%I #64 Aug 13 2024 09:01:06
%S 3,7,8,9,15,17,25,26,27,31,48,49,55,63,64,80,81,97,99,125,127,161,224,
%T 225,242,243,244,251,255,288,289,325,343,351,361,449,485,487,511,512,
%U 513,575,577,624,625,649,675,676,687,721,728,729,783,960,961,999
%N Positive integers n such that (1, n^2 - 1, n^2) is an abc triple.
%C If a number appears in this sequence, so do all of its powers. This immediately implies that this sequence is infinite.
%C All the terms A216323(n), A216323(n)+1, and 2*A216323(n)+1 appear in this sequence.
%C The highest known quality abc triple of this form occurs with n = 49, with quality 1.4557, for the triple (1, 2400, 2401).
%H William Hu, <a href="/A375019/b375019.txt">Table of n, a(n) for n = 1..1816</a>
%H Bart de Smit, <a href="https://www.math.leidenuniv.nl/~smitbde/abc/">ABC-triples</a>
%H Wikipedia, <a href="http://en.wikipedia.org/wiki/Abc_conjecture">abc conjecture</a>
%e 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.
%o (Python)
%o """
%o Note that this code generates all terms <= n, not the nth term.
%o This code can be further optimized with an O(n log n) sieve, which we do not write here.
%o """
%o n = 10**5 # replace this number with whatever limit
%o from sympy import primefactors, prod
%o def rad(n): return 1 if n < 2 else prod(primefactors(n))
%o # Function to help determine whether a value is a term.
%o def is_term(k: int):
%o # Calculate rad((k^2-1)*k^2) = rad((k-1)*k*(k+1)).
%o rad_abc = rad(k-1) * rad(k) * rad(k+1)
%o if k % 2 == 1:
%o rad_abc //= 2 # 2 is double-counted as a prime factor. No other multiple-counts are possible.
%o return rad_abc < k**2
%o # The final sequence.
%o a = list(filter(is_term, range(2, n+1))) # _William Hu_, Aug 09 2024
%o (PARI) is_a375019(n) = factorback(factorint((n-1)*n*(n+1))[,1]) < n^2 \\ _Hugo Pfoertner_, Aug 09 2024
%Y Cf. A007531, A007947, A216323.
%K nonn
%O 1,1
%A _William Hu_, Aug 09 2024