A075829 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 + c(n))/(d(n)*x + a(n)) (in lowest terms) and a(n), b(n), c(n), d(n) are positive integers.

1, 0, 1, 1, 5, 13, 23, 101, 307, 641, 893, 7303, 9613, 97249, 122989, 19793, 48595, 681971, 818107, 13093585, 77107553, 66022193, 76603673, 1529091919, 1752184789, 7690078169, 8719737569, 23184641107, 3721854001, 96460418429

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]

Difference between the denominator and the numerator of the (n-1)-th alternating harmonic number Sum_{k=1..n-1} (-1)^(k+1)*1/k = A058313(n-1)/A058312(n-1). - Alexander Adamchuk, Jul 22 2006

From Petros Hadjicostas, May 06 2020: (Start)

Inspired by Michael Somos's result below, we established the following formulas (valid for n >= 2). All the denominators in the first three formulas are equal to A334958(n).

b(n) = A024167(n)/gcd(A024167(n-1), A024167(n)).

c(n) = A024168(n)/gcd(A024168(n-1), A024168(n)).

d(n) = A024167(n-1)/gcd(A024167(n-1), A024167(n)).

b(n) + c(n) = n*(d(n) + a(n)).

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

Links

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

Crossrefs

Cf. A075827 (= b), A075828 (= c), A075830 (= d).

Cf. A024167, A024168, A058312, A058313, A334958.

Programs

Mathematica

Denominator[Table[Sum[(-1)^(k+1)*1/k,{k,1,n-1}],{n,1,30}]]-Numerator[Table[Sum[(-1)^(k+1)*1/k,{k,1,n-1}],{n,1,30}]] (* Alexander Adamchuk, Jul 22 2006 *)

PARI

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

Formula

a(n) = A024168(n-1)/gcd(A024168(n-1), A024168(n)). - Michael Somos, Oct 29 2002

From Alexander Adamchuk, Jul 22 2006: (Start)

a(n) = A058312(n-1) - A058313(n-1) for n > 1 with a(1) = 1.

a(n) = denominator(Sum_{k=1..n-1} (-1)^(k+1)*1/k) - numerator(Sum_{k=1..n-1}(-1)^(k+1)*1/k). (End)

Extensions

Name edited by Petros Hadjicostas, May 06 2020