A376143 Numbers c such that a + b + c = d are abcd quadruples in the "abcd-conjecture" with a < b < c < d, all |a|, b, c, d are pairwise coprime, the quality q of the quadruple has q > 1, term a = +/- 1 = A376149(n) and term b = A376144(n) (with repetitions and sorted by c then b).

2187, 2375, 2401, 11881, 14641, 14641, 15125, 16807, 16807, 19321, 22201, 28561, 28561, 42875, 50625, 72171, 77763, 78003, 83349, 83521, 85169, 101761, 116281, 121801, 127253, 128125, 146689, 152881, 177023, 177147, 177147, 249001, 303601, 369603, 390625, 390625

Offset: 1

Author

Frank M Jackson, Sep 11 2024

Keywords

• nonn

Comments

An abcd quadruple is defined as (a, b, c, d) with a+b+c+d = 0, all |a|, |b|, |c|, |d| are pairwise coprime, and radical of a*b*c*d, rad(|a|*|b|*|c|*|d|) < max (|a|, |b|, |c|, |d|).

The quality of an abcd quadruple is q = log(max(|a|,|b|,|c|,|d|))/log(rad(|a|*|b|*|c|*|d|)).

This sequence considers quadruples of the form a = +/- 1 and a+b+c = d with a < b < c < d.

Corresponding numbers for b can be found at A376144 and the sequence indicating whether a is 1 or -1 can be found at A376149.

From David A. Corneth, Sep 17 2024: (Start)

All of a, b, c, d in a valid quadruple are odd. Proof: As a + b + c = d, a + b + c + d = 2d is even. Hence an even number of numbers in {a, b, c, d} is odd and as there is an even number of numbers in {a, b, c, d} consequently an even number of them is even.

If more than 0 of them is even then at least two of them are even and not all of {a, b, c, d} are pairwise coprime. Therefore exactly 0 of them are even i.e. they are all odd. (End)

Examples

			a(2) = 2375 because the second occurrence of these abcd quadruples with a = +/- 1 is (-1, 27, 2375, 2401) with c = 2375. As prime factors of the form a+d = b+c, we have 1 + 7^4 = 3^3 + 5^3 * 19.
a(4) = 11881 because the fourth occurrence of these abcd quadruples with a = +/- 1 is (1, 25, 11881, 11907) with c = 11881. As prime factors of the form a+b+c = d, we have 1 + 5^2 + 109^2 = 3^5 * 7^2.
From _David A. Corneth_, Sep 16 2024: (Start)
Suppose we look for terms <= 2500. We could iterate over squarefree numbers <= 2500. One of the numbers we will see is 285.
List its prime factors; {3,5,19}.
Make all pairs of sets of distinct prime factors.
These are {{{3}, {5, 19}}, {{5}, {3, 19}}, {{19}, {3,5}}}. One of the pairs we will check is {{3}, {5, 19}}. For each set in it, list all numbers of the form 3^k <= 2500 and 5^m*19^t <= 2500 with k, m, t >= 1. These are (3, 9, 27, 81, 243, 729, 2187) and (95, 475, 1805, 2375). Check all pairs from the Cartesian product of these numbers. One pair is (27, 2375).
The smallest is a candidate for b, the largest for c. See if either (or both) of candidate values a in {-1, 1} give a quadruple (a, b, c, d) (where d = a + b + c meeting the quality requirements and pairwise coprimeness (and d <= 2500)). For this example we find the quadruple (-1, 27, 2375, 2401). (End)
		

Links

• David A. Corneth, Table of n, a(n) for n = 1..161
• C. F. W. Ramaekers, The abc-Conjecture and the n-conjecture, Eindhoven University of technology Nov 12, 2009.
• David A. Corneth, quadruples (a, b, c, d) with b + c <= 10^7

Crossrefs

Cf. A007947, A216323, A376144, A376149.

Programs

Mathematica

Rad[n_] := Module[{lst=FactorInteger[n]}, Times@@(First/@lst)]; lst={}; Do[Do[If[d=b+c+a; AllTrue[{{Abs[a],b},{a,c},{a,d},{b,c},{b,d},{c,d}}, Apply[CoprimeQ]]&&d>Rad[Abs[a]*b*c*d], AppendTo[lst,{a,b,c}]], {c, 1, 3000}, {b, 2, c}], {a, {-1, 1}}]; Last/@SortBy[lst,{#[[2]]&,#[[3]]&}]

Extensions

a(15) corrected by David A. Corneth, Sep 16 2024