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%I #12 Jul 30 2018 19:35:06
%S 1,11,23,241,505,5291,11087,116161,243409,2550251,5343911,55989361,
%T 117322633,1229215691,2575754015,26986755841,56549265697,592479412811,
%U 1241508091319,13007560326001
%N Numerators of continued fraction convergents to sqrt(6/5).
%C In general, denominators, a(k,n) and numerators, b(k,n), of continued
%C fraction convergents to sqrt((k+1)/k) may be found as follows:
%C a(k,0) = 1, a(k,1) = 2k; for n>0, a(k,2n)=2*a(k,2n-1)+a(k,2n-2)
%C and a(k,2n+1)=(2k)*a(k,2n)+a(k,2n-1);
%C b(k,0) = 1, b(k,1) = 2k+1; for n>0, b(k,2n)=2*b(k,2n-1)+b(k,2n-2)
%C and b(k,2n+1)=(2k)*b(k,2n)+b(k,2n-1).
%C For example, the convergents to sqrt(4/3) start 1/1, 11/10, 23/21,
%C 241/220, 505/461.
%C In general, if a(k,n) and b(k,n) are the denominators and numerators,
%C respectively, of continued fraction convergents to sqrt((k+1)/k)
%C as defined above, then
%C k*a(k,2n)^2-a(k,2n-1)*a(k,2n+1)=k=k*a(k,2n-2)*a(k,2n)-a(k,2n-1)^2 and
%C b(k,2n-1)*b(k,2n+1)-k*b(k,2n)^2=k+1=b(k,2n-1)^2-k*b(k,2n-2)*b(k,2n);
%C for example, if k=5 and n=3, then b(5,n)=a(n) and
%C 5*a(5,6)^2-a(5,5)*a(5,7)=5*10121^2-4830*106040=5;
%C 5*a(5,4)*a(5,6)-a(5,5)^2=5*461*10121-4830^2=5;
%C b(5,5)*b(5,7)-5*b(5,6)^2=5291*116161-5*11087^2=6;
%C b(5,5)^2-5*b(5,4)*b(5,6)=5291^2-5*505*11087=6.
%H <a href="/index/Rec#order_04">Index entries for linear recurrences with constant coefficients</a>, signature (0, 22, 0, -1).
%F For n>0, a(2n)=2a(2n-1)+a(2n-2) and a(2n+1)=10a(2n)+a(2n-1).
%F Empirical G.f.: (1+11*x+x^2-x^3)/(1-22*x^2+x^4) [Colin Barker, Jan 01 2012]
%e The initial convergents are 1, 11/10, 23/21, 241/220,
%e 505/461, 5291/4830, 11087/10121, 116161/106040,
%e 243409/222201, 2550251/2328050, 55989361/4878301,
%t Numerator[Convergents[Sqrt[6/5],20]] (* or *) LinearRecurrence[{0,22,0,-1},{1,11,23,241},20] (* _Harvey P. Dale_, Jul 30 2018 *)
%Y Cf. A000129, A001333, A142238-A142239, A153313-153318.
%K nonn
%O 0,2
%A _Charlie Marion_, Jan 07 2009
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