A376240 a(n) = number of {x,y,z} in Z/nZ such that x+y+z = x*y*z is nonzero.

0, 1, 0, 1, 4, 2, 6, 6, 6, 21, 18, 7, 30, 30, 22, 28, 50, 37, 56, 63, 30, 86, 84, 42, 120, 145, 72, 87, 144, 124, 154, 140, 98, 245, 188, 123, 234, 274, 176, 274, 286, 169, 300, 253, 270, 410, 360, 196, 336, 607, 308, 437, 476, 388, 520, 378, 344, 709, 570, 429, 630, 758, 372, 600, 882, 517, 736, 751, 524, 968, 828, 546, 900, 1153, 810, 839, 720, 912, 1026, 1140, 738, 1413, 1134, 585, 1502, 1482, 930

Offset: 1

Author

M. F. Hasler, Sep 24 2024

Keywords

• nonn

Comments

Similar to A376183, but the sum equal to the product is restricted to be nonzero, as to exclude many rather trivial solutions.

Counting sets {x, y, z} means that the order of the three (not necessarily distinct) values is irrelevant. Equivalently, one could use representatives with, e.g., x <= y <= z.

Examples

			For n=1, we can't have x+y+z = x*y*z nonzero because zero is the only element, so there are a(1) = 0 solutions.
For n=2, x+y+z = x*y*z != 0 implies x = y = z = 1 (in Z/2Z), so there is only one unique solution, and a(2) = 1.
For n=3, x+y+z = x*y*z != 0 is impossible: if x = y = z, the sum is zero (in Z/3Z), and 1 + 2 + z = 1*2*z <=> z = 0, so there is no solution, a(3) = 0.
For n=4, since 2*2 = 0 in Z/4Z, at most one among {x, y, z}, say z, can equal 2. In this case, x = -y = +-1, gives a solution {x, y, z} = {1, 2, 3}. One then checks that x = +-y = z = +-1 can't yield a solution, so a(4) = 1.
For n=5, we see that (x, y) = (2, 3) gives z = z, so we have a solution for any nonzero z, and exhaustive verification shows that these are no other solutions: a(5) = 4.
		

Links

• Chai Wah Wu, Table of n, a(n) for n = 1..7542

Crossrefs

Cf. A376183 (same without restriction to nonzero sum/product).

Programs

PARI

apply( {A376240(n)=sum(x=1,n-1,sum(y=1,x,sum(z=1,y, (x+y+z-x*y*z)%n==0 && x*y*z%n)))}, [1..99])

Python

def A376240(n):
    c = 0
    for x in range(1,n):
        for y in range(x,n):
            xy,xyp = x*y%n-1,n-(x+y)%n
            c += sum(not (z==xyp or (xy*z+xyp)%n) for z in range(y,n))
    return c # Chai Wah Wu, Sep 30 2024