|
|
A075830
|
|
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))/(a(n)*x + d(n)) (in lowest terms) and a(n), b(n), c(n), d(n) are positive integers.
|
|
6
|
|
|
0, 1, 1, 5, 7, 47, 37, 319, 533, 1879, 1627, 20417, 18107, 263111, 237371, 52279, 95549, 1768477, 1632341, 33464927, 155685007, 166770367, 156188887, 3825136961, 3602044091, 19081066231, 18051406831, 57128792093, 7751493599
(list;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
OFFSET
|
1,4
|
|
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]
Given x > 0, u(n) = (A075827(n)*x + A075828(n))/(a(n)*x + A075829(n)) = (b(n)*x + c(n))/(a(n)*x + d(n)) with gcd(gcd(b(n), c(n)), gcd(a(n), d(n))) = 1 for each n >= 1.
Conjecture 1: Define the sequences (A(n): n >= 1) and (B(n): n >= 1) by A(n+1) = n^2/A(n) + 1 for n >= 2 with A(1) = infinity and A(2) = 1, and B(n+1) = n^2/B(n) + 1 for n >= 3 with B(1) = 0, B(2) = infinity, and B(3) = 1. Then a(n) = denominator(A(n)), b(n) = numerator(A(n)), c(n) = numerator(B(n)), and d(n) = denominator(B(n)) (assuming infinity = 1/0). Also, gcd(a(n), d(n)) = 1.
Conjecture 3: b(p) = a(p+1) for p = 1 or prime. In general, it seems that b(n) = A048671(n)*a(n+1) for all n for which A048671(n) < n.
Conjecture 4: c(n) = n*(a(n) + d(n)) - b(n) for n >= 1. (End)
All conjectures are proved in the link below except for the second part of Conjecture 3. - Petros Hadjicostas, May 21 2020
|
|
LINKS
|
|
|
PROG
|
(PARI) u(n)=if(n<2, x, (n-1)^2/u(n-1)+1);
a(n)=polcoeff(denominator(u(n)), 1, x);
|
|
CROSSREFS
|
Apart from the leading term, same as A058313.
|
|
KEYWORD
|
nonn
|
|
AUTHOR
|
|
|
EXTENSIONS
|
|
|
STATUS
|
approved
|
|
|
|