A129184 Shift operator, right.

0, 1, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0

Offset: 1

Author

Gary W. Adamson, Apr 01 2007

Keywords

• nonn
• tabl
• less
• easy

Comments

Let A129184 = matrix M, then M*V, (V a vector); shifts V to the right, preceded by zeros. Example: M*V, V = [1, 2, 3, ...] = [0, 1, 2, 3, ...]. A129185 = left shift operator.

Given a polynomial sequence p_n(x) with p_0(x)=1 and the lowering and raising operators L and R defined by L P_n(x)= n * P_(n-1)(x) and R P_n(x)= P_(n+1)(x), the matrix T represents the action of R in the p_n(x) basis. For p_n(x) = x^n, L = D = d/dx and R = x. For p_n(x)= x^n/n!, L= DxD and R=D^(-1). - Tom Copeland, Nov 10 2012

Examples

			First few rows of the triangle:
  0;
  1, 0;
  0, 1, 0;
  0, 0, 1, 0;
  0, 0, 0, 1, 0;
  ...
		

Links

• Andrew Howroyd, Table of n, a(n) for n = 1..1275

Crossrefs

Cf. A129185, A129186.

Cf. A010054.

Formula

Infinite lower triangular matrix with all 1's in the subdiagonal and the rest zeros.

From Tom Copeland, Nov 10 2012: (Start)

Let M(t) = I/(I-t*T) = I + t*T + (t*T)^2 + ... where T is the shift operator matrix and I the Identity matrix. Then the inverse matrix is MI(t)=(I-tT) and M(t) is A000012 with each n-th diagonal multiplied by t^n. M(1)=A000012 with inverse MI(1)=A167374. Row sums of M(2), M(3), and M(4) are A000225, A003462, and A002450.

Let E(t)=exp(t*T) with inverse E(-t). Then E(t) is A000012 with each n-th diagonal multiplied by t^n/n! and each row represents e^t truncated at the n+1 term.

The matrix operation b = T*a can be characterized in several ways in terms of the coefficients a(n) and b(n), their o.g.f.s A(x) and B(x), or e.g.f.s EA(x) and EB(x):

1) b(0) = 0, b(n) = a(n-1),

2) B(x) = x A(x), or

3) EB(x) = D^(-1) EA(x), where D^(-1)x^j/j! = x^(j+1)/(j+1)!.

The operator M(t) can be characterized as

4)M(t)EA(x)= sum(n>=0)a(n)[e^(x*t)-[1+x*t+...+ (x*t)^(n-1)/(n-1)!]]/t^n

= exp(a*D_y)[t*e^(x*t)-y*e(x*y)]/(t-y)

= [t*e^(x*t)-a*e(x*a)]/(t-a), umbrally where (a)^k=a_k,

5)[M(t) * a]_n = a(0)t^n +a(1)t^(n-1)+a(2)t^(n-2)+...+a(n).

The exponentiated operator can be characterized as

6) E(t) A(x) = exp(t*x) A(x),

7) E(t) EA(x) = exp(t*D^(-1)) EA(x)

8) [E(t) * a]_n = a(0)t^n/n! + a(1)t^(n-1)/(n-1)! + ... + a(n).

(End)

a(n) = A010054(n+1). - Andrew Howroyd, Feb 02 2020

Extensions

Terms a(46) and beyond from Andrew Howroyd, Feb 02 2020