A144229 The numerators of the convergents to the recursion x=1/(x^2+1).

1, 1, 4, 25, 1681, 5317636, 66314914699609, 8947678119828215014722891025, 178098260698995011212395018312912894502905113202338936836

Offset: 0

Author

Cino Hilliard, Sep 15 2008

Keywords

• frac
• nonn

Comments

The recursion converges to the real root of 1/(x^2+1) - x = 0, 0.682327803...

An interesting consequence of this result occurs if we multiply by x^2+1 to get 1-x-x^3=0. These different equations intersect at the same root 0.682327803... Note also that a(n) is a square. The square roots form sequence A076725.

a(n) is the number of (0,1)-labeled perfect binary trees of height n such that no adjacent nodes have 1 as the label and the root is labeled 0. - Ran Pan, May 22 2015

Links

• Ran Pan, Exercise R, Project P.

Crossrefs

Cf. A076725, A088559.

Programs

Mathematica

f[n_]:=(n+1/n)/n;Prepend[Denominator[NestList[f,2,7]],1] (* Vladimir Joseph Stephan Orlovsky, Nov 19 2010 *)
RecurrenceTable[{a[n]==(a[n-2]^2 + a[n-1])^2, a[0]==1, a[1]==1},a,{n,0,10}] (* Vaclav Kotesovec, May 22 2015 after Ran Pan *)

PARI

x=0;for(j=1,10,x=1/(x^2+1);print1((numerator(x))","))

Formula

a(n+2) = (a(n)^2 + a(n+1))^2. - Ran Pan, May 22 2015

a(n) ~ c * d^(2^n), where c = A088559 = 0.465571231876768... is the root of the equation c*(1+c)^2 = 1, d = 1.6634583970724267140029... . - Vaclav Kotesovec, May 22 2015