A356360 Denominator of the continued fraction 1/(2-3/(3-4/(4-5/(...(n-1)-n/(n+1))))).

5, 7, 3, 11, 13, 1, 17, 19, 1, 23, 1, 1, 29, 31, 1, 1, 37, 1, 41, 43, 1, 47, 1, 1, 53, 1, 1, 59, 61, 1, 1, 67, 1, 71, 73, 1, 1, 79, 1, 83, 1, 1, 89, 1, 1, 1, 97, 1, 101, 103, 1, 107, 109, 1, 113, 1, 1, 1, 1, 1, 1, 127, 1, 131, 1, 1, 137, 139, 1, 1, 1, 1, 149, 151, 1, 1, 157, 1, 1, 163, 1, 167

Offset: 3

Author

Mohammed Bouras, Oct 15 2022

Keywords

• nonn

Comments

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

Similar continued fraction to A356247.

Same as A128059(n), A145737(n-1) and A097302(n-2) for n > 5.

a(n) = 1 positions appear to correspond to A104275(m), m > 2. Conjecture: all odd primes are seen in order after 11. - Bill McEachen, Aug 05 2024

Links

• Mohammed Bouras, The Distribution Of Prime Numbers And The Continued Fractions, (2022).

Crossrefs

Cf. A051403, A051417, A097302, A128059, A145737, A356247.

Cf. A104275.

Programs

Python

from math import gcd, factorial
def A356360(n): return (a:=(n<<1)-1)//gcd(a, a*sum(factorial(k) for k in range(n-2))+n*factorial(n-2)>>1) # Chai Wah Wu, Feb 26 2024

Formula

For n >= 3, the formula of the continued fraction is as follows:

(A051403(n-2) + A051403(n-3))/(2n - 1) = 1/(2-3/(3-4/(4-5/(...(n-1)-n/(n+1))))).

a(n) = (2n - 1)/gcd(2n - 1, A051403(n-2) + A051403(n-3)).

From the conjecture: Except for n = 5, a(n)= 2n - 1 if 2n-1 is prime, 1 otherwise.