A075828 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.

0, 1, 1, 10, 13, 138, 101, 1228, 1923, 8930, 7303, 115356, 97249, 1721846, 1484475, 388760, 681971, 14725926, 13093585, 308430212, 1386466053, 1685280806, 1529091919, 42052434936, 38450390845, 226713176794, 208661769963

Offset: 1

Author

Benoit Cloitre, Oct 14 2002

Keywords

• nonn

Comments

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]

Links

• Petros Hadjicostas, Proofs of various results about the sequence u(n), 2020.

Crossrefs

Cf. A075827 (= b), A075829 (= d), A075830 (= c).

Cf. A024167, A024168, A334958.

Programs

PARI

u(n) = if(n<2, x, (n-1)^2/u(n-1)+1);
a(n) = polcoeff(numerator(u(n)), 0 ,x)

Formula

From Petros Hadjicostas, May 18 2020: (Start)

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.)

u(n) = (A024167(n)*x + A024168(n))/(A024167(n-1)*x + A024168(n-1)) for n >= 2. (End)

Extensions

Name edited by Petros Hadjicostas, May 06 2020