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A099867
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a(n) = 5*a(n-1) - a(n-2) for n>1, a(0)=1, a(1)=9.
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2
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1, 9, 44, 211, 1011, 4844, 23209, 111201, 532796, 2552779, 12231099, 58602716, 280782481, 1345309689, 6445765964, 30883520131, 147971834691, 708975653324, 3396906431929, 16275556506321, 77980876099676, 373628823992059, 1790163243860619, 8577187395311036
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OFFSET
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0,2
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COMMENTS
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For any two terms (a(n), a(n+1)) = (x, y), x^2 - 5*x*y + y^2 = 37 = A082111(4). This is valid in general for all recursive sequences (t) with constant coefficients (5,-1) and t(0) = 1: x^2 - 5*x*y + y^2 = A082111(t(1)-5). This includes and interprets the Feb 04 2014 comment in A004253 by Colin Barker.
By analogy to all this, for three consecutive terms (x, y, z) of any sequence (t) of the form (5,-1) with t(0) = 1: y^2 - x*z = A082111(t(1)-5). (End)
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LINKS
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A. F. Horadam, Pell Identities, Fib. Quart., Vol. 9, No. 3, 1971, pp. 245-252.
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FORMULA
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G.f.: (1+4*x) / (1-5*x+x^2).
(End)
a(n) = 2^(-1-n)*((5-sqrt(21))^n*(-13+sqrt(21)) + (5+sqrt(21))^n*(13+sqrt(21))) / sqrt(21). - Colin Barker, Mar 31 2017
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MATHEMATICA
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a[0] = 1; a[1] = 9; a[n_] := a[n] = 5 a[n - 1] - a[n - 2]; Table[ a[n], {n, 0, 21}] (* Robert G. Wilson v, Dec 14 2004 *)
LinearRecurrence[{5, -1}, {1, 9}, 30] (* or *) CoefficientList[Series[(1 + 4 x)/(1 - 5 x + x^2), {x, 0, 30}], x] (* Harvey P. Dale, Jun 26 2011 *)
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PROG
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(Magma) I:=[1, 9]; [n le 2 select I[n] else 5*Self(n-1)-Self(n-2): n in [1..30]]; // Vincenzo Librandi, Jul 30 2015
(PARI) Vec((1+4*x) / (1-5*x+x^2) + O(x^30)) \\ Colin Barker, Mar 31 2017
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CROSSREFS
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KEYWORD
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nonn,easy
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AUTHOR
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STATUS
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approved
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