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%I #24 Sep 06 2020 13:44:42
%S 6,234,8886,337434,12813606,486579594,18477210966,701647437114,
%T 26644125399366,1011775117738794,38420810348674806,
%U 1458979018131903834,55402781878663670886,2103846732371087589834,79890773048222664742806
%N Numbers k such that 10*k^2 + 40 is a square.
%C Sequence of all positive integers k such that the continued fraction [k,k,k,k,k,k,...] belongs to Q(sqrt(10)).
%C As 10*n^2 + 40 = 10 * (n^2 + 4), n == 6 (mod 10) or n == 4 (mod 10) alternately. - _Bernard Schott_, Jul 24 2019
%H Colin Barker, <a href="/A309330/b309330.txt">Table of n, a(n) for n = 1..600</a>
%H <a href="/index/Rec#order_02">Index entries for linear recurrences with constant coefficients</a>, signature (38,-1).
%F a(n) = 38*a(n-1) - a(n-2); a(1) = 6, a(2) = 234.
%F a(n) = 2*sqrt(10*A097315(n-1)^2-1).
%F 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
%F G.f.: 6*x*(1 + x) / (1 - 38*x + x^2). - _Colin Barker_, Jul 24 2019
%F a(n) = 6*A097314(n-1). - _R. J. Mathar_, Sep 06 2020
%e 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)).
%t LinearRecurrence[{38, -1}, {6, 234}, 15]
%o (PARI) Vec(6*x*(1 + x) / (1 - 38*x + x^2) + O(x^20)) \\ _Colin Barker_, Jul 24 2019
%Y Cf. A097315.
%K nonn,easy
%O 1,1
%A _Greg Dresden_, Jul 23 2019
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