cp's OEIS Frontend

A129184 Shift operator, right.

%I A129184 #23 Feb 18 2022 22:35:43
%S A129184 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,
%T A129184 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,
%U A129184 0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,0,1,0
%N A129184 Shift operator, right.
%C A129184 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.
%C A129184 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
%H A129184 Andrew Howroyd, <a href="/A129184/b129184.txt">Table of n, a(n) for n = 1..1275</a>
%F A129184 Infinite lower triangular matrix with all 1's in the subdiagonal and the rest zeros.
%F A129184 From _Tom Copeland_, Nov 10 2012: (Start)
%F A129184 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.
%F A129184 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.
%F A129184 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):
%F A129184   1) b(0) = 0, b(n) = a(n-1),
%F A129184   2) B(x) = x A(x), or
%F A129184   3) EB(x) = D^(-1)  EA(x), where D^(-1)x^j/j! = x^(j+1)/(j+1)!.
%F A129184 The operator M(t) can be characterized as
%F A129184   4)M(t)EA(x)= sum(n>=0)a(n)[e^(x*t)-[1+x*t+...+ (x*t)^(n-1)/(n-1)!]]/t^n
%F A129184     = exp(a*D_y)[t*e^(x*t)-y*e(x*y)]/(t-y) <evaluated at y=0>
%F A129184     = [t*e^(x*t)-a*e(x*a)]/(t-a), umbrally where (a)^k=a_k,
%F A129184   5)[M(t) * a]_n = a(0)t^n +a(1)t^(n-1)+a(2)t^(n-2)+...+a(n).
%F A129184 The exponentiated operator can be characterized as
%F A129184   6) E(t) A(x) = exp(t*x) A(x),
%F A129184   7) E(t) EA(x) = exp(t*D^(-1)) EA(x)
%F A129184   8) [E(t) * a]_n = a(0)t^n/n! + a(1)t^(n-1)/(n-1)! + ... + a(n).
%F A129184   (End)
%F A129184 a(n) = A010054(n+1). - _Andrew Howroyd_, Feb 02 2020
%e A129184 First few rows of the triangle:
%e A129184   0;
%e A129184   1, 0;
%e A129184   0, 1, 0;
%e A129184   0, 0, 1, 0;
%e A129184   0, 0, 0, 1, 0;
%e A129184   ...
%Y A129184 Cf. A129185, A129186.
%Y A129184 Cf. A010054.
%K A129184 nonn,tabl,less,easy
%O A129184 1,1
%A A129184 _Gary W. Adamson_, Apr 01 2007
%E A129184 Terms a(46) and beyond from _Andrew Howroyd_, Feb 02 2020