, for (k=1, n, print1(m[n,k], ", ");); print(););} \\ _Michel Marcus_, Jul 03 2018
%Y Cf. A127804.
%K sign,tabl
%O 0,3
%A _Paul Barry_, Jan 29 2007