A142239 Denominators of continued fraction convergents to sqrt(3/2).

1, 4, 9, 40, 89, 396, 881, 3920, 8721, 38804, 86329, 384120, 854569, 3802396, 8459361, 37639840, 83739041, 372596004, 828931049, 3688320200, 8205571449, 36510605996, 81226783441, 361417739760, 804062262961, 3577666791604, 7959395846169, 35415250176280

Offset: 0

Author

N. J. A. Sloane, Oct 05 2008, following a suggestion from Rob Miller (rmiller(AT)AmtechSoftware.net)

Keywords

• nonn
• frac
• easy

Comments

sqrt(3/2) = 1.224744871... = 2/2 + 2/9 + 2/(9*89) + 2/(89*881) + 2/(881*8721) + 2/(8721*86329) + ... - Gary W. Adamson, Oct 08 2008

From Charlie Marion, Jan 07 2009: (Start)

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(3/2) start 1/1, 5/4, 11/9, 49/40, 109/89.

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=2 and n=3, then a(2,n)=a(n) and

2*a(2,6)^2 - a(2,5)*a(2,7) = 2*881^2 - 396*3920 = 2;

2*a(2,4)*a(2,6) - a(2,5)^2 = 2*89*881 - 396^2 = 2;

b(2,5)*b(2,7) - 2*b(2,6)^2 = 485*4801 - 2*1079^2 = 3;

b(2,5)^2 - 2*b(2,4)*b(2,6) = 485^2 - 2*109*1079 = 3.

(End)

For n > 0, a(n) equals the permanent of the n X n tridiagonal matrix with the main diagonal alternating sequence [4, 2, 4, 2, 4, ...] and 1's along the superdiagonal and the subdiagonal. - Rogério Serôdio, Apr 01 2018

Examples

			The initial convergents are 1, 5/4, 11/9, 49/40, 109/89, 485/396, 1079/881, 4801/3920, 10681/8721, 47525/38804, 105731/86329, ...
		

Links

• Vincenzo Librandi, Table of n, a(n) for n = 0..200
• Index entries for linear recurrences with constant coefficients, signature (0,10,0,-1).

Crossrefs

Cf. A115754, A142238.

Cf. A000129, A001333, A142238, A153315, A153316, A153317, A153318, A382713.

Programs

Magma

I:=[1,4,9,40]; [n le 4 select I[n] else 10*Self(n-2)-Self(n-4): n in [1..30]]; // Vincenzo Librandi, Feb 01 2014

Maple

with(numtheory): cf := cfrac (sqrt(3)/sqrt(2),100): [seq(nthnumer(cf,i), i=0..50)]; [seq(nthdenom(cf,i), i=0..50)]; [seq(nthconver(cf,i), i=0..50)];

Mathematica

Table[Denominator[FromContinuedFraction[ContinuedFraction[Sqrt[3/2], n]]], {n, 1, 50}] (* Vladimir Joseph Stephan Orlovsky, Jun 23 2011 *)
Denominator[Convergents[Sqrt[3/2], 30]] (* Bruno Berselli, Nov 11 2013 *)

Formula

G.f.'s for numerators and denominators are -(1+5*x+x^2-x^3)/(-1-x^4+10*x^2) and -(1+4*x-x^2)/(-1-x^4+10*x^2).

a(n) = 10*a(n-2) - a(n-4) for n > 3. - Vincenzo Librandi, Feb 01 2014

From: Rogério Serôdio, Apr 02 2018: (Start)

Recurrence formula: a(n) = (3-(-1)^n)*a(n-1) + a(n-2), a(0) = 1, a(1) = 4;

Some properties:

(1) a(n)^2 - a(n-2)^2 = (3-(-1)^n)*a(2*n-1), for n > 1;

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

(3) a(2*n) = A041007(2*n);

(4) a(2*n+1) = 2*A041007(2*n+1). (End)