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The (1,1)-term of the 3 X 3 matrix M^n, where M = [1,2,3 / 4,7,11 / 6,10,16].
1

%I #17 Mar 02 2017 13:15:31

%S 1,27,649,15603,375121,9018507,216819289,5212681443,125321173921,

%T 3012920855547,72435421707049,1741463041824723,41867548425500401,

%U 1006562625253834347,24199370554517524729,581791455933674427843,13987194312962703792961,336274454967038565458907

%N The (1,1)-term of the 3 X 3 matrix M^n, where M = [1,2,3 / 4,7,11 / 6,10,16].

%H Colin Barker, <a href="/A095898/b095898.txt">Table of n, a(n) for n = 1..700</a>

%H Tanya Khovanova, <a href="http://www.tanyakhovanova.com/RecursiveSequences/RecursiveSequences.html">Recursive Sequences</a>

%H <a href="/index/Rec#order_02">Index entries for linear recurrences with constant coefficients</a>, signature (24,1).

%F a(n) = 24*a(n-1) + a(n-2) for n>=3; a(1)=1, a(2)=27 (follows from the minimal polynomial of the matrix M).

%F G.f.: (x+3*x^2) / (1-24*x-x^2). - _Philippe Deléham_, Nov 21 2008

%F a(n) = (-12 - sqrt(145))^(-n)*(87+7*sqrt(145) + (-289-24*sqrt(145))^n*(87-7*sqrt(145))) / 58. - _Colin Barker_, Mar 02 2017

%e a(4)=15603 because M^4 = [15603,26590,42193 / 56642,96527,153169 / 82078,139874,221952]. Alternatively, a(4) = 24*649+27 = 15603.

%p a[1]:=1: a[2]:=27: for n from 3 to 18 do a[n]:=24*a[n-1]+a[n-2] od: seq(a[n],n=1..18);

%o (PARI) Vec(x*(1 + 3*x) / (1 - 24*x - x^2) + O(x^30)) \\ _Colin Barker_, Mar 02 2017

%Y Cf. A083412, A035513, A003622, A001950, A000201.

%K nonn,easy

%O 1,2

%A _Gary W. Adamson_, Jun 12 2004

%E Corrected by _T. D. Noe_, Nov 07 2006

%E Edited by _N. J. A. Sloane_, Dec 16 2006