0,5

Matrix inverse is triangle A118800. Row sums are: (1-n). Unsigned row sums equal A007051(n) = (3^n + 1)/2. Row squared sums equal A118802. Antidiagonal sums equal A080956(n) = (n+1)(2-n)/2. Unsigned antidiagonal sums form A024537 (with offset).

T = C^2*D^-1 where matrix product D = C^-1*T*C = T^-1*C^2 has only 2 nonzero diagonals: D(n,n)=-D(n+1,n)=(-1)^n, with zeros elsewhere. Also, [B^-1]*T*[B^-1] = B*[T^-1]*B forms a self-inverse matrix, where B^2 = C and B(n,k) = C(n,k)/2^(n-k). - Paul D. Hanna (pauldhanna(AT)juno.com), May 04 2006

T(n,k) = 1 + (-1)^k*2^(n-k+1)*Sum_{j=0..[k/2]} C(n-2j-2,k-2j-1) for n>=k>=0 with T(0,0) = 1.

For k>0, T(n,k) = -T(n-1,k-1) + 2*T(n-1,k). - Gerald McGarvey (gerald.mcgarvey(AT)comcast.net), Aug 05 2006

Formulas for initial columns are, for n>=0:

T(n+1,1) = 1 - 2^(n+1);

T(n+2,2) = 1 + 2^(n+1)*n;

T(n+3,3) = 1 - 2^(n+1)*(n*(n+1)/2 + 1);

T(n+4,4) = 1 + 2^(n+1)*(n*(n+1)*(n+2)/6 + n);

T(n+5,5) = 1 - 2^(n+1)*(n*(n+1)*(n+2)*(n+3)/24 + n*(n+1)/2 + 1).

Triangle begins:

1;

1,-1;

1,-3,1;

1,-7,5,-1;

1,-15,17,-7,1;

1,-31,49,-31,9,-1;

1,-63,129,-111,49,-11,1;

1,-127,321,-351,209,-71,13,-1;

1,-255,769,-1023,769,-351,97,-15,1;

1,-511,1793,-2815,2561,-1471,545,-127,17,-1;

1,-1023,4097,-7423,7937,-5503,2561,-799,161,-19,1; ...

The matrix square, T^2, starts:

1;

0,1;

-1,0,1;

-2,-1,0,1;

-3,-2,-1,0,1;

-4,-3,-2,-1,0,1; ...

where all columns are the same.

The matrix product C^-1*T*C = T^-1*C^2 is:

1;

-1,-1;

0, 1, 1;

0, 0,-1,-1;

0, 0, 0, 1, 1; ...

where C(n,k) = n!/(n-k)!/k!.

(PARI) {T(n, k)=if(n==0&k==0, 1, 1+(-1)^k*2^(n-k+1)*sum(j=0, k\2, binomial(n-2*j-2, k-2*j-1)))}

Cf. A118800 (inverse), A007051 (unsigned row sums), A118802 (Row squared sums), A080956 (antidiagonal sums), A024537 (unsigned antidiagonal sums).

Sequence in context: A136621 A108625 A112857 this_sequence A080936 A094507 A065625

Adjacent sequences: A118798 A118799 A118800 this_sequence A118802 A118803 A118804

sign,tabl

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