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

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