OFFSET
0,2
COMMENTS
Suppose that a positive irrational number r has continued fraction [a(0), a(1), ...]. Define sequences p(i), q(i), P(i), Q(i) from the numerators and denominators of finite continued fractions as follows:
p(i)/q(i) = [a(0), a(1), ... a(i)] and
P(i)/Q(i) = [a(0), a(1), ..., a(i) + 1].
The fractions p(i)/q(i) are the convergents to r, and the fractions P(i)/Q(i) are introduced here as the "other-side convergents" to r, because p(2k)/q(2k) < r < P(2k)/Q(2k) and P(2k+1)/Q(2k+1) < r < p(2k+1)/q(2k+1), for k >= 0.
Closeness of P(i)/Q(i) to r is indicated by |r - P(i)/Q(i)| < |p(i)/q(i) - P(i)/Q(i)| = 1/(q(i)Q(i)), for i >= 0.
LINKS
Clark Kimberling, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (0,4,0,-1).
FORMULA
p(i)*Q(i) - P(i)*q(i) = (-1)^(i+1), for i >= 0, where a(i) = Q(i).
a(n) = 4*a(n-2) - a(n-4) for n>3. - Colin Barker, Jul 21 2015
G.f.: -(x+1)*(x^2-x-1) / (x^4-4*x^2+1). - Colin Barker, Jul 21 2015
a(n) = 3^(n/2 + 1/2 - t)*((2 + sqrt(3))^t - (-1)^n*(2 - sqrt(3))^t)/2, where t = floor(n/2) + 1. - Ridouane Oudra, Aug 03 2021
EXAMPLE
For r = sqrt(3), the first 7 other-side convergents are 4, 25/8, 355/113, 688/219, 104348/33215, 208341/66317, 312689/99532. A comparison of convergents with other-side convergents:
i p(i)/q(i) P(i)/Q(i) p(i)*Q(i) - P(i)*q(i)
0 1/1 < sqrt(3) < 2/1 -1
1 2/1 > sqrt(3) > 3/2 1
2 5/3 < sqrt(3) < 7/4 -1
3 7/4 > sqrt(3) > 12/7 1
4 19/11 < sqrt(3) < 26/15 -1
5 26/15 > sqrt(3) > 45/26 1
MATHEMATICA
r = Sqrt[3]; a[i_] := Take[ContinuedFraction[r, 35], i];
b[i_] := ReplacePart[a[i], i -> Last[a[i]] + 1];
t = Table[FromContinuedFraction[b[i]], {i, 1, 35}]
u = Denominator[t]
PROG
(PARI) Vec(-(x+1)*(x^2-x-1)/(x^4-4*x^2+1) + O(x^50)) \\ Colin Barker, Jul 21 2015
CROSSREFS
KEYWORD
nonn,easy,frac
AUTHOR
Clark Kimberling, Jul 20 2015
STATUS
approved