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%I #28 May 21 2020 10:15:40
%S 0,1,1,10,13,138,101,1228,1923,8930,7303,115356,97249,1721846,1484475,
%T 388760,681971,14725926,13093585,308430212,1386466053,1685280806,
%U 1529091919,42052434936,38450390845,226713176794,208661769963
%N Let u(1) = x and u(n+1) = (n^2/u(n)) + 1 for n >= 1; then a(n) is such that u(n) =(b(n)*x + a(n))/(c(n)*x + d(n)) (in lowest terms) and a(n), b(n), c(n), d(n) are positive integers.
%C For x real <> 1 - 1/log(2), Lim_{n -> infinity} abs(u(n) - n) = abs((x - 1)/(1 + (x - 1)*log(2))). [Corrected by _Petros Hadjicostas_, May 18 2020]
%H Petros Hadjicostas, <a href="/A075829/a075829.pdf">Proofs of various results about the sequence u(n)</a>, 2020.
%F From _Petros Hadjicostas_, May 18 2020: (Start)
%F a(n) = A024168(n)/gcd(A024168(n), A024168(n-1)) = A024168(n)/A334958(n) for n >= 2. (Cf. Michael Somos's claim for d = A075829 using A024168.)
%F u(n) = (A024167(n)*x + A024168(n))/(A024167(n-1)*x + A024168(n-1)) for n >= 2. (End)
%o (PARI) u(n) = if(n<2, x, (n-1)^2/u(n-1)+1);
%o a(n) = polcoeff(numerator(u(n)), 0 ,x)
%Y Cf. A075827 (= b), A075829 (= d), A075830 (= c).
%Y Cf. A024167, A024168, A334958.
%K nonn
%O 1,4
%A _Benoit Cloitre_, Oct 14 2002
%E Name edited by _Petros Hadjicostas_, May 06 2020
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