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A129184
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Shift operator, right.
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7
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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
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OFFSET
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1,1
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COMMENTS
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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
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LINKS
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FORMULA
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Infinite lower triangular matrix with all 1's in the subdiagonal and the rest zeros.
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) <evaluated at y=0>
= [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)
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EXAMPLE
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First few rows of the triangle:
0;
1, 0;
0, 1, 0;
0, 0, 1, 0;
0, 0, 0, 1, 0;
...
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CROSSREFS
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KEYWORD
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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