OFFSET
2,1
COMMENTS
For n = 1, any (x_1) would be a solution, therefore the offset is 2.
For any n, the equation would also admit the trivial zero solution (0,...,0). If any x_k = 0, then all x_i must be zero, so there is no other solution in the nonnegative integers.
This sequence gives row sums of A328911 which lists the number of solutions with given number of components > 1.
The Shiu paper gives different values for f(2), f(3) and f(8) than those given here, but the author has confirmed (personal communication) that 2, 6 and 19 are correct.
The multiset [1^(n-2),n+1,n*(2*n-1)] is the solution with maximum product. - Hugo Pfoertner, Nov 08 2019
From David A. Corneth, Nov 08 2019: (Start)
Given x_1,...,x_{n-1}, one can find the value x_n by solving the resulting linear univariate equation.
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*log_2(n)). - Max Alekseyev, Nov 10 2019
REFERENCES
Lars Blomberg, Posting to the Sequence Fans Mailing List, Nov 07 2019, seems to have been the first person to notice that there were problems with the published values (given in A328980). - N. J. A. Sloane, Nov 08 2019
R. K. Guy, Sum equals product. in: Unsolved Problems in Number Theory, 3rd ed. New York: Springer-Verlag, chapter D24, (2004), 299-301 (citing Erdős's question dated Aug 19, 1996)
LINKS
David A. Corneth, Table of n, a(n) for n = 2..6250
David A. Corneth, List of solutions and product of variables of solutions for n = 2..160, omitting ones
Peter Shiu, On Erdős's Last Equation, Amer. Math. Monthly, 126 (2019), 802-808; correction, 127:5 (2020), 478.
EXAMPLE
For n = 2 variables, we have the equation x1*x2 = 2*(x1 + x2) with positive integer solutions (3,6) and (4,4).
For n = 3, we have 3 solutions in C_1(3) = {(1, 4, 15), (1, 5, 9), (1, 6, 7)} (with 2 components > 1), and 3 others in C_2(3) = {(2, 2, 12), (2, 3, 5), (3,3,3)} (with 3 components > 1), for a total of a(3) = 6.
For n = 4 we have the 8 solutions (1, 1, 5, 28), (1, 1, 6, 16), (1, 1, 7, 12), (1, 1, 8, 10), (1, 2, 3, 12), (1, 2, 4, 7), (1, 3, 4, 4) and (2, 2, 2, 6).
For n = 5, the solutions are, omitting initial components x_i = 1: {(6, 45), (7, 25), (9, 15), (10, 13), (2, 3, 35), (2, 5, 9), (3, 3, 10), (3, 5, 5)}.
For n = 6, the solutions are (omitting x_i = 1): {(7, 66), (8, 36), (9, 26), (10, 21), (11, 18), (12, 16), (2, 4, 27), (2, 5, 15), (2, 6, 11), (2, 7, 9), (3, 3, 18), (3, 4, 10), (3, 6, 6), (2, 2, 2, 24), (2, 2, 3, 9), (2, 2, 4, 6), (2, 3, 3, 5)}.
For n = 9, the 27 solutions are (omitting '1's): {(10, 153), (11, 81), (12, 57), (13, 45), (15, 33), (17, 27), (18, 25), (21, 21), (2, 5, 117), (2, 6, 42), (2, 7, 27), (2, 9, 17), (2, 12, 12), (3, 4, 39), (3, 5, 21), (3, 6, 15), (3, 7, 12), (3, 9, 9), (4, 4, 18), (5, 5, 9), (6, 6, 6), (2, 2, 3, 36), (2, 2, 6, 9), (2, 3, 3, 13), (3, 3, 3, 7), (3, 3, 4, 5), (2, 3, 3, 3, 3)}.
PROG
(PARI) a(n, show=1)={my(s=0, d); forvec(x=vector(n-1, i, [1, n\(sqrt(2)-1)]), 0<(d=vecprod(x)-n) && n*vecsum(x)%d==0 && n*vecsum(x)\d >= x[n-1] &&s++ &&show &&printf("%d, ", concat(x, n*vecsum(x)\d)), 1); s}
(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
CROSSREFS
KEYWORD
nonn
AUTHOR
M. F. Hasler, Nov 07 2019
EXTENSIONS
More terms from David A. Corneth, Nov 07 2019
STATUS
approved