%I #37 Jul 23 2024 10:49:26
%S 1,1,-1,1,-3,1,1,-7,5,-1,1,-15,17,-7,1,1,-31,49,-31,9,-1,1,-63,129,
%T -111,49,-11,1,1,-127,321,-351,209,-71,13,-1,1,-255,769,-1023,769,
%U -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
%N Triangle T that satisfies the matrix products: C*[T^-1]*C = T and T*[C^-1]*T = C, where C is Pascal's triangle.
%C 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).
%C 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
%C 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
%H M. Shattuck and T. Waldhauser, <a href="http://www.fq.math.ca/Papers1/48-4/Shattuck_Waldhauser.pdf">Proofs of some binomial identities using the method of last squares</a>, Fib. Q., 48 (2010), 290-297.
%F 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.
%F For k>0, T(n,k) = -T(n-1,k-1) + 2*T(n-1,k). - _Gerald McGarvey_, Aug 05 2006
%F 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
%F From _Tom Copeland_, Nov 17 2016: (Start)
%F Let M = A200139^(-1) = (unsigned A118800)^(-1) and NpdP be the signed padded Pascal matrix defined in A097805. Then T(n,k) = (-1)^n* M(n,k) and T = P*NpdP = (A239473)^(-1)*P^(-1) = P*A167374*P^(-1) = A156644*P^(-1), where P is the Pascal matrix A007318 with inverse A130595. Cf. A112857.
%F Signed P^2 = signed A032807 = T*A167374. (End)
%e Formulas for initial columns are, for n>=0:
%e T(n+1,1) = 1 - 2^(n+1);
%e T(n+2,2) = 1 + 2^(n+1)*n;
%e T(n+3,3) = 1 - 2^(n+1)*(n*(n+1)/2 + 1);
%e T(n+4,4) = 1 + 2^(n+1)*(n*(n+1)*(n+2)/6 + n);
%e T(n+5,5) = 1 - 2^(n+1)*(n*(n+1)*(n+2)*(n+3)/24 + n*(n+1)/2 + 1).
%e Triangle begins:
%e 1;
%e 1,-1;
%e 1,-3,1;
%e 1,-7,5,-1;
%e 1,-15,17,-7,1;
%e 1,-31,49,-31,9,-1;
%e 1,-63,129,-111,49,-11,1;
%e 1,-127,321,-351,209,-71,13,-1;
%e 1,-255,769,-1023,769,-351,97,-15,1;
%e 1,-511,1793,-2815,2561,-1471,545,-127,17,-1;
%e 1,-1023,4097,-7423,7937,-5503,2561,-799,161,-19,1; ...
%e The matrix square, T^2, starts:
%e 1;
%e 0,1;
%e -1,0,1;
%e -2,-1,0,1;
%e -3,-2,-1,0,1;
%e -4,-3,-2,-1,0,1; ...
%e where all columns are the same.
%e The matrix product C^-1*T*C = T^-1*C^2 is:
%e 1;
%e -1,-1;
%e 0, 1, 1;
%e 0, 0,-1,-1;
%e 0, 0, 0, 1, 1; ...
%e where C(n,k) = n!/(n-k)!/k!.
%t 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 *)
%o (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)))}
%Y Cf. A118800 (inverse), A007051 (unsigned row sums), A118802 (Row squared sums), A080956 (antidiagonal sums), A024537 (unsigned antidiagonal sums).
%Y A145661, A119258 and A118801 are all essentially the same (see the Shattuck and Waldhauser paper). - Tamas Waldhauser, Jul 25 2011
%Y Cf. A007318, A032807, A097805, A112857, A130595, A156644, A167374, A200139, A239473.
%K sign,tabl
%O 0,5
%A _Paul D. Hanna_, May 02 2006