A375019 Positive integers n such that (1, n^2 - 1, n^2) is an abc triple.

3, 7, 8, 9, 15, 17, 25, 26, 27, 31, 48, 49, 55, 63, 64, 80, 81, 97, 99, 125, 127, 161, 224, 225, 242, 243, 244, 251, 255, 288, 289, 325, 343, 351, 361, 449, 485, 487, 511, 512, 513, 575, 577, 624, 625, 649, 675, 676, 687, 721, 728, 729, 783, 960, 961, 999

Offset: 1

Author

William Hu, Aug 09 2024

Keywords

• nonn

Comments

If a number appears in this sequence, so do all of its powers. This immediately implies that this sequence is infinite.

All the terms A216323(n), A216323(n)+1, and 2*A216323(n)+1 appear in this sequence.

The highest known quality abc triple of this form occurs with n = 49, with quality 1.4557, for the triple (1, 2400, 2401).

Examples

			For a(1) = 3: (a,b,c) = (1,8,9) is an abc triple. Reason: rad(1*8*9) = rad(72) = 6. Since 6 < 3^2, 3 is a term of this sequence.
		

Links

• William Hu, Table of n, a(n) for n = 1..1816
• Bart de Smit, ABC-triples
• Wikipedia, abc conjecture

Crossrefs

Cf. A007531, A007947, A216323.

Programs

PARI

is_a375019(n) = factorback(factorint((n-1)*n*(n+1))[,1]) < n^2 \\ Hugo Pfoertner, Aug 09 2024

Python

"""
Note that this code generates all terms <= n, not the nth term.
This code can be further optimized with an O(n log n) sieve, which we do not write here.
"""
n = 10**5  # replace this number with whatever limit
from sympy import primefactors, prod
def rad(n): return 1 if n < 2 else prod(primefactors(n))
# Function to help determine whether a value is a term.
def is_term(k: int):
    # Calculate rad((k^2-1)*k^2) = rad((k-1)*k*(k+1)).
    rad_abc = rad(k-1) * rad(k) * rad(k+1)
    if k % 2 == 1:
        rad_abc //= 2  # 2 is double-counted as a prime factor. No other multiple-counts are possible.
    return rad_abc < k**2
# The final sequence.
a = list(filter(is_term, range(2, n+1))) # William Hu, Aug 09 2024