A142238 Numerators of continued fraction convergents to sqrt(3/2).

1, 5, 11, 49, 109, 485, 1079, 4801, 10681, 47525, 105731, 470449, 1046629, 4656965, 10360559, 46099201, 102558961, 456335045, 1015229051, 4517251249, 10049731549, 44716177445, 99482086439, 442644523201, 984771132841, 4381729054565, 9748229241971

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

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 b(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.

Cf. A000129, A001333, A142239, A153315-A153318. (End)

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

• Index entries for linear recurrences with constant coefficients, signature (0,10,0,-1).

Crossrefs

Cf. A115754, A142239, A382713.

Programs

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

Numerator[Convergents[Sqrt[3/2], 30]] (* Bruno Berselli, Nov 11 2013 *)
LinearRecurrence[{0,10,0,-1},{1,5,11,49},30] (* Harvey P. Dale, Dec 30 2017 *)

PARI

a(n)=([0,1,0,0; 0,0,1,0; 0,0,0,1; -1,0,10,0]^n*[1;5;11;49])[1,1] \\ Charles R Greathouse IV, Jun 21 2015

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(2n) = A041006(2n)/2 = A054320(n), a(2n-1) = A041006(2n-1) = A041038(2n-1) = A001079(n). - M. F. Hasler, Feb 14 2009