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 A084219 Inverse binomial transform of A053088. 2
 1, -1, 4, -8, 20, -44, 100, -220, 484, -1052, 2276, -4892, 10468, -22300, 47332, -100124, 211172, -444188, 932068, -1951516, 4077796, -8505116, 17709284, -36816668, 76429540, -158451484, 328087780, -678545180, 1401829604 (list; graph; refs; listen; history; text; internal format)
 OFFSET 0,3 COMMENTS Contribution from Gary W. Adamson, Jan 05 2009: (Start) Unsigned, starting with offset 1: generated from iterates of M * [1,1,1,...] where M = a tridiagonal matrix with [0,1,1,1,...] as the main diagonal, [1,1,1,...] as the uperdiagonal and [2,0,0,0,...] as the subdiagonal. (End) Define a triangle via T(n,0) = T(n,n) = A001045(n) and T(r,c) = T(r-1,c-1) + T(r-1,c). The row sums of the triangle are s(n) = 0, 2, 4, 12, ... = 2*A059570(n), and their first differences are s(n+1)-s(n) = 2*|a(n)|. J. M. Bergot, May 15 2013 LINKS Table of n, a(n) for n=0..28. Roland Bacher, Chebyshev polynomials, quadratic surds and a variation of Pascal's triangle, arXiv:1509.09054 [math.CO], 2015. Index entries for linear recurrences with constant coefficients, signature (-3,0,4). FORMULA a(n) = (4 - 3*n*(-2)^(n-1) + 5*(-2)^n)/9. a(n) = (1/4) + Sum_{k=0..n} (-2)^k*(k+3)/4. G.f.: (1+x)^2/((1-x)(1+2x)^2). MATHEMATICA LinearRecurrence[{-3, 0, 4}, {1, -1, 4}, 30] (* Harvey P. Dale, Dec 16 2016 *) CROSSREFS Sequence in context: A009889 A334706 A095804 * A190589 A009916 A203167 Adjacent sequences: A084216 A084217 A084218 * A084220 A084221 A084222 KEYWORD easy,sign AUTHOR Paul Barry, May 20 2003 STATUS approved

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Last modified May 27 16:57 EDT 2024. Contains 372880 sequences. (Running on oeis4.)