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%I #38 Jan 17 2023 21:24:30
%S 1,5,11,49,109,485,1079,4801,10681,47525,105731,470449,1046629,
%T 4656965,10360559,46099201,102558961,456335045,1015229051,4517251249,
%U 10049731549,44716177445,99482086439,442644523201,984771132841,4381729054565,9748229241971
%N Numerators of continued fraction convergents to sqrt(3/2).
%C From _Charlie Marion_, Jan 07 2009: (Start)
%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(3/2) start 1/1, 5/4, 11/9,
%C 49/40, 109/89.
%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=2 and n=3, then b(2,n)=a(n) and
%C 2*a(2,6)^2-a(2,5)*a(2,7)=2*881^2-396*3920=2;
%C 2*a(2,4)*a(2,6)-a(2,5)^2=2*89*881-396^2=2;
%C b(2,5)*b(2,7)-2*b(2,6)^2=485*4801-2*1079^2=3;
%C b(2,5)^2-2*b(2,4)*b(2,6)=485^2-2*109*1079=3.
%C Cf. A000129, A001333, A142239, A153313-153318. (End)
%H <a href="/index/Rec#order_04">Index entries for linear recurrences with constant coefficients</a>, signature (0,10,0,-1).
%F G.f.'s for numerators and denominators are -(1+5*x+x^2-x^3)/(-1-x^4+10*x^2) and -(1+4*x-x^2)/(-1-x^4+10*x^2).
%F a(2n) = A041006(2n)/2 = A054320(n), a(2n-1) = A041006(2n-1) = A041038(2n-1) = A001079(n). - _M. F. Hasler_, Feb 14 2009
%e The initial convergents are 1, 5/4, 11/9, 49/40, 109/89, 485/396, 1079/881, 4801/3920, 10681/8721, 47525/38804, 105731/86329, ...
%p with(numtheory): cf := cfrac (sqrt(3)/sqrt(2),100): [seq(nthnumer(cf,i), i=0..50)]; [seq(nthdenom(cf,i), i=0..50)]; [seq(nthconver(cf,i), i=0..50)];
%t Numerator[Convergents[Sqrt[3/2], 30]] (* _Bruno Berselli_, Nov 11 2013 *)
%t LinearRecurrence[{0,10,0,-1},{1,5,11,49},30] (* _Harvey P. Dale_, Dec 30 2017 *)
%o (PARI) a(n)=([0,1,0,0; 0,0,1,0; 0,0,0,1; -1,0,10,0]^n*[1;5;11;49])[1,1] \\ _Charles R Greathouse IV_, Jun 21 2015
%Y Cf. A115754, A142239.
%K nonn,frac,easy
%O 0,2
%A _N. J. A. Sloane_, Oct 05 2008, following a suggestion from Rob Miller (rmiller(AT)AmtechSoftware.net)
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