A309330 Numbers k such that 10*k^2 + 40 is a square.

6, 234, 8886, 337434, 12813606, 486579594, 18477210966, 701647437114, 26644125399366, 1011775117738794, 38420810348674806, 1458979018131903834, 55402781878663670886, 2103846732371087589834, 79890773048222664742806

Offset: 1

Author

Greg Dresden, Jul 23 2019

Keywords

• nonn
• easy

Comments

Sequence of all positive integers k such that the continued fraction [k,k,k,k,k,k,...] belongs to Q(sqrt(10)).

As 10*n^2 + 40 = 10 * (n^2 + 4), n == 6 (mod 10) or n == 4 (mod 10) alternately. - Bernard Schott, Jul 24 2019

Examples

			a(2) = 234, and 10*234^2 + 40 is indeed a perfect square (it's 740^2) and furthermore the continued fraction [234, 234, 234, 234, ...] equals 117 + 37*sqrt(10), which is indeed in Q(sqrt(10)).
		

Links

• Colin Barker, Table of n, a(n) for n = 1..600
• Index entries for linear recurrences with constant coefficients, signature (38,-1).

Crossrefs

Cf. A097315.

Programs

Mathematica

LinearRecurrence[{38, -1}, {6, 234}, 15]

PARI

Vec(6*x*(1 + x) / (1 - 38*x + x^2) + O(x^20)) \\ Colin Barker, Jul 24 2019

Formula

a(n) = 38*a(n-1) - a(n-2); a(1) = 6, a(2) = 234.

a(n) = 2*sqrt(10*A097315(n-1)^2-1).

a(n) = (3-sqrt(10))*(19-6*sqrt(10))^(n-1) + (3+sqrt(10))*(19+6*sqrt(10))^(n-1). - Jinyuan Wang, Jul 24 2019

G.f.: 6*x*(1 + x) / (1 - 38*x + x^2). - Colin Barker, Jul 24 2019

a(n) = 6*A097314(n-1). - R. J. Mathar, Sep 06 2020