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A105794
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Inverse of a generalized Stirling number triangle of first kind.
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4
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1, -1, 1, 1, -1, 1, -1, 1, 0, 1, 1, -1, 1, 2, 1, -1, 1, 0, 5, 5, 1, 1, -1, 1, 10, 20, 9, 1, -1, 1, 0, 21, 70, 56, 14, 1, 1, -1, 1, 42, 231, 294, 126, 20, 1, -1, 1, 0, 85, 735, 1407, 924, 246, 27, 1, 1, -1, 1, 170, 2290, 6363, 6027, 2400, 435, 35, 1
(list; table; graph; refs; listen; history; internal format)
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OFFSET
| 0,14
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COMMENTS
| Inverse of number triangle A105793. Row sums are the generalized Bell numbers A000296.
From Wolfdieter Lang Jun 20 2011: (Start))
T(n,k)*(-1)^(n-k) gives the inverse Sheffer matrix of A094645. In the umbral notation (cf. Roman, p. 17, quoted in A094645) this is Sheffer for (1-t,-ln(1-t)). (End)
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FORMULA
| Term k in row n is given by {(-1)^(k+n) * [sum from j=0 to j=k of (-1)^j * binomial(k,j) * (1-j)^n] / k! }; i.e. a finite difference. - Tom Copeland (tcjpn(AT)msn.com), Jun 05 2006
O.G.F. for row n = n-th finite difference of the Touchard (Bell) polynomials, T(x,j) and so the E.G.F. for these finite differences and therefore the sequence = exp{x*[exp(t)-1]-t} = exp{t*[T(x,.)-1]} umbrally. - Tom Copeland (tcjpn(AT)msn.com), Jun 05 2006
The e.g.f. A(x,t) = exp(x*(exp(t)-1)-t) satisfies the partial differential equation x*dA/dx - dA/dt = (1-x)*A.
Recurrence relation: T(n+1,k) = T(n,k-1)+(k-1)*T(n,k).
Let f(x) = ((x-1)/x)*exp(x). For n >= 1, the n-th row polynomial R(n,x) = x*exp(-x)*(x*d/dx)^(n-1)(f(x)) and satisfies the recurrence equation R(n+1,x) = (x-1)*R(n,x) + x*R'(n,x). - Peter Bala Oct 28 2011.
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EXAMPLE
| The triangle starts with
n=0: 1;
n=1: -1, 1;
n=2: 1, -1, 1;
n=3: -1, 1, 0, 1;
n=4: 1, -1, 1, 2, 1;
n=5: -1, 1, 0, 5, 5, 1;
... [From W. Lang, June 2011]
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CROSSREFS
| Sequence in context: A153462 A126310 A109086 * A160380 A122433 A056977
Adjacent sequences: A105791 A105792 A105793 * A105795 A105796 A105797
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KEYWORD
| easy,sign,tabl
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AUTHOR
| Paul Barry (pbarry(AT)wit.ie), Apr 20 2005
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