A362086 Denominator of the continued fraction 1/(2-3/(3-4/(4-5/(...(n-1)-n/(-3))))).

3, 17, 9, 13, 53, 23, 29, 107, 43, 17, 179, 23, 79, 269, 101, 113, 29, 139, 1, 503, 61, 199, 647, 233, 251, 809, 17, 103, 43, 1, 373, 1187, 419, 443, 61, 1, 173, 1637, 191, 601, 1889, 659, 53, 127, 751, 1, 2447, 283, 883, 2753, 953, 1, 181, 1063, 367, 263, 131

Offset: 3

Author

Mohammed Bouras, May 28 2023

Keywords

• nonn

Comments

Conjecture: Except for 9, every term of this sequence is either a prime or 1.

Conjecture: Record values correspond to A248697 (n>3). - Bill McEachen, Mar 06 2024

Examples

			For n=3, 1/(2 - 3/(-3)) = 1/3, so a(3) = 3.
For n=4, 1/(2 - 3/(3 - 4/(-3))) = 13/17, so a(4) = 17.
For n=5, 1/(2 - 3/(3 - 4/(4 - 5/(-3)))) = 13/9, so a(5) = 9.
a(4) = a(12) = 4 + 12 + 1 = 17.
a(7) = a(45) = 7 + 45 + 1 = 53.
		

Links

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

Crossrefs

Cf. A006530, A014209, A051403, A248697.

Formula

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

If gpf(n^2 + n - 3) > n, then we have:

a(n) = gpf(n^2 + n - 3), where gpf = "greatest prime factor".

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

a(n) = n + m + 1.

a(n) divides gcd(n^2 + n - 3, m^2 + m - 3).