A072879 a(n) = 5*a(n-1)*a(n-2)*a(n-3)*a(n-4) - a(n-5) with a(1) = a(2) = a(3) = a(4) = a(5) = 1.
1, 1, 1, 1, 1, 4, 19, 379, 144019, 20741616379, 107553662508585672001, 608831069421618273050865038881215685876, 978035016076705458999330010986670207956236476587064788804921180339451725001
Offset: 1
Links
- Seiichi Manyama, Table of n, a(n) for n = 1..16
- Arthur Baragar, Integral solutions of the Markoff-Hurwitz equations, J. Number Theory 49 (1994), 27-44.
- 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
nxt[{a_,b_,c_,d_,e_}]:={b,c,d,e,(5b c d e)-a}; NestList[nxt,{1,1,1,1,1},20][[All,1]] (* Harvey P. Dale, Nov 07 2016 *)
Formula
a(1) = a(2) = a(3) = a(4) = a(5) = 1; a(n) = (a(n-1)^2+a(n-2)^2+a(n-3)^2+a(n-4)^2)/a(n-5) for n >= 6.
From the recurrence a(n) = 5*a(n-1)*a(n-2)*a(n-3)*a(n-4) - a(n-5), any five successive terms satisfy the five-variable Hurwitz equation a(n)^2+a(n-1)^2+a(n-2)^2+a(n-3)^2+a(n-4)^2 = 5*a(n)*a(n-1)*a(n-2)*a(n-3)*a(n-4). As n tends to infinity, the limit of log(log(a(n)))/n is log x = 0.6562559790..., where x=1.927561975... is the largest real root of the quartic x^4-x^3-x^2-x-1=0. - Andrew Hone, Nov 16 2005
Extensions
Entry revised Nov 19 2005, based on comments from Andrew Hone
Name clarified by Petros Hadjicostas, May 11 2019
Comments