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A127803
Inverse of number triangle A(n,k) = 1/(2*2^n-1) if k <= n <= 2k, 0 otherwise.
2
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, 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, -511, 1023, 0, 0, 0, 0, 0, 0, 0, 0, 0, -1023, 2047
OFFSET
0,3
COMMENTS
Row sums are A127804.
EXAMPLE
Triangle begins
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, 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, -511, 1023;
0, 0, 0, 0, 0, 0, 0, 0, 0, -1023, 2047;
...
Inverse of
1;
0, 1/3;
0, 1/7, 1/7;
0, 0, 1/15, 1/15;
0, 0, 1/31, 1/31, 1/31;
0, 0, 0, 1/63, 1/63, 1/63;
0, 0, 0, 1/127, 1/127, 1/127, 1/127;
0, 0, 0, 0, 1/255, 1/255, 1/255, 1/255;
0, 0, 0, 0, 1/511, 1/511, 1/511, 1/511, 1/511;
0, 0, 0, 0, 0, 1/1023, 1/1023, 1/1023, 1/1023, 1/1023;
0, 0, 0, 0, 0, 1/2047, 1/2047, 1/2047, 1/2047, 1/2047, 1/2047;
...
MAPLE
A127803 := proc(n, k)
A := Matrix(n+1, n+1) ;
for r from 0 to n do
for c from 0 to n do
if c <= r and r <= 2*c then
A[r+1, c+1] := 1/(2*2^r-1) ;
else
A[r+1, c+1] := 0 ;
end if;
end do:
end do:
Ainv := LinearAlgebra[MatrixInverse](A) ;
Ainv[n+1, k+1] ;
end proc:
seq(seq( A127803(n, k), k=0..n), n=0..12) ; # R. J. Mathar, Feb 12 2024
MATHEMATICA
rows = 11;
A[n_, k_] := If[k <= n, If[n <= 2 k, 1/(2*2^n - 1), 0], 0];
T = Table[A[n, k], {n, 0, rows-1}, {k, 0, rows-1}] // Inverse;
Table[T[[n, k]], {n, 1, rows}, {k, 1, n}] // Flatten (* Jean-François Alcover, Jul 03 2018 *)
PROG
(PARI) B(n, k) = if(k<=n, if(n<=2*k, 1/(2*2^n-1), 0), 0);
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
CROSSREFS
Cf. A127804.
Sequence in context: A165012 A198126 A074694 * A021771 A154853 A139214
KEYWORD
sign,tabl
AUTHOR
Paul Barry, Jan 29 2007
STATUS
approved