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A094185
Number of solutions to n == xy (mod z) == yz (mod x) == zx (mod y) with 0 < x < y < z.
1
1, 6, 12, 24, 24, 49, 41, 59, 61, 100, 56, 132, 76, 128, 122, 165, 87, 223, 95, 208, 176, 190, 110, 318, 187, 232, 211, 342, 138, 440, 127, 327, 275, 327, 242, 511, 188, 313, 292, 581, 180, 587, 177, 467, 489, 321, 179, 727, 327, 520, 344, 580, 193, 713, 408, 640, 410, 442, 222, 1055, 288
OFFSET
1,2
COMMENTS
Possner (see also Knuth) asks for solutions when n=2. There is at least one solution for all positive n: (x,y,z) = (n+1, n^2+2n, n^3+3n^2+n). All solutions appear to be in the polytope n < x <= 2n^2 + n, x < y <= 2n^3 + 2n^2 - n, y < z <= n^5 + 2n^4 + 2n^3 + n^2 - n. Many solutions, especially for prime n, are such that n divides x, y and z. See A094595.
Following the linked solution by Silvia Fernández, it can be shown that all solutions satisfy n < x <= 3n^2 - n, x < y <= 2nx - n, and y < z <= xy - n. Contrary to the above comment, there are solutions satisfying x > 2n^2 + n. The first example is given by (x,y,z) = (434,630,826) when n = 14. - Robin Visser, Dec 18 2023
LINKS
M. F. Possner, Problem 11021, Amer. Math. Monthly, 110 (2003), p. 542.
Donald Knuth, Silvia Fernández and Gerry Myerson, A Modular Triple: 11021, Amer. Math. Monthly, 112 (2005), p. 279.
EXAMPLE
a(2) = 6 because there are 6 solutions: (x,y,z) = (3, 8, 22), (3, 10, 14), (4, 5, 18), (4, 6, 11), (6, 14, 82) and (6, 22, 26).
PROG
(Sage)
def a(n):
ans = 0
for x in range(n+1, 3*n^2-n+1):
for y in range(x+1, 2*n*x-n+1):
for z in Integer(x*y-n).divisors():
if (z > y) and ((x*y)%z)==n and ((y*z)%x)==n and ((z*x)%y)==n:
ans += 1
return ans # Robin Visser, Dec 12 2023
CROSSREFS
Cf. A094595 (number of solutions to 1 = nxy (mod z) = nyz (mod x) = nzx (mod y) with 0<x<y<z).
Sequence in context: A172079 A081512 A096387 * A074902 A096366 A247145
KEYWORD
nonn
AUTHOR
T. D. Noe, May 06 2004, revised May 13 2004
EXTENSIONS
Corrected and a(33)-a(46) added by Robin Visser, Dec 18 2023
a(47)-a(61) from Robin Visser, Mar 18 2024
STATUS
approved