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A075827 Let u(1) = x and u(n+1) = (n^2/u(n)) + 1 for n >= 1; then a(n) is such that u(n) =(a(n)*x + b(n))/(c(n)*x + d(n)) (in lowest terms) and a(n), b(n), c(n), d(n) are positive integers.

%I A075827 #34 Aug 01 2025 07:12:42
%S A075827 1,1,5,14,47,222,319,2132,5637,16270,20417,217284,263111,3323194,
%T A075827 3920925,764392,1768477,29382138,33464927,622740028,3502177707,
%U A075827 3436155514,3825136961,86449058184,95405331155,469336577606,514159128837,1519292745404,236266661971,6755272778730,7313175618421
%N A075827 Let u(1) = x and u(n+1) = (n^2/u(n)) + 1 for n >= 1; then a(n) is such that u(n) =(a(n)*x + b(n))/(c(n)*x + d(n)) (in lowest terms) and a(n), b(n), c(n), d(n) are positive integers.
%C A075827 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 A075827 Petros Hadjicostas, <a href="/A075829/a075829.pdf">Proofs of various results about the sequence u(n)</a>, 2020.
%F A075827 From _Petros Hadjicostas_, May 18 2020: (Start)
%F A075827 a(n) = A024167(n)/gcd(A024167(n), A024167(n-1)) = A024167(n)/A334958(n-1) for n >= 2. (Cf. _Michael Somos_'s result for d = A075829 using A024168.)
%F A075827 u(n) = (A024167(n)*x + A024168(n))/(A024167(n-1)*x + A024168(n-1)) for n >= 2. (End)
%o A075827 (PARI) u(n) = if(n<2, x, (n-1)^2/u(n-1)+1);
%o A075827 a(n) = polcoeff(numerator(u(n)), 1, x);
%o A075827 for(n=1, 30, print1(a(n)", ")) \\ _Petros Hadjicostas_, May 06 2020
%Y A075827 Cf. A075828 (= b), A075829 (= d), A075830 (= c).
%Y A075827 Cf. A024167, A024168, A334958.
%K A075827 nonn
%O A075827 1,3
%A A075827 _Benoit Cloitre_, Oct 14 2002
%E A075827 Name edited by _Petros Hadjicostas_, May 06 2020
%E A075827 More terms from _Michel Marcus_, Aug 01 2025