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 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_(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 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/(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 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

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Last modified April 19 22:40 EDT 2021. Contains 343117 sequences. (Running on oeis4.)