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

11, 5, 31, 11, 59, 19, 19, 29, 139, 41, 191, 1, 251, 71, 29, 89, 79, 109, 479, 131, 571, 31, 61, 181, 41, 1, 179, 239, 1019, 271, 1151, 61, 1291, 1, 1439, 379, 1, 419, 1759, 461, 1931, 101, 2111, 1, 1, 599, 499, 59, 2699, 701, 71, 151, 101, 811

Offset

3,1

Comments

Conjecture 1: Every term of this sequence is either a prime or 1.

Conjecture 2: The sequence contains all prime numbers which end with a 1 or 9.

Conjecture 3: Except for 5, the primes all appear exactly twice.

Conjecture: The sequence of record values is A028877. - Bill McEachen, May 20 2024

Links

Table of n, a(n) for n=3..56.

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

Formula

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

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).

If n != m and a(n) = a(m) != 1, then we have:

a(n) = n + m + 2.

a(n) = gcd(n^2 + 2*n - 4, m^2 + 2*m - 4).

Example

For n=3, 1/(2 - 3/(-4)) = 4/11, so a(3) = 11.
For n=4, 1/(2 - 3/(3 - 4/(-4))) = 4/5, so a(4) = 5.
For n=5, 1/(2 - 3/(3 - 4/(4 -5/(-4)))) = 47/31, so a(5) = 31.
a(3) = a(6) = 3 + 6 + 2 = 11.
a(5) = a(24) = 5 + 24 + 2 = 31.
a(7) = a(50) = 7 + 50 + 2 = 59.

Crossrefs

Cf. A006530, A051403, A229525, A356247, A028877.

Sequence in context: A120831 A253254 A373730 * A229525 A174103 A038319

Adjacent sequences: A363344 A363345 A363346 * A363348 A363349 A363350

Keyword

nonn

Author

Mohammed Bouras, May 28 2023

Status

approved