A363347 - OEIS

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

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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

Conjectures 1 and 2 were proved by an autonomous AI agent, see the Lean file. The proof uses the fact that the continued-fraction denominator A363347(n) reduces to |n^2+2n-4| divided by its gcd with the numerator. Since p is congruent (+/-)1 (mod 10) makes 5 a quadratic residue, picking n=x-1 with x^2 congruent 5 forces p to divide n^2+2n-4; divisibility lemmas show the gcd cancels only the cofactor, leaving exactly p. - Ralf Stephan, Jun 09 2026

LINKS

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

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

Google Deepmind, AlphaProof Nexus: A363347 Lean file.

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.