A113310 Riordan array ((1+x)/(1-x),x/(1+x)).

1, 2, 1, 2, 1, 1, 2, 1, 0, 1, 2, 1, 1, -1, 1, 2, 1, 0, 2, -2, 1, 2, 1, 1, -2, 4, -3, 1, 2, 1, 0, 3, -6, 7, -4, 1, 2, 1, 1, -3, 9, -13, 11, -5, 1, 2, 1, 0, 4, -12, 22, -24, 16, -6, 1, 2, 1, 1, -4, 16, -34, 46, -40, 22, -7, 1, 2, 1, 0, 5, -20, 50, -80, 86, -62, 29, -8, 1, 2, 1, 1, -5, 25, -70, 130, -166, 148, -91, 37, -9, 1

Offset

0,2

Comments

Row sums are A113311. Diagonal sums are A113312. Inverse is A113313. The family of Riordan arrays ((1+x)/(1-(q-1)x),x/(1+x)) allow one to calculate the weight distribution of MDS codes.

References

F.J. MacWilliams, N. J. A. Sloane, The Theory of Error-Correcting Codes, North-Holland, 2003, p. 321.

Links

Table of n, a(n) for n=0..90.

Formula

T(n, k)=sum{j=0..n-k, (-1)^j*C(j+k-2, i)}; T(n, k)=sum{j=0..n-k, (-1)^(n-k-j)C(n-j-2, n-j-k); T(n, k)=sum{j=k..n, (-1)^(n-j)*C(n, j)(2^(j-k+1)-1).

Example

Triangle begins
1;
2,1;
2,1,1;
2,1,0,1;
2,1,1,-1,1;
2,1,0,2,-2,1;

Mathematica

T[n_, k_] := Sum[(-1)^(n-j) Binomial[n, j] (2^(j-k+1) - 1), {j, k, n}]; Table[T[n, k], {n, 0, 12}, {k, 0, n}] // Flatten (* Jean-François Alcover, Apr 27 2017 *)

Crossrefs

Sequence in context: A351819 A076348 A263835 * A359537 A081653 A249615

Adjacent sequences: A113307 A113308 A113309 * A113311 A113312 A113313

Keyword

easy,sign,tabl

Author

Paul Barry, Oct 25 2005

Status

approved