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%I #45 Jul 25 2024 22:55:34
%S 1,1,3,5,9,15,27,45,81,135,243,405,729,1215,2187,3645,6561,10935,
%T 19683,32805,59049,98415,177147,295245,531441,885735,1594323,2657205,
%U 4782969,7971615,14348907,23914845,43046721,71744535,129140163,215233605,387420489
%N a(n) = a(n-1) + a(n-2) + gcd(a(n-1), a(n-2)) for n > 1; a(0)=1, a(1)=1.
%C Record high values in A003961 (except for the duplicated 1). - _Nicolas Bělohoubek_, Jun 18 2022
%C Apart from a(0), this sequence is the answer to Question 21 in the 2022 Shanghai College Entrance Mathematics Examination: a(1) = 1, a(2*m) = 3^m for all m; for any n >= 2, there exists 1 <= i <= n-1 such that a(n+1) = 2*a(n)-a(i). Find a(n). - _Yifan Xie_, Jul 20 2022
%C a(n) n>1 are a subset of the record values formed by the odd composite numbers (A071904) divided by their largest prime factor. For example, A071904[2434] = 6561 with largest pf = 3. 6561/3 = 2187 and appears in A083658. - _Bill McEachen_, Jul 06 2024
%H Michael De Vlieger, <a href="/A083658/b083658.txt">Table of n, a(n) for n = 0..4191</a>
%H Yulu Education, <a href="http://sh.yuloo.com/gaokao/shiti/shuxue/273453.shtml">2022 Shanghai College Entrance Examination Mathematics Paper and Answer Analysis</a> (Examinee Recall Version) (In Chinese)
%H <a href="/index/Rec#order_02">Index entries for linear recurrences with constant coefficients</a>, signature (0,3).
%F a(2n) = 3^n, a(2n+1) = 5*3^(n-1) for n>0; a(0)=1, a(1)=1.
%F G.f.: (2*x^3+1+x)/(1-3*x^2). - _R. J. Mathar_, Feb 27 2010
%F a(n) = 3 * a(n-2), n>3, a(2)=3, a(3)=5. - _Bill McEachen_, Jul 06 2024
%t CoefficientList[Series[(-2*x^3 - x - 1)/(3*x^2 - 1), {x, 0, 200}], x] (* _Vladimir Joseph Stephan Orlovsky_, Jun 10 2011 *)
%Y Cf. A003961.
%K nonn,easy
%O 0,3
%A _Paul D. Hanna_, Jun 13 2003