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Bisection of the numerators of the convergents of cf (1,1,6,1,6,1,...,6,1).
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%I #50 Mar 25 2021 05:13:51

%S 1,2,15,118,929,7314,57583,453350,3569217,28100386,221233871,

%T 1741770582,13712930785,107961675698,849980474799,6691882122694,

%U 52685076506753,414788729931330,3265624762943887,25710209373619766,202416050226014241,1593618192434494162,12546529489249939055,98778617721565018278

%N Bisection of the numerators of the convergents of cf (1,1,6,1,6,1,...,6,1).

%C 15*a(n)^2 - 11 is a square for all terms.

%C x = a(n) and y = a(n+1) satisfy the equation x^2 + y^2 - 8*x*y = -11.

%C x = a(n) and y = a(n+2) satisfy x^2 + y^2 - 62*x*y = -704.

%H <a href="/index/Rec#order_02">Index entries for linear recurrences with constant coefficients</a>, signature (8,-1).

%F a(n) = 8*a(n-1) - a(n-2) for n >= 2.

%F a(n) = A237262(2*n) for n >= 1.

%F G.f.: (1 - 6*x)/(1 - 8*x + x^2). - _Stefano Spezia_, Mar 01 2021

%e a(3) = 8*15 - 2 = 118.

%t LinearRecurrence [{8, -1}, {1,2}, 15]

%o (PARI) my(p=Mod('x,'x^2-8*'x+1)); a(n) = subst(lift(p^n),'x,2); \\ _Kevin Ryde_, Feb 27 2021

%Y After a(0), bisection of A237262.

%Y Cf. A341927.

%K nonn,easy

%O 0,2

%A _John O. Oladokun_, Feb 23 2021