0,4

The matrix square, T^2, consists of columns that are all the same. Matrix inverse is triangle A118801. Row sums form {0^n, n>=0}. Unsigned row sums equal A025192(n) = 2*3^(n-1), n>=1. Row squared sums equal A051708. Antidiagonal sums equals all 1's. Unsigned antidiagonal sums form A078057 (with offset). Antidiagonal squared sums form A002002(n) = Sum_{k=0..n-1} C(n,k+1)*C(n+k,k), n>=1.

T(n,k) = (-1)^k * 2^(n-k) * ( C(n,k) + C(n-1,k-1) )/2 for n>=k>=0 with T(0,0) = 1. Antidiagonals form the coefficients of Chebyshev polynomials: T(n,k) = [x^(2*n)] [(1+sqrt(1-x^2))^(n+k) + (1-sqrt(1-x^2))^(n+k)]/2.

Rows of the triangle are generated by taking successive iterates of (A135387)^n * [1, 1, 0, 0, 0,...]. - Gary W. Adamson (qntmpkt(AT)yahoo.com), Dec 09 2007

Triangle begins:

1;

1,-1;

2,-3,1;

4,-8,5,-1;

8,-20,18,-7,1;

16,-48,56,-32,9,-1;

32,-112,160,-120,50,-11,1;

64,-256,432,-400,220,-72,13,-1;

128,-576,1120,-1232,840,-364,98,-15,1;

256,-1280,2816,-3584,2912,-1568,560,-128,17,-1;

512,-2816,6912,-9984,9408,-6048,2688,-816,162,-19,1; ...

The matrix square, T^2, equals:

1;

0,1;

1,0,1;

2,1,0,1;

4,2,1,0,1;

8,4,2,1,0,1;

16,8,4,2,1,0,1; ...

where all columns are the same.

(PARI) {T(n, k)=if(n==0&k==0, 1, (-1)^k*2^(n-k)*(binomial(n, k)+binomial(n-1, k-1))/2)} (PARI) /* Chebyshev Polynomials as Antidiagonals: */ {T(n, k)=local(Ox=x*O(x^(2*k))); polcoeff(((1+sqrt(1-x^2+Ox))^(n+k)+(1-sqrt(1-x^2+Ox))^(n+k))/2, 2*k, x)}

Cf. A118801 (inverse), A025192 (unsigned row sums), A051708 (row squared sums), A078057 (unsigned antidiagonal sums), A002002 (antidiagonal squared sums).

Cf. A135387.

Sequence in context: A129566 A073809 A021436 this_sequence A075297 A057597 A121340

Adjacent sequences: A118797 A118798 A118799 this_sequence A118801 A118802 A118803

sign,tabl

Paul D. Hanna (pauldhanna(AT)juno.com), May 02 2006