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URL: https://oeis.org/A375019

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A375019
Positive integers n such that (1, n^2 - 1, n^2) is an abc triple.
2
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
1,1
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).
LINKS
Bart de Smit, ABC-triples
Wikipedia, abc conjecture
EXAMPLE
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.
PROG
(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
(PARI) is_a375019(n) = factorback(factorint((n-1)*n*(n+1))[, 1]) < n^2 \\ Hugo Pfoertner, Aug 09 2024
CROSSREFS
KEYWORD
nonn
AUTHOR
William Hu, Aug 09 2024
STATUS
approved