

A129184


Shift operator, right.


7



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
(list;
table;
graph;
refs;
listen;
history;
text;
internal format)



OFFSET

1,1


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_(n1)(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


LINKS

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


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/(It*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)=(ItT) and M(t) is A000012 with each nth 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 nth 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(n1),
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)^(n1)/(n1)!]]/t^n
= exp(a*D_y)[t*e^(x*t)y*e(x*y)]/(ty) <evaluated at y=0>
= [t*e^(x*t)a*e(x*a)]/(ta), umbrally where (a)^k=a_k,
5)[M(t) * a]_n = a(0)t^n +a(1)t^(n1)+a(2)t^(n2)+...+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^(n1)/(n1)! + ... + a(n).
(End)
a(n) = A010054(n+1).  Andrew Howroyd, Feb 02 2020


EXAMPLE

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


CROSSREFS

Cf. A129185, A129186.
Cf. A010054.
Sequence in context: A068716 A188020 A179828 * A129185 A283683 A118605
Adjacent sequences: A129181 A129182 A129183 * A129185 A129186 A129187


KEYWORD

nonn,tabl,less,easy


AUTHOR

Gary W. Adamson, Apr 01 2007


EXTENSIONS

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


STATUS

approved



