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