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A072880 A recurrence of order 6: a(1)=a(2)=a(3)=a(4)=a(5)=a(6)=1; a(n) = (a(n-1)^2 + a(n-2)^2 + a(n-3)^2 + a(n-4)^2 + a(n-5)^2)/a(n-6). 9

%I #50 May 16 2019 08:50:38

%S 1,1,1,1,1,1,5,29,869,756029,571580604869,326704387862983487112029,

%T 21347151409785350408171299054974277225256721769,

%U 15713823217665540462976624783900822313284439536736221766688609460305249837839107387688348185

%N A recurrence of order 6: a(1)=a(2)=a(3)=a(4)=a(5)=a(6)=1; a(n) = (a(n-1)^2 + a(n-2)^2 + a(n-3)^2 + a(n-4)^2 + a(n-5)^2)/a(n-6).

%C Any six successive terms satisfy the Markoff-Hurwitz equation a^2 + b^2 + c^2 + d^2 + e^2 + f^2 = 6*a*b*c*d*e*f. - _Bruno Langlois_, Aug 09 2016

%H Seiichi Manyama, <a href="/A072880/b072880.txt">Table of n, a(n) for n = 1..17</a>

%H Andrew N. W. Hone, <a href="https://arxiv.org/abs/math/0601324">Diophantine non-integrability of a third order recurrence with the Laurent property</a>, arXiv:math/0601324 [math.NT], 2006.

%H Andrew N. W. Hone, <a href="https://doi.org/10.1088/0305-4470/39/12/L01">Diophantine non-integrability of a third order recurrence with the Laurent property</a>, J. Phys. A: Math. Gen. 39 (2006), L171-L177.

%H Matthew Christopher Russell, <a href="https://doi.org/doi:10.7282/T3MC926D">Using experimental mathematics to conjecture and prove theorems in the theory of partitions and commutative and non-commutative recurrences</a>, PhD Dissertation, Mathematics Department, Rutgers University, May 2016.

%F a(n) = 6*a(n-1)*a(n-2)*a(n-3)*a(n-4)*a(n-5) - a(n-6). - _Bruno Langlois_, Aug 09 2016

%t RecurrenceTable[{a[n] == (a[n - 1]^2 + a[n - 2]^2 + a[n - 3]^2 + a[n - 4]^2 + a[n - 5]^2)/a[n - 6], a[1] == a[2] == a[3] == a[4] == a[5] == a[6] == 1}, a, {n, 1, 14}] (* _Michael De Vlieger_, Aug 11 2016 *)

%t nxt[{a_, b_, c_, d_, e_, f_}] := {b, c, d, e, f,(b^2+c^2+d^2+e^2+f^2)/a}; NestList[ nxt, Table[1,6],20][[All,1]] (* _Harvey P. Dale_, Mar 18 2018 *)

%Y Cf. A006720, A064098, A072878, A072879.

%K easy,nonn

%O 1,7

%A _Benoit Cloitre_, Jul 28 2002

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