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A309330 Numbers k such that 10*k^2 + 40 is a square. 1
6, 234, 8886, 337434, 12813606, 486579594, 18477210966, 701647437114, 26644125399366, 1011775117738794, 38420810348674806, 1458979018131903834, 55402781878663670886, 2103846732371087589834, 79890773048222664742806 (list; graph; refs; listen; history; text; internal format)
OFFSET

1,1

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

LINKS

Colin Barker, Table of n, a(n) for n = 1..600

Index entries for linear recurrences with constant coefficients, signature (38,-1).

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

EXAMPLE

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)).

MATHEMATICA

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

PROG

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

CROSSREFS

Cf. A097315.

Sequence in context: A324232 A307888 A194482 * A266657 A145180 A256275

Adjacent sequences:  A309327 A309328 A309329 * A309331 A309332 A309333

KEYWORD

nonn,easy

AUTHOR

Greg Dresden, Jul 23 2019

STATUS

approved

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Last modified October 6 12:35 EDT 2022. Contains 357264 sequences. (Running on oeis4.)