cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

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

Original entry on oeis.org

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

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Author

Mohammed Bouras, Jun 04 2023

Keywords

Comments

Conjecture: Except for 49, every term of this sequence is either a prime or 1.
The conjecture holds through n=10000. The set of ordered primes appear to match A038901. - Bill McEachen, Jul 08 2025

Examples

			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.
		

Crossrefs

Programs

  • 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

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).
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