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.

From Paul Barry, Nov 10 2008: (Start)

T is [1,1,0,0,0,...] DELTA [ -1,0,0,0,0,...] or C(1,n) DELTA -C(0,n). (DELTA defined in A084938).

The positive matrix T_p is [1,1,0,0,0,...] DELTA [1,0,0,0,0,...]. T_p*C^-1 is

[0,1,0,0,0,....] DELTA [1,0,0,0,0,...] which is C(n-1,k-1) for n,k>=1. (End)

The triangle formed by deleting the minus signs is the mirror of the self-fusion of Pascal's triangle; see Comments at A081277 and A193722. - Clark Kimberling, Aug 04 2011

Riordan array ( (1 - x)/(1 - 2*x), -x/(1 - 2*x) ). Cf. A209149. The matrix square is the Riordan array ( (1 - x)^2/(1 - 2*x), x ), which belongs to the Appell subgroup of the Riordan group. See the Example section below. - Peter Bala, Jul 17 2013

From Peter Bala, Feb 23 2019: (Start)

There is a 1-parameter family of solutions to the simultaneous equations C*T*C = T^-1 and T*C*T = C^-1, where C is Pascal's triangle. Let T(k) denote the Riordan array ( (1 - k*x)/(1 - (k + 1)*x), -x/(1 - (k + 1)*x) ) so that T(1) = T. Then C*T(k)*C = T(k)^-1 and T(k)*C*T(k) = C^-1, for arbitrary k. For arbitrary m, the Riordan arrays (T(k)*C^m)^2 and (C^m*T(k))^2 both belong to the Appell subgroup of the Riordan group.

More generally, given a fixed m, we can ask for a lower triangular array X solving the simultaneous equations (C^m)*X*(C^m) = X^-1 and X*(C^m)*X = C^(-m). A 1-parameter family of solutions is given by the Riordan arrays X = ( (1 - m*k*x)/(1 - m*(k + 1)*x), -x/(1 - m*(k + 1)*x) ). The Riordan arrays X^2 , (X*C^n)^2 and (C^n*X)^2, for arbitrary n, all belong to the Appell subgroup of the Riordan group. (End)

Paul D. Hanna, Rows 0..45 of triangle, flattened.

Florian Luca, Attila Pethő, and László Szalay, Duplications in the k-generalized Fibonacci sequences, New York J. Math. (2021) Vol. 27, 1115-1133.

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, Dec 09 2007

O.g.f.: (1 - t)/(1 + t*(x - 2)) = 1 + (1 - x)*t + (2 - 3*x + x^2)^t^2 + (4 - 8*x + 5*x^2 - x^3)*t^3 + ....  Row polynomial R(n,x) = (1 - x)*(2 - x)^(n-1) for n >= 1. - Peter Bala, Jul 17 2013

T(n,k)=2*T(n-1,k)-T(n-1,k-1) with T(0,0)=T(1,0)=1, T(1,1)=-1, T(n,k)=0 if k<0 or if k>n. - Philippe Deléham, Nov 25 2013

G.f. for row n (n>=1): Sum_{k=0..n} T(n,k)*x^k = (1-x)*(2-x)^(n-1). - Philippe Deléham, Nov 25 2013

From Tom Copeland, Nov 15 2016: (Start)

E.g.f. is [1 + (1-x)e^((2-x)t)]/(2-x), so the row polynomials are p_n(x) = (1-q,(x))^n, umbrally, where (q.(x))^k = q_k(x) are the row polynomials of A239473, or, equivalently, T = M*A239473, where M is the inverse Pascal matrix C^(-1) = A130595 with the odd rows negated, i.e., M(n,k) = (-1)^n C^(-1)(n,k) with e.g.f. exp[(1-x)t]. Cf. A200139: A200139(n,k) = (-1)^k* A118800(n,k).

TCT = C^(-1) = A130595 and A239473 = A000012*C^(-1) = S*C^(-1) imply (M*S)^2 = Identity matrix, i.e., M*S = (M*S)^(-1) = S^(-1)*M^(-1) = A167374*M^(-1). Note that M = M^(-1). Cf. A097805. (End)

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;

  1024, -6144, 16640, -26880, 28800, -21504, 11424, -4320, 1140, -200, 21, -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;

  32, 16,  8,  4,  2,  1,  0,  1;

  64, 32, 16,  8,  4,  2,  1,  0,  1; ...

where all columns are the same.

(* This program generates A118800 as the mirror of the self-fusion of Pascal's triangle. *)

z = 8; a = 1; b = 1; c = 1; d = 1;

p[n_, x_] := (a*x + b)^n ; q[n_, x_] := (c*x + d)^n;

t[n_, k_] := Coefficient[p[n, x], x^k]; t[n_, 0] := p[n, 0];

w[n_, x_] := Sum[t[n, k]*q[n + 1 - k, x], {k, 0, n}]; w[-1, _] = 1;

g[n_] := CoefficientList[w[n, -x], x];

TableForm[Table[Reverse[Abs@g[n]], {n, -1, z}]]

Flatten[Table[Reverse[Abs@g[n]], {n, -1, z}]] (* A081277 *)

TableForm[Table[g[n], {n, -1, z}]]

Flatten[Table[g[n], {n, -1, z}]] (* A118800 *)

(* Clark Kimberling, Aug 04 2011 *)

T[ n_, k_] := If[ n<0 || k<0, 0, (-1)^k 2^(n-k) (Binomial[ n, k] + Binomial[ n-1, n-k)) / 2; (* Michael Somos, Nov 25 2016 *)

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

for(n=0, 12, for(k=0, n, print1(T(n, k), ", ")); print(""))

(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)}

for(n=0, 12, for(k=0, n, print1(T(n, k), ", ")); print(""))

(Sage) # uses[riordan_square from A321620]

# Computes the unsigned triangle.

riordan_square((1-x)/(1-2*x), 8) # Peter Luschny, Jan 03 2019

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

Cf. A135387, A209149, A000012, A097805, A130595, A167374, A200139, A239473, A321620.

Sequence in context: A179738 A187889 A353593 * A200139 A321621 A321629

Adjacent sequences:  A118797 A118798 A118799 * A118801 A118802 A118803

sign,tabl

Paul D. Hanna, May 02 2006

approved