login

Year-end appeal: Please make a donation to the OEIS Foundation to support ongoing development and maintenance of the OEIS. We are now in our 61st year, we have over 378,000 sequences, and we’ve reached 11,000 citations (which often say “discovered thanks to the OEIS”).

A144229
The numerators of the convergents to the recursion x=1/(x^2+1).
1
1, 1, 4, 25, 1681, 5317636, 66314914699609, 8947678119828215014722891025, 178098260698995011212395018312912894502905113202338936836
OFFSET
0,3
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
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
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 *)
PROG
(PARI) x=0; for(j=1, 10, x=1/(x^2+1); print1((numerator(x))", "))
CROSSREFS
Sequence in context: A132553 A305679 A317061 * A316926 A317699 A284106
KEYWORD
frac,nonn
AUTHOR
Cino Hilliard, Sep 15 2008
STATUS
approved