cp's OEIS Frontend

A363347 Denominator of the continued fraction 1/(2-3/(3-4/(4-5/(...(n-1)-n/(-4))))).

%I A363347 #29 May 21 2024 07:02:58
%S A363347 11,5,31,11,59,19,19,29,139,41,191,1,251,71,29,89,79,109,479,131,571,
%T A363347 31,61,181,41,1,179,239,1019,271,1151,61,1291,1,1439,379,1,419,1759,
%U A363347 461,1931,101,2111,1,1,599,499,59,2699,701,71,151,101,811
%N A363347 Denominator of the continued fraction 1/(2-3/(3-4/(4-5/(...(n-1)-n/(-4))))).
%C A363347 Conjecture 1: Every term of this sequence is either a prime or 1.
%C A363347 Conjecture 2: The sequence contains all prime numbers which end with a 1 or 9.
%C A363347 Conjecture 3: Except for 5, the primes all appear exactly twice.
%C A363347 Conjecture: The sequence of record values is A028877. - _Bill McEachen_, May 20 2024
%H A363347 Mohammed Bouras, <a href="https://doi.org/10.5281/zenodo.10992128">The Distribution Of Prime Numbers And Continued Fractions</a>, (ppt) (2022)
%F A363347 a(n) = (n^2 + 2*n - 4)/gcd(n^2 + 2*n - 4, 4*A051403(n-3) + n*A051403(n-4)).
%F A363347 a(n) = gpf(n^2 + 2*n - 4) if gpf(n^2 + 2*n - 4) > n, otherwise a(n) = 1 (where gpf(n) denotes the greatest prime factor of n).
%F A363347 If n != m and a(n) = a(m) != 1, then we have:
%F A363347 a(n) = n + m + 2.
%F A363347 a(n) = gcd(n^2 + 2*n - 4, m^2 + 2*m - 4).
%e A363347 For n=3, 1/(2 - 3/(-4)) = 4/11, so a(3) = 11.
%e A363347 For n=4, 1/(2 - 3/(3 - 4/(-4))) = 4/5, so a(4) = 5.
%e A363347 For n=5, 1/(2 - 3/(3 - 4/(4 -5/(-4)))) = 47/31, so a(5) = 31.
%e A363347 a(3) = a(6) = 3 + 6 + 2 = 11.
%e A363347 a(5) = a(24) = 5 + 24 + 2 = 31.
%e A363347 a(7) = a(50) = 7 + 50 + 2 = 59.
%Y A363347 Cf. A006530, A051403, A229525, A356247, A028877.
%K A363347 nonn
%O A363347 3,1
%A A363347 _Mohammed Bouras_, May 28 2023