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A138041 a(1) = 1, a(2) = 10; for n>2, a(n+1) = 4*a(n) + 6*a(n-1). Also a(n) = upper left term in the 2 X 2 matrix [1,3; 3,3].

%I A138041 #16 Jan 02 2024 08:59:29
%S A138041 1,10,46,244,1252,6472,33400,172432,890128,4595104,23721184,122455360,
%T A138041 632148544,3263326336,16846196608,86964744448,448936157440,
%U A138041 2317533096448,11963749330432,61760195900416,318823279584256
%N A138041 a(1) = 1, a(2) = 10; for n>2, a(n+1) = 4*a(n) + 6*a(n-1). Also a(n) = upper left term in the 2 X 2 matrix [1,3; 3,3].
%H A138041 Harvey P. Dale, <a href="/A138041/b138041.txt">Table of n, a(n) for n = 1..1000</a>
%H A138041 <a href="/index/Rec#order_02">Index entries for linear recurrences with constant coefficients</a>, signature (4, 6).
%F A138041 a(n)/a(n-1) tends to (2 + sqrt(10)) = 5.16227766... (a root of x^2 - 4*x - 6 and an eigenvalue of the matrix).
%F A138041 a(n) mod 9 == 1.
%F A138041 O.g.f.: -x*(1+6*x)/(-1+4*x+6*x^2). a(n) = A085939(n)+6*A085939(n-1). - _R. J. Mathar_, Mar 03 2008
%F A138041 From the characteristic polynomial of the matrix we get g.f.: (6*x + 1)/(-6*x^2 - 4*x + 1), with roots a=-(2+sqrt(10))/6, b=-(2-sqrt(10))/6. Let A=3+3*sqrt(10)/10 and B=3-3*sqrt(10)/10. Then a(n) = (A*(1/a)^n + B*(1/b)^n)/6. - Lambert Herrgesell (zero815(AT)googlemail.com), Apr 04 2008
%e A138041 a(4) = 244 = 4*46 + 6*10 = 4*a(3) + 6*a(2).
%e A138041 a(4) = 244 = upper left term in [1,3; 3,3]^4.
%t A138041 a = {1, 10}; Do[AppendTo[a, 4*a[[ -1]] + 6*a[[ -2]]], {25}]; a (* _Stefan Steinerberger_ *)
%t A138041 LinearRecurrence[{4,6},{1,10},30] (* _Harvey P. Dale_, Mar 09 2014 *)
%K A138041 nonn
%O A138041 1,2
%A A138041 _Gary W. Adamson_, Mar 02 2008
%E A138041 More terms from _Stefan Steinerberger_ and _R. J. Mathar_, Mar 02 2008
%E A138041 Definition corrected by _Paolo P. Lava_, Jun 03 2008