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
KEYWORD
nonn,easy
AUTHOR
Greg Dresden, Jul 23 2019
STATUS
approved