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Denominator of the continued fraction 1/(2-3/(3-4/(4-5/(...(n-1)-n/(-2))))).
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%I #43 Aug 06 2024 22:00:31

%S 7,7,23,17,47,31,79,7,17,71,167,97,223,127,41,23,359,199,439,241,31,

%T 41,89,337,727,1,839,449,137,73,1087,577,1223,647,1367,103,1,47,73,

%U 881,1847,967,1,151,2207,1151,2399,1249,113,193,401,1,3023,1567,191,41,71,257,3719,113,3967,89,103,311

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

%C Conjecture 1: The sequence contains only 1's and primes.

%C Conjecture 2: All prime numbers appear either twice (same as A356247 and A357127) or three times.

%C Similar terms of A164314.

%C Conjecture: Record values correspond to A028871(m), m > 1. - _Bill McEachen_, Mar 06 2024

%C a(n) = 1 positions appear to correspond to A060515(m), m > 2. - _Bill McEachen_, Aug 05 2024

%H Bill McEachen, <a href="/A363102/b363102.txt">Table of n, a(n) for n = 3..10002</a>

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

%F a(n) = A164314(n) if A164314(n) > n.

%F If a(n) = a(m) and n < m < a(n), then a(n) = n + m.

%e a(5) = (5^2 - 2)/gcd(5^2 - 2, 2*A051403(5-3) + 5*A051403(5-4))= 23.

%e a(6) = a(11) = 6 + 11 = 17.

%e a(7) = a(40) = 7 + 40 = 47.

%o (PARI) a051403(n) = (n+2)*sum(k=0, n, k!)/2;

%o a(n) = (n^2 - 2)/gcd(n^2 - 2, 2*a051403(n-3) + n*a051403(n-4)); \\ _Michel Marcus_, May 24 2023

%Y Cf. A008865, A051403, A059772, A164314, A356247, A357127.

%K nonn

%O 3,1

%A _Mohammed Bouras_, May 19 2023