A064098 a(n+1) = (a(n)^2 + a(n-1)^2)/a(n-2), with a(1) = a(2) = a(3) = 1.

1, 1, 1, 2, 5, 29, 433, 37666, 48928105, 5528778008357, 811537892743746482789, 13460438563050022083842073547074914, 32770967840592833551621556305285371426044732591005957081

Offset: 1

Author

Santi Spadaro, Sep 16 2001

Keywords

• nonn

Comments

This sequence was introduced by Dana Scott but possibly studied earlier by others. - James Propp, Jan 27 2005

Sequence gives the upper-left entries of the respective matrices

[1 1] [1 0] [2 1] [5 2] [29 12] [433 179] [37666 15571]

[1 2] [0 1] [1 1] [2 1] [12 5] [179 74] [15571 6437], ...

satisfying the recurrence M(n) = M(n-1) M(n-3)^(-1) M(n-1) (note that the Fibonacci numbers satisfy the additive version of this recurrence). - James Propp, Jan 27 2005

Define b(1) = b(2) = b(3) = 1; b(n) = (b(n-1) + b(n-2))^2/b(n-3); then a(n) = sqrt(b(n)). - Benoit Cloitre, Jul 28 2002

Any 3 successive terms of the sequence satisfy the Markov equation x^2 + y^2 + z^2 = 3 xyz. Therefore from the 3rd term on this is a subsequence of the Markov numbers, A002559. Also, we conjecture that the limit of log(log(a(n)))/n is log((sqrt(5) + 1)/2). - Martin Giese (martin.giese(AT)oeaw.ac.at), Oct 13 2005

A subsequence of the Markoff numbers A002559. - Andrew Hone, Jan 16 2006

The recursion exhibits the Laurent phenomenon. Let F(n) = Fibonacci(n), e(n) = F(n) - 1, a(1) = a1, a(2) = a2, a(3) = a3, a(n) = (a(n-1)^2 + a(n-3)^2) / a(n-3), b(n) = a(n) * a1^e(n-1) * a2^e(n-2) * a3^e(n-3). Then b(n) for n > 1 is an irreducible polynomial in {a1^2, a2^2, a3^2}, b(n) = (b(n-1)^2 + (b(n-2) * a1^F(n-4) * a2^F(n-5) * a3^F(n-6))^2) / b(n-3), and a(n) = a(n-1) * a(n-2) * (a1^2 + a2^2 + a3^2) / (a1 * a2 * a3) - a(n-3). - Michael Somos, Jan 12 2013

Starting with n = 5, a(n) is the largest number in row n - 5 of the Markov tree, A368546. These numbers are obtained by descending the tree in alternating right and left steps; their Farey indices (see A368546 for the definition) are ratios of successive Fibonacci numbers, 1/2, 2/3, 3/5, 5/8, ... See Aigner, Proposition 10.2. - Wouter Meeussen and William P. Orrick, Feb 11 2024

Examples

			G.f. = x + x^2 + x^3 + 2*x^4 + 5*x^5 + 29*x^6 + 433*x^7 + 37666*x^8 + ...
		

References

• Martin Aigner, Markov's theorem and 100 years of the uniqueness conjecture. A mathematical journey from irrational numbers to perfect matchings. Springer, 2013. x+257 pp. ISBN: 978-3-319-00887-5; 978-3-319-00888-2 MR3098784

Links

• Harry J. Smith, Table of n, a(n) for n = 1..18
• Martin Aigner, Markov's theorem and 100 years of the uniqueness conjecture. A mathematical journey from irrational numbers to perfect matchings, [archive.org copy of the book].
• S. Fomin and A. Zelevinsky, The Laurent Phenomenon, Advances in Applied Mathematics, 28 (2002), 119-144.
• 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.
• Andrew N. W. Hone, Growth of Mahler measure and algebraic entropy of dynamics with the Laurent property, arXiv:2109.08217 [math.NT], 2021.
• KöMaL-Mathematical and Physical Journal for Secondary Schools, New advanced problems: proposed problem A265, April 2001.
• L. J. Mordell, On the integer solutions of the equation x^2+y^2+z^2+2xyz=n, J. Lond. Math. Soc. 28 (1953), 500-510.
• J. Propp, The combinatorics of frieze patterns and Markoff numbers, arXiv:math/0511633 [math.CO], 2005-2008.
• 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.

Crossrefs

Cf. A002559, A072878, A072879, A072880.

Markov tree: A327345, A368546.

Programs

Magma

[n le 3 select 1 else (Self(n-1)^2 + Self(n-2)^2)/Self(n-3): n in [1..16]]; // G. C. Greubel, Nov 07 2024

Maple

f:=proc(n) option remember; global K; local i;
if n <= K then 1
else add(f(n-i)^2,i=1..K-1)/f(n-K); fi; end;
K:=3;
[seq(f(n),n=1..10)]; # N. J. A. Sloane, Mar 17 2017

Mathematica

a[n_]:= (a[n-1]^2 +a[n-2]^2)/a[n-3]; a[1]=a[2]=a[3]=1; Array[a, 13] (* Or *)
a[n_]:= 3*a[n-1]*a[n-2] - a[n-3]; a[1]= a[2]= a[3]= 1; Array[a, 13] (* Robert G. Wilson v, Dec 26 2012 *)
nxt[{a_,b_,c_}]:={b,c,(c^2+b^2)/a}; NestList[nxt,{1,1,1},15][[;;,1]] (* Harvey P. Dale, Jul 07 2025 *)

PARI

{a(n) = if( n<1, n = 4-n); if( n<4, 1, 3 * a(n-1) * a(n-2) - a(n-3))}; /* Michael Somos, Jan 12 2013 */

PARI

{ a=a3=a2=a1=1; for (n = 1, 18, if (n>3, a=(a1^2 + a2^2)/a3; a3=a2; a2=a1; a1=a); write("b064098.txt", n, " ", a) ) } /* Harry J. Smith, Sep 06 2009 */

SageMath

def A064098(n):
    def a(n): return 1 if n<4 else (a(n-1)^2 + a(n-2)^2)/a(n-3)
    return a(n)
[A064098(n) for n in range(16)] # G. C. Greubel, Nov 07 2024

Formula

Conjecture: lim_{n -> infinity} log(log(a(n)))/n exists = 0.48.... - Benoit Cloitre, Aug 07 2002. This is true - see below.

For this subsequence of the Markoff numbers, we have 2^(F(n-1) - 1) < a(n) < 3^(F(n-1) - 1) for n > 4, where F(n) are the Fibonacci numbers with F(0)=0, F(1)=1, F(n+1) = F(n) + F(n-1). Hence log(log(a(n)))/n tends to log((1 + sqrt(5))/2) as previously conjectured. - Andrew Hone, Jan 16 2006

a(n) = 3 * a(n-1) * a(n-2) - a(n-3). a(4-n) = a(n) for all n in Z. - Michael Somos, Jan 12 2013

a(n) ~ 1/3 * c^(((1 + sqrt(5))/2)^n), where c = 1.2807717799265504005186306582930649245... . - Vaclav Kotesovec, May 06 2015

Extensions

Entry improved by comments from Michael Somos, Sep 25 2001