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 A091141 a(n) = 2*a(n-1) + 4*a(n-2) - 2*a(n-3) with initial terms 1, 4, 13. 4
 1, 4, 13, 40, 124, 382, 1180, 3640, 11236, 34672, 107008, 330232, 1019152, 3145216, 9706576, 29955712, 92447296, 285304288, 880486336, 2717295232, 8385927232, 25880062720, 79869243904, 246486884224, 760690618624, 2347590286336, 7244969278720, 22358918465536 (list; graph; refs; listen; history; text; internal format)
 OFFSET 1,2 COMMENTS One of 3 related sequences generated from finite difference operations. Let r(1)=s(1)=t(1)=1. Given r(n), s(n) and t(n), let f(x) = r(n) x^2 + s(n) x + t(n) and let r(n+1), s(n+1) and t(n+1) be the 0th, 1st and 2nd differences of f(x) at x=1. I.e., r(n+1) = f(1) = r(n)+s(n)+t(n), s(n+1) = f(2)-f(1) = 3r(n)+s(n) and t(n+1) = f(3)-2f(2)+f(1) = 2r(n). This sequence gives s(n). LINKS Colin Barker, Table of n, a(n) for n = 1..1000 Index entries for linear recurrences with constant coefficients, signature (2,4,-2). FORMULA Let v(n) be the column vector with elements r(n), s(n), t(n); then v(n) = [1 1 1 / 3 1 0 / 2 0 0] v(n-1). The limit as n->infinity of a(n+1)/a(n) is the largest root of x^3 - 2x^2 - 4x + 2 = 0, which is about 3.086130197651494. G.f.: x*(x+1)^2 / (2*x^3-4*x^2-2*x+1). - Colin Barker, May 21 2015 MATHEMATICA a[n_] := (MatrixPower[{{1, 1, 1}, {3, 1, 0}, {2, 0, 0}}, n-1].{{1}, {1}, {1}})[[2, 1]] LinearRecurrence[{2, 4, -2}, {1, 4, 13}, 30] (* Harvey P. Dale, Jun 19 2018 *) PROG (PARI) Vec(x*(x+1)^2/(2*x^3-4*x^2-2*x+1) + O(x^100)) \\ Colin Barker, May 21 2015 CROSSREFS Cf. r(n) = A091140(n), t(n) = A091142(n). Sequence in context: A003462 A076040 A261547 * A098183 A171556 A227747 Adjacent sequences:  A091138 A091139 A091140 * A091142 A091143 A091144 KEYWORD nonn,easy AUTHOR Gary W. Adamson, Dec 21 2003 STATUS approved

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Last modified July 29 08:46 EDT 2021. Contains 346340 sequences. (Running on oeis4.)