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 A093406 A sequence converging to 1 + 2^(1/4). 2
 1, 3, 11, 31, 71, 145, 289, 601, 1321, 2979, 6683, 14743, 32111, 69697, 151777, 332113, 728689, 1598883, 3503627, 7668079, 16774775, 36704017, 80343361, 175916521, 385196761, 843365379, 1846290395, 4041672871, 8847607391, 19368919297, 42403014721, 92830645537 (list; graph; refs; listen; history; text; internal format)
 OFFSET 1,2 COMMENTS a(n)/a(n-1) tends to 2.189207115... = 1 + 2^(1/4). Example: a(18)/a(17) = 1598883/728689 = 2.19419... A052101 is the series derived from analogous 3rd order operations, with lim a(n)/a(n-1) as n approaches inf. = 1 + 2^(1/3). REFERENCES E. J. Barbeau, Polynomials, Springer-Verlag NY Inc, 1989, p. 136. LINKS Index entries for linear recurrences with constant coefficients, signature (4,-6,4,1). FORMULA We use a 4 X 4 matrix corresponding to the characteristic polynomial (x - 1)^4 - 2 = 0 = x^4 - 4x^3 + 6x^2 - 4x - 1 = 0, being [0 1 0 0 / 0 0 1 0 / 0 0 0 1 / 1 4 -6 4]. Let the matrix = M. Perform M^n * [1, 1, 1, 1]. a(n) = the third term from the left, (the other 3 terms being offset members of the series). a(n) = 4*a(n-1)-6*a(n-2)+4*a(n-3)+a(n-4). G.f.: -x*(x^3+5*x^2-x+1)/ (x^4+4*x^3-6*x^2+4*x-1). [Colin Barker, Oct 21 2012] EXAMPLE a(4) = 31, since M^4 * [1,1,1,1] = [3, 11, 31, 71]. MATHEMATICA LinearRecurrence[{4, -6, 4, 1}, {1, 3, 11, 31}, 40] (* Harvey P. Dale, Jul 22 2013 *) CROSSREFS Cf. A052101. Sequence in context: A261148 A071568 A097081 * A107587 A245931 A190590 Adjacent sequences:  A093403 A093404 A093405 * A093407 A093408 A093409 KEYWORD nonn,easy AUTHOR Gary W. Adamson, Mar 28 2004 EXTENSIONS Corrected by T. D. Noe, Nov 08 2006 STATUS approved

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Last modified January 26 01:48 EST 2020. Contains 331270 sequences. (Running on oeis4.)