A356247 Denominator of the continued fraction 1/(2 - 3/(3 - 4/(4 - 5/(...(n-1) - n/(-1))))).

1, 5, 11, 19, 29, 41, 11, 71, 89, 109, 131, 31, 181, 19, 239, 271, 61, 31, 379, 419, 461, 101, 29, 599, 59, 701, 151, 811, 79, 929, 991, 211, 59, 41, 1259, 1, 281, 1481, 1559, 149, 1721, 1, 61, 1979, 2069, 2161, 1, 2351, 79, 2549, 241, 1, 2861, 2969, 3079, 3191

Offset: 2

Author

Mohammed Bouras, Jul 30 2022

Keywords

• nonn
• easy
• frac
• changed

Comments

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

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

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

a(n) divides n^2 - n - 1, which is the unreduced denominator.

Conjecture: The ordered sequence of prime values is A038872. - Bill McEachen, Jul 28 2025

For a proof of Conjectures 1-3, see Cloitre (2025). - Sean A. Irvine, Aug 25 2025

Examples

			For n=2, 1/(2 - 3) = -1, so a(2)=1.
For n=3, 1/(2 - 3/(3 - 4)) = 1/5, so a(3)=5.
For n=4, 1/(2 - 3/(3 - 4/(4 - 5))) = 7/11, so a(4)=11.
For n=5, 1/(2 - 3/(3 - 4/(4 - 5/(5 - 6)))) = 23/19, so a(5)=19.
For n=6, 1/(2 - 3/(3 - 4/(4 - 5/(5 - 6/(6 - 7))))) = 73/29, so a(6)=29.
a(23) = a(79) = 23 + 79 - 1 = 101.
a(26) = a(34) = gcd(26^2 - 26 -1, 34^2 - 34 - 1) = gcd(649, 1121) = 59.
		

Links

• Michael De Vlieger, Table of n, a(n) for n = 2..10001
• Mohammed Bouras, A new sequence of prime numbers, Romanian Math. Mag. (15 August 2022). See Table 2 p. 2.
• Mohammed Bouras, A New Sequence of Prime Numbers, EasyChair Preprint 8686, 2022.
• Mohammed Bouras, The Distribution Of Prime Numbers And Continued Fractions, (ppt) (2022).
• Benoit Cloitre, On the sequence A356247, 2025.

Crossrefs

Cf. A002327 (primes of the form k^2-k-1), A028387, A051403, A165900, A356684.

Programs

Mathematica

a[n_] := ContinuedFractionK[-i-1, If[i == n, 1, i+1], {i, 1, n}] //
   Denominator;
Table[a[n], {n, 2, 100}] (* Jean-François Alcover, Aug 11 2022 *)

PARI

a(n) = if (n==1, 1, n--; my(v = vector(2*n, k, (k+4)\2)); my(q = 1/(v[2*n-1] - v[2*n])); forstep(k=2*n-3, 1, -2, q = v[k] - v[k+1]/q; ); denominator(1/q)); \\ Michel Marcus, Aug 07 2022

Python

from fractions import Fraction
def A356247(n):
    k = -1
    for i in range(n-1,1,-1):
        k = i-Fraction(i+1,k)
    return abs(k.numerator) # Chai Wah Wu, Aug 23 2022

Formula

a(n) = (n^2 - n - 1)/gcd(n^2 - n - 1, A356684(n)).

If conjecture 3 is true, then we have:

a(n) = a(m) = n + m - 1.

a(n) = a(m) = gcd(n^2 - n - 1, m^2 - m - 1).

a(n) = a(a(n) - n + 1).