A295333 - OEIS

%I #18 Jun 04 2019 11:51:52

%S 1,2,3,8,11,19,49,68,117,302,419,721,1861,2582,4443,11468,15911,27379,

%T 70669,98048,168717,435482,604199,1039681,2683561,3723242,6406803,

%U 16536848,22943651,39480499,101904649,141385148,243289797,627964742,871254539,1499219281,3869693101,5368912382,9238605483

%N Numerators of continued fraction convergents to sqrt(10)/2 = sqrt(5/2) = A020797 + 1.

%C The denominators are given in A295334.

%C The regular continued fraction expansion of sqrt(10)/2 is [1, repeat(1, 1, 2)].

%H Robert Israel, <a href="/A295333/b295333.txt">Table of n, a(n) for n = 0..3794</a>

%H <a href="/index/Rec#order_06">Index entries for linear recurrences with constant coefficients</a>, signature (0,0,6,0,0,1).

%F G.f.: G(x) = (1 + 2*x + 3*x^2 + 2*x^3 - x^4 + x^5)/(1 - 6*x^3 - x^6). From the recurrence a(n) = b(n)*a(n-1) + a(n-2), with the trisection b(3*(k+1)) = 2, b(3*k+1) = 1 = b(3*k+2), k >= 0, b(0) = 1, and the input a(0) = 1 = a(-1). With G_j(x) = Sum_{k>=0} a(3*k+j)*x^k, for j = 0,1,2, one finds (omitting here the G_j arguments) G_0 = 1 + 2*x*G_2 + x*G_1, G_1 = G_0 + 1 + x*G_2, G_2 = G_1 + G_0. This can be solved and leads to the given formula for G(x) = Sum_{j=0..2} x^j*G_j(x^3).

%F a(n) = 6*a(n-3) + a(n-6), for n >= 6, with inputs a(0)..a(5).

%e The convergents a(n)/A295334(n) begin: 1, 2, 3/2, 8/5, 11/7, 19/12, 49/31, 68/43, 117/74, 302/191, 419/265, 721/456, 1861/1177, 2582/1633, 4443/2810, 11468/7253, 15911/10063, 27379/17316, 70669/44695, 98048/62011, ...

%p numtheory:-cfrac(sqrt(5/2),100,'con'):

%p map(numer,con[1..-2]); # _Robert Israel_, Nov 22 2017

%t Numerator[Convergents[Sqrt[5/2], 50]] (* _Vaclav Kotesovec_, Nov 22 2017 *)

%t LinearRecurrence[{0,0,6,0,0,1},{1,2,3,8,11,19},40] (* _Harvey P. Dale_, Apr 08 2019 *)

%Y Cf. A020797, A295334.

%K nonn,frac,cofr,easy

%O 0,2

%A _Wolfdieter Lang_, Nov 21 2017