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.

1, 1, 5, 14, 47, 222, 319, 2132, 5637, 16270, 20417, 217284, 263111, 3323194, 3920925, 764392, 1768477, 29382138, 33464927, 622740028, 3502177707, 3436155514, 3825136961, 86449058184, 95405331155, 469336577606

Offset

1,3

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

Table of n, a(n) for n=1..26.

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

Formula

From Petros Hadjicostas, May 18 2020: (Start)

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.)

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

Prog

(PARI) u(n) = if(n<2, x, (n-1)^2/u(n-1)+1);
a(n) = polcoeff(numerator(u(n)), 1, x);
for(n=1, 30, print1(a(n), ", ")) (* Petros Hadjicostas, May 06 2020 *)

Crossrefs

Cf. A075828 (= b), A075829 (= d), A075830 (= c).

Cf. A024167, A024168, A334958.

Sequence in context: A081496 A152051 A220563 * A134418 A272147 A272223

Adjacent sequences: A075824 A075825 A075826 * A075828 A075829 A075830

Keyword

nonn

Author

Benoit Cloitre, Oct 14 2002

Extensions

Name edited by Petros Hadjicostas, May 06 2020

Status

approved