%I #23 May 05 2024 19:46:45

%S 13,23,7,49,13,83,103,5,149,1,29,233,53,23,67,373,59,1,499,109,593,

%T 643,139,107,1,863,71,197,1049,223,1,179,53,1399,59,1553,71,1,257,1,

%U 1973,2063,431,173,67,349,2543,1,2749,571,2963,439,1,3299,683,3533,281,151,557,1,4153

%N Denominator of the continued fraction 1/(2-3/(3-4/(4-5/(...(n-1)-n/(-5))))).

%C Conjecture: Except for 49, every term of this sequence is either a prime or 1.

%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 + 3*n - 5)/gcd(n^2 + 3*n - 5, 5*A051403(n-3) + n*A051403(n-4)).

%F Except for n=6, if gpf(n^2 + 3*n - 5) > n, then we have:

%F a(n) = gpf(n^2 + 3*n - 5), where gpf = "greatest prime factor".

%F If a(n) = a(m) and n < m < a(n), then we have:

%F a(n) = n + m + 3.

%F a(n) divides gcd(n^2 + 3*n - 5, m^2 + 3*m - 5).

%e For n=3, 1/(2 - 3/(-5)) = 5/13, so a(3) = 13.

%e For n=4, 1/(2 - 3/(3 - 4/(-5))) = 19/23, so a(4) = 23.

%e For n=5, 1/(2 - 3/(3 - 4/(4 - 5/(-5)))) = 11/7, so a(5) = 7.

%o (PARI) lf(n) = sum(k=0, n-1, k!); \\ A003422

%o f(n) = (n+2)*lf(n+1)/2; \\ A051403

%o a(n) = (n^2 + 3*n - 5)/gcd(n^2 + 3*n - 5, 5*f(n-3) + n*f(n-4)); \\ _Michel Marcus_, Jun 06 2023

%Y Cf. A006530, A051403.

%Y Cf. A362086, A363102.

%K nonn

%O 3,1

%A _Mohammed Bouras_, Jun 04 2023