OFFSET
0,5
COMMENTS
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, May 04 2006
Riordan array ( 1/(1 - x), -x/(1 - 2*x) ) The matrix square is the Riordan array ( (1 - 2*x)/(1 - x)^2, x ), which belongs to the Appell subgroup of the Riordan group. See the Example section below. - Peter Bala, Jul 17 2013
LINKS
M. Shattuck and T. Waldhauser, Proofs of some binomial identities using the method of last squares, Fib. Q., 48 (2010), 290-297.
FORMULA
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, Aug 05 2006
O.g.f.: (1 - 2*t)/(1 - t) * 1/(1 + t*(x - 2)) = 1 + (1 - x)*t + (1 - 3*x + x^2)*t^2 + (1 - 7*x + 5*x^2 - x^3)*t^3 + .... - Peter Bala, Jul 17 2013
From Tom Copeland, Nov 17 2016: (Start)
EXAMPLE
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!.
MATHEMATICA
Table[(1 + (-1)^k*2^(n - k + 1)*Sum[ Binomial[n - 2 j - 2, k - 2 j - 1], {j, 0, Floor[k/2]}]) - 4 Boole[And[n == 1, k == 0]], {n, 0, 10}, {k, 0, n}] // Flatten (* Michael De Vlieger, Nov 24 2016 *)
PROG
(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)))}
CROSSREFS
KEYWORD
sign,tabl
AUTHOR
Paul D. Hanna, May 02 2006
STATUS
approved