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A082585 a(1)=1, a(n) = ceiling(r(5)*a(n-1)) where r(5) = (1/2)*(5 + sqrt(29)) is the positive root of X^2 = 5*X + 1.

%I A082585 #34 Jan 02 2024 08:53:00
%S A082585 1,6,32,167,868,4508,23409,121554,631180,3277455,17018456,88369736,
%T A082585 458867137,2382705422,12372394248,64244676663,333595777564,
%U A082585 1732223564484,8994713599985,46705791564410,242523671422036
%N A082585 a(1)=1, a(n) = ceiling(r(5)*a(n-1)) where r(5) = (1/2)*(5 + sqrt(29)) is the positive root of X^2 = 5*X + 1.
%H A082585 <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (6,-4,-1).
%F A082585 For n > 3, a(n) = 6*a(n-1) - 4*a(n-2) - a(n-3); a(n) = floor(t(5)*r(5)^n) where t(5) = (1/10)*(1 + 7/sqrt(29)) is the positive root of 145*X^2 = 29*X + 1.
%F A082585 G.f.: x/((x-1)*(x^2+5*x-1)). - _Colin Barker_, Jan 27 2013
%F A082585 G.f.: 1/(1/Q(0) + 3*x^3 - 3*x) where Q(k) = 1 + k*(2*x+1) + 8*x - 2*x*(k+1)*(k+5)/Q(k+1) ; (continued fraction). - _Sergei N. Gladkovskii_, Mar 15 2013
%p A082585 a:=n->sum(fibonacci(i,5), i=0..n): seq(a(n), n=1..21); # _Zerinvary Lajos_, Mar 20 2008
%t A082585 Join[{a=1,b=6},Table[c=5*b+1*a+1;a=b;b=c,{n,60}]] (* _Vladimir Joseph Stephan Orlovsky_, Feb 06 2011*)
%t A082585 LinearRecurrence[{6,-4,-1},{1,6,32},30] (* _Harvey P. Dale_, Jan 06 2012 *)
%Y A082585 Cf. A000071, A048739, A049652, A082574.
%K A082585 nonn,easy
%O A082585 1,2
%A A082585 _Benoit Cloitre_, May 07 2003