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For example, for n = 4, if we are given (1, 1, 7, x_4) then we can solve 4*(9 + x_4) = 7*x_4, getting 36 = 3*x_4 , i.e. , x_4 = 12. As 12 is integer and >= x_3 = 7, we have a new solution: (1, 1, 7, 12). (End)
In any solution, we have x_1*...*x_{n-1} <= n^2, implying that a(n) is finite for all n > 1. Furthermore, x_1 = x_2 = ... = x_k = 1 for k = n - 1 - floor(2*log2log_2(n)). - Max Alekseyev, Nov 10 2019
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Peter Shiu, <a href="https://doi.org/10.1080/00029890.2019.1639466">On Erdős's Last Equation</a>, Amer. Math. Monthly, 126 (2019), 802-808; correction, loc. cit, 127:5 (2020), 478.
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Peter Shiu, <a href="https://doi.org/10.1080/00029890.2019.1639466">On Erdős's Last Equation</a>, Amer. Math. Monthly, 126 (2019), 802-808; correction, loc. cit, 127:5 (2020), 478.
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(PARI) { A328910(n, k=n-1, m=n^2, p=1, s=0, y=1) = if(k==0, return( p>n && Mod(n*s, p-n)==0 && n*s>=(p-n)*y ) ); sum(x=y, sqrtnint(m, k), A328910(n, k-1, m\x, p*x, s+x, x) ); } \\ Max Alekseyev, Nov 10 2019
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