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A112373 a(n+2) = (a(n+1)^3 + a(n+1)^2)/a(n) with a(0)=1, a(1)=1.

%I A112373 #54 Sep 07 2024 08:55:09
%S A112373 1,1,2,12,936,68408496,342022190843338960032,
%T A112373 584861200495456320274313200204390612579749188443599552
%N A112373 a(n+2) = (a(n+1)^3 + a(n+1)^2)/a(n) with a(0)=1, a(1)=1.
%C A112373 As n tends to infinity, log(log(a(n)))/n tends to log((3+sqrt(5))/2) = A104457.
%C A112373 The Laurent property is satisfied by any second-order recurrence of the form a(n+2)=f(a(n+1))/a(n) where f is a polynomial of the form f(x)=x^m*p(x) with m a positive integer >= 2 and p arbitrary. In that case a(0)=a(1)=1 generates a sequence of integers and the ratios a(n+1)/a(n) and a(n+1)*a(n-1)/a(n)^2 are integers for all n. - _Andrew Hone_, Dec 12 2005
%C A112373 Also denominators of Sum_{k=0..n} 1/a(k) with numerators = A259644. - _Reinhard Zumkeller_, Jul 02 2015
%C A112373 The next term (a(8)) has 141 digits. - _Harvey P. Dale_, Apr 05 2019
%H A112373 Reinhard Zumkeller, <a href="/A112373/b112373.txt">Table of n, a(n) for n = 0..10</a>
%H A112373 Joshua Alman, Cesar Cuenca, and Jiaoyang Huang, <a href="https://doi.org/10.1007/s10801-015-0647-5">Laurent phenomenon sequences</a>, Journal of Algebraic Combinatorics, Vol. 43, No. 3 (2015), pp. 589-633.
%H A112373 Sergey Fomin and Andrei Zelevinsky, <a href="http://dx.doi.org/10.1006/aama.2001.0770">The Laurent Phenomenon</a>, Advances in Applied Mathematics, Vol. 28, No. 2 (2002) pp. 119-144.
%H A112373 Andrew N. W. Hone, <a href="https://cs.uwaterloo.ca/journals/JIS/VOL18/Hone/hone3.html">Curious continued fractions, nonlinear recurrences and transcendental numbers</a>Journal of Integer Sequences, Vol. 18 (2015), Article #15.8.4; <a href="http://arxiv.org/abs/1507.00063">arXiv preprint</a>, arXiv:1507.00063 [math.NT], 2015.
%H A112373 Andrew N. W. Hone, <a href="http://dx.doi.org/10.1007/s00605-015-0844-2">Continued fractions for some transcendental numbers</a>, Monatshefte für Mathematik, Vol. 182 (2017), pp. 33-38; <a href="http://arxiv.org/abs/1509.05019">arXiv preprint</a>, arXiv:1509.05019 [math.NT], 2015-2016.
%F A112373 a(n+1) / a(n) = A114552(n), a(n) = a(1-n) for all n in Z. - _Michael Somos_, Apr 19 2017
%F A112373 Sum_{n>=0} 1/a(n) = A114550. - _Amiram Eldar_, Nov 13 2020
%p A112373 a[0]:=1; a[1]:=1; f(x):=x^3+x^2; for n from 0 to 8 do a[n+2]:=simplify(subs(x=a[n+1],f(x))/a[n]) od; s[3]:=ln(2^2*3); s[4]:=ln(2^3*3^2*13); for n from 3 to 10000 do s[n+2]:=evalf(3*s[n+1]+ln(1+exp(-s[n+1]))-s[n]): od: print(evalf(ln(s[10002])/(10002))): evalf(ln((3+sqrt(5))/2));
%p A112373 # s[n]=ln(a[n]); ln(s[n])/n converges slowly to 0.962...
%t A112373 RecurrenceTable[{a[n] == (a[n - 1]^3 + a[n - 1]^2)/a[n - 2], a[0] == a[1] == 1}, a, {n, 0, 7}] (* _Michael De Vlieger_, Jul 02 2015 *)
%t A112373 nxt[{a_,b_}]:={b,(b^3+b^2)/a}; NestList[nxt,{1,1},8][[All,1]] (* _Harvey P. Dale_, Apr 05 2019 *)
%o A112373 (Magma) I:=[1,1]; [n le 2 select I[n] else (Self(n-1)^3+Self(n-1)^2)/Self(n-2): n in [1..10]]; // _Vincenzo Librandi_, Jul 02 2015
%o A112373 (Haskell)
%o A112373 a112373 n = a112373_list !! n
%o A112373 a112373_list = 1 : 1 : zipWith (\u v -> (u^3 + u^2) `div` v)
%o A112373                                (tail a112373_list) a112373_list
%o A112373 -- _Reinhard Zumkeller_, Jul 02 2015
%o A112373 (PARI) {a(n) = my(a=self()); if(n<0, a(1-n), n<2, 1, a(n-1)^2 * (1 + a(n-1)) / a(n-2))}; /* _Michael Somos_, Apr 19 2017 */
%Y A112373 Cf. A114550, A114551, A114552.
%Y A112373 Cf. A259644.
%K A112373 nonn
%O A112373 0,3
%A A112373 _Andrew Hone_, Dec 02 2005