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).
1, 1, 1, 1, 1, 1, 5, 29, 869, 756029, 571580604869, 326704387862983487112029, 21347151409785350408171299054974277225256721769, 15713823217665540462976624783900822313284439536736221766688609460305249837839107387688348185
Offset: 1
Links
- Seiichi Manyama, Table of n, a(n) for n = 1..17
- Andrew N. W. Hone, Diophantine non-integrability of a third order recurrence with the Laurent property, arXiv:math/0601324 [math.NT], 2006.
- Andrew N. W. Hone, Diophantine non-integrability of a third order recurrence with the Laurent property, J. Phys. A: Math. Gen. 39 (2006), L171-L177.
- Matthew Christopher Russell, Using experimental mathematics to conjecture and prove theorems in the theory of partitions and commutative and non-commutative recurrences, PhD Dissertation, Mathematics Department, Rutgers University, May 2016.
Programs
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Mathematica
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 *) 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 *)
Formula
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
Comments