A153316 Numerators of continued fraction convergents to sqrt(5/4).

1, 9, 19, 161, 341, 2889, 6119, 51841, 109801, 930249, 1970299, 16692641, 35355581, 299537289, 634430159, 5374978561, 11384387281, 96450076809, 204284540899, 1730726404001, 3665737348901, 31056625195209, 65778987739319, 557288527109761, 1180356041958841, 10000136862780489

Offset: 0

Author

Charlie Marion, Jan 07 2009

Keywords

• nonn
• easy
• changed

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 b(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, 1841/46368, 109801/98209, 930249/832040, 1970299/1762289.
		

Links

• Colin Barker, Table of n, a(n) for n = 0..1000
• Index entries for linear recurrences with constant coefficients, signature (0,18,0,-1).

Crossrefs

Cf. A000129, A001333, A142238, A142239, A153315, A153317, A153318.

Programs

Mathematica

Numerator[Convergents[Sqrt[5/4], 30]] (* Paolo Xausa, Jan 16 2025 *)

PARI

Vec((1+9*x+x^2-x^3)/((1+4*x-x^2)*(1-4*x-x^2)) + O(x^30)) \\ Colin Barker, Mar 27 2016

Formula

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

G.f.: (1 + 9*x + x^2 - x^3) / ((1 + 4*x - x^2)*(1 - 4*x - x^2)). - Colin Barker, Jan 01 2012

From Colin Barker, Mar 27 2016: (Start)

a(n) = ((5*(-2+sqrt(5))^n - 2*sqrt(5)*(-2+sqrt(5))^n + 15*(2+sqrt(5))^n + 6*sqrt(5)*(2+sqrt(5))^n + 3*(2-sqrt(5))^n*(-5+2*sqrt(5)) - (-2-sqrt(5))^n*(5+2*sqrt(5))))/(8*sqrt(5)).

a(n) = 18*a(n-2) - a(n-4) for n > 3. (End)

a(n) = (3 - (-1)^n)*Lucas(3*(n + 1))/8. - Ehren Metcalfe, Apr 04 2019