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%I #59 Aug 28 2023 14:37:40
%S 3,8,23,68,203,608,1823,5468,16403,49208,147623,442868,1328603,
%T 3985808,11957423,35872268,107616803,322850408,968551223,2905653668,
%U 8716961003,26150883008,78452649023,235357947068,706073841203,2118221523608
%N a(n) = (5*3^(n-1)+1)/2.
%C It appears that if s(n) is a first-order rational sequence of the form s(0)=4, s(n) = (2*s(n-1)+1)/(s(n-1)+2), n > 0, then s(n) = a(n)/(a(n)-1), n > 0.
%H Vincenzo Librandi, <a href="/A057198/b057198.txt">Table of n, a(n) for n = 1..1000</a>
%H Nathan Fox, <a href="https://arxiv.org/abs/1611.08244">A Slow Relative of Hofstadter's Q-Sequence</a>, arXiv:1611.08244 [math.NT], 2016.
%H <a href="/index/Rec#order_02">Index entries for linear recurrences with constant coefficients</a>, signature (4,-3).
%F a(n+1) = 3*a(n) - 1 for n > 1. - _Reinhard Zumkeller_, Jan 22 2011
%F G.f.: (5/2)*U(0) where U(k) = 1 + 2/(5*3^k + 5*3^k/(1 - 30*x*3^k/(15*x*3^k - 1/U(k+1)))); (continued fraction, 4-step). - _Sergei N. Gladkovskii_, Nov 01 2012
%F E.g.f.: (5/2)*U(0) where U(k) = 1 + 2/(5*3^k + 5*3^k/(1 - 30*x*3^k/(15*x*3^k - (k+1)/U(k+1)))); (continued fraction, 4-step). - _Sergei N. Gladkovskii_, Nov 01 2012
%F G.f.: x*(3-4*x) / ( (3*x-1)*(x-1) ). - _R. J. Mathar_, Jan 25 2015
%F E.g.f.: (5*exp(3*x) + 3*exp(x) - 8)/6. - _Stefano Spezia_, Aug 28 2023
%e G.f. = 3*x + 8*x^2 + 23*x^3 + 68*x^4 + 203*x^5 + 608*x^6 + 1823*x^7 + 5468*x^8 + ...
%t Table[(5*3^(n-1) + 1)/2, {n, 30}] (* _T. D. Noe_, Oct 11 2012 *)
%o (PARI) a(n)=(5*3^(n-1)+1)/2 \\ _Charles R Greathouse IV_, Oct 11 2012
%o (Magma) [(5*3^(n-1) + 1)/2: n in [1..30]]; // _Vincenzo Librandi_, Oct 11 2012
%Y Related to A046901.
%Y Equals A060816 + 1.
%Y Cf. A135423 (bisection), A191450 (2nd row).
%K nonn,easy
%O 1,1
%A _Colin Mallows_ and _N. J. A. Sloane_, Sep 16 2000
%E Incorrect zeroth term removed by _Jon Perry_, Oct 11 2012
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