|
|
A144229
|
|
The numerators of the convergents to the recursion x=1/(x^2+1).
|
|
1
|
|
|
|
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
|
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
|
|
|
KEYWORD
|
frac,nonn
|
|
AUTHOR
|
|
|
STATUS
|
approved
|
|
|
|