A363482 Denominator of the continued fraction 1/(2-3/(3-4/(4-5/(...(n-1)-n/(-5))))).

13, 23, 7, 49, 13, 83, 103, 5, 149, 1, 29, 233, 53, 23, 67, 373, 59, 1, 499, 109, 593, 643, 139, 107, 1, 863, 71, 197, 1049, 223, 1, 179, 53, 1399, 59, 1553, 71, 1, 257, 1, 1973, 2063, 431, 173, 67, 349, 2543, 1, 2749, 571, 2963, 439, 1, 3299, 683, 3533, 281, 151, 557, 1, 4153

Offset

3,1

Comments

Conjecture: Except for 49, every term of this sequence is either a prime or 1.

Links

Table of n, a(n) for n=3..63.

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

Formula

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

Except for n=6, if gpf(n^2 + 3*n - 5) > n, then we have:

a(n) = gpf(n^2 + 3*n - 5), where gpf = "greatest prime factor".

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

a(n) = n + m + 3.

a(n) divides gcd(n^2 + 3*n - 5, m^2 + 3*m - 5).

Example

For n=3, 1/(2 - 3/(-5)) = 5/13, so a(3) = 13.
For n=4, 1/(2 - 3/(3 - 4/(-5))) = 19/23, so a(4) = 23.
For n=5, 1/(2 - 3/(3 - 4/(4 - 5/(-5)))) = 11/7, so a(5) = 7.

Prog

(PARI) lf(n) = sum(k=0, n-1, k!); \\ A003422
f(n) = (n+2)*lf(n+1)/2; \\ A051403
a(n) = (n^2 + 3*n - 5)/gcd(n^2 + 3*n - 5, 5*f(n-3) + n*f(n-4)); \\ Michel Marcus, Jun 06 2023

Crossrefs

Cf. A006530, A051403.

Cf. A362086, A363102.

Sequence in context: A298235 A111676 A372056 * A058030 A347344 A102498

Adjacent sequences: A363479 A363480 A363481 * A363483 A363484 A363485

Keyword

nonn

Author

Mohammed Bouras, Jun 04 2023

Status

approved