cp's OEIS Frontend

A363482 Denominator of the continued fraction 1/(2-3/(3-4/(4-5/(...(n-1)-n/(-5))))).

%I A363482 #29 Jul 13 2025 19:53:25
%S A363482 13,23,7,49,13,83,103,5,149,1,29,233,53,23,67,373,59,1,499,109,593,
%T A363482 643,139,107,1,863,71,197,1049,223,1,179,53,1399,59,1553,71,1,257,1,
%U A363482 1973,2063,431,173,67,349,2543,1,2749,571,2963,439,1,3299,683,3533,281,151,557,1,4153
%N A363482 Denominator of the continued fraction 1/(2-3/(3-4/(4-5/(...(n-1)-n/(-5))))).
%C A363482 Conjecture: Except for 49, every term of this sequence is either a prime or 1.
%C A363482 The conjecture holds through n=10000. The set of ordered primes appear to match A038901. - _Bill McEachen_, Jul 08 2025
%H A363482 Bill McEachen, <a href="/A363482/b363482.txt">Table of n, a(n) for n = 3..10000</a>
%H A363482 Mohammed Bouras, <a href="https://doi.org/10.5281/zenodo.10992128">The Distribution Of Prime Numbers And Continued Fractions</a>, (ppt) (2022)
%F A363482 a(n) = (n^2 + 3*n - 5)/gcd(n^2 + 3*n - 5, 5*A051403(n-3) + n*A051403(n-4)).
%F A363482 Except for n=6, if gpf(n^2 + 3*n - 5) > n, then we have:
%F A363482 a(n) = gpf(n^2 + 3*n - 5), where gpf = "greatest prime factor".
%F A363482 If a(n) = a(m) and n < m < a(n), then we have:
%F A363482 a(n) = n + m + 3.
%F A363482 a(n) divides gcd(n^2 + 3*n - 5, m^2 + 3*m - 5).
%e A363482 For n=3, 1/(2 - 3/(-5)) = 5/13, so a(3) = 13.
%e A363482 For n=4, 1/(2 - 3/(3 - 4/(-5))) = 19/23, so a(4) = 23.
%e A363482 For n=5, 1/(2 - 3/(3 - 4/(4 - 5/(-5)))) = 11/7, so a(5) = 7.
%o A363482 (PARI) lf(n) = sum(k=0, n-1, k!); \\ A003422
%o A363482 f(n) = (n+2)*lf(n+1)/2; \\ A051403
%o A363482 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 A363482 Cf. A006530, A038901, A051403.
%Y A363482 Cf. A362086, A363102.
%K A363482 nonn
%O A363482 3,1
%A A363482 _Mohammed Bouras_, Jun 04 2023