%I #35 May 09 2024 07:31:51
%S 3,17,9,13,53,23,29,107,43,17,179,23,79,269,101,113,29,139,1,503,61,
%T 199,647,233,251,809,17,103,43,1,373,1187,419,443,61,1,173,1637,191,
%U 601,1889,659,53,127,751,1,2447,283,883,2753,953,1,181,1063,367,263,131
%N Denominator of the continued fraction 1/(2-3/(3-4/(4-5/(...(n-1)-n/(-3))))).
%C Conjecture: Except for 9, every term of this sequence is either a prime or 1.
%C Conjecture: Record values correspond to A248697 (n>3). - _Bill McEachen_, Mar 06 2024
%H Mohammed Bouras, <a href="https://doi.org/10.5281/zenodo.10992128">The Distribution Of Prime Numbers And Continued Fractions</a>, (ppt) (2022)
%F a(n) = (n^2 + n - 3)/gcd(n^2 + n - 3, 3*A051403(n-3) + n*A051403(n-4)).
%F If gpf(n^2 + n - 3) > n, then we have:
%F a(n) = gpf(n^2 + n - 3), where gpf = "greatest prime factor".
%F If a(n) = a(m) and n < m < a(n), then we have:
%F a(n) = n + m + 1.
%F a(n) divides gcd(n^2 + n - 3, m^2 + m - 3).
%e For n=3, 1/(2 - 3/(-3)) = 1/3, so a(3) = 3.
%e For n=4, 1/(2 - 3/(3 - 4/(-3))) = 13/17, so a(4) = 17.
%e For n=5, 1/(2 - 3/(3 - 4/(4 - 5/(-3)))) = 13/9, so a(5) = 9.
%e a(4) = a(12) = 4 + 12 + 1 = 17.
%e a(7) = a(45) = 7 + 45 + 1 = 53.
%Y Cf. A006530, A014209, A051403, A248697.
%K nonn
%O 3,1
%A _Mohammed Bouras_, May 28 2023