A153315 Denominators of continued fraction convergents to sqrt(5/4).

1, 8, 17, 144, 305, 2584, 5473, 46368, 98209, 832040, 1762289, 14930352, 31622993, 267914296, 567451585, 4807526976, 10182505537, 86267571272, 182717648081, 1548008755920, 3278735159921, 27777890035288, 58834515230497, 498454011879264, 1055742538989025, 8944394323791464

Offset: 0

Author

Charlie Marion, Jan 07 2009

Keywords

• nonn
• frac
• easy

Comments

In general, denominators, a(k,n) and numerators, b(k,n), of continued fraction convergents to sqrt((k+1)/k) may be found as follows:

a(k,0) = 1, a(k,1) = 2k; for n > 0, a(k,2n) = 2*a(k,2n-1) + a(k,2n-2)

and a(k,2n+1) = (2k)*a(k,2n) + a(k,2n-1);

b(k,0) = 1, b(k,1) = 2k+1; for n > 0, b(k,2n) = 2*b(k,2n-1) + b(k,2n-2)

and b(k,2n+1) = (2k)*b(k,2n) + b(k,2n-1).

For example, the convergents to sqrt(4/3) start 1/1, 9/8, 19/17, 161/144, 341/305.

In general, if a(k,n) and b(k,n) are the denominators and numerators, respectively, of continued fraction convergents to sqrt((k+1)/k) as defined above, then

k*a(k,2n)^2 - a(k,2n-1)*a(k,2n+1) = k = k*a(k,2n-2)*a(k,2n) - a(k,2n-1)^2 and

b(k,2n-1)*b(k,2n+1) - k*b(k,2n)^2 = k+1 = b(k,2n-1)^2 - k*b(k,2n-2)*b(k,2n);

for example, if k=4 and n=3, then a(4,n)=a(n) and

4*a(4,6)^2 - a(4,5)*a(4,7) = 4*5473^2 - 2584*46368 = 4;

4*a(4,4)*a(4,6) - a(4,5)^2 = 4*305*5473 - 2584^2 = 4;

b(4,5)*b(4,7) - 4*b(4,6)^2 = 2889*51841 - 4*6119^2 = 5;

b(4,5)^2 - 4*b(4,4)*b(4,6) = 2889^2 - 4*341*6119 = 5.

Examples

			The initial convergents are 1, 9/8, 19/17, 161/144, 341/305, 2889/2584, 6119/5473, 51841/46368, 109801/98209, 930249/832040, 1970299/1762289, ...
		

Links

• Paolo Xausa, Table of n, a(n) for n = 0..1000 (terms 0..200 from Vincenzo Librandi)
• Index entries for linear recurrences with constant coefficients, signature (0,18,0,-1).

Crossrefs

Cf. A000129, A001333, A142238, A142239, A153316, A153317, A153318.

Programs

Mathematica

Denominator[Convergents[Sqrt[5/4], 30]] (* Harvey P. Dale, Aug 17 2012 *)

Formula

For n > 0, a(2n) = 2a(2n-1) + a(2n-2) and a(2n+1) = 8a(2n) + a(2n-1).

Empirical g.f.: (1 + 8*x - x^2)/(1 - 18*x^2 + x^4). - Colin Barker, Jan 01 2012

a(n) = (3 - (-1)^n)*Fibonacci(3*(n + 1))/4. - Ehren Metcalfe, Apr 04 2019

Extensions

Corrected and extended by Harvey P. Dale, Aug 17 2012