OFFSET
1,1
COMMENTS
Solving for z gives z = (x*y) / (x+y), so x*y == 0 (mod x+y).
All known terms are from A025487:
a(1) = 2 = 2;
a(2) = 12 = 2^2 * 3;
a(3) = 60 = 2^2 * 3 * 5;
a(4) = 840 = 2^3 * 3 * 5 * 7;
a(5) = 9240 = 2^3 * 3 * 5 * 7 * 11.
If a solution to the equation 1/z = 1/x + 1/y is found such that gcd(x,y,z) is a square, then x+y, x*y*z, and (x-y)^2 + (2*z)^2 are also squares.
For all solutions, x^2 + y^2 + z^2 is a square.
The sequence is indeed a subsequence of A025487, and likely of A126098 as well. - Max Alekseyev, Mar 01 2023
a(n) < 5*10^(n-1). - Max Alekseyev, Mar 01 2023
LINKS
Max Alekseyev, Table of n, a(n) for n = 1..30
Project Euler, Diophantine reciprocals I, Problem 108.
EXAMPLE
For n=1, we have the following, where r = (x*y) mod (x+y). (In the last four columns, each number marked by an asterisk is a square.)
.
r z x y x*y x+y x*y*z x^2+y^2+z^2
- - - - --- --- ----- -----------
0 1 2 2 4* 4* 4* 9* (solution)
2 1 2 4 8 6 8 21
4 1 2 6 12 8 12 41
6 1 2 8 16* 10 16* 69
3 1 3 3 9 6 9* 19
0 2 3 6 18* 9* 36* 49* (solution)
3 2 3 9 27 12 54 94
0 2 4 4 16* 8 32 36* (solution)
8 2 4 8 32 12 64* 84
5 2 5 5 25* 10 50 54
0 3 6 6 36* 12 108 81* (solution)
7 3 7 7 49* 14 147 107
0 4 8 8 64* 16* 256* 144* (solution)
9 4 9 9 81* 18 324* 178
.
z = 2 has the largest number of solutions, so a(1) = 2.
The number of solutions for the resulting z cannot exceed A018892(z).
PROG
(Python)
def a(n):
# k=x*y and d=x+y
bc, bk, L = 0, None, 10**n
for k in range(1, L):
c, k2 = 0, k * k
for d in range(max(1, k2 // (L - k) + 1), k + 1):
if k2 % d == 0: c += 1
if c > bc:
bc, bk = c, k
return bk
(PARI)
a(n)=my(bc=0, bk=0, L=10^n); for(k=1, L-1, my(c=0, k2=k^2); for(d=max(1, k2\(L-k)+1), k, if(k2%d==0, c++); ); if(c>bc, bc=c; bk=k); ); return(bk); \\ Darío Clavijo, Mar 03 2025
CROSSREFS
KEYWORD
nonn,changed
AUTHOR
Darío Clavijo, Apr 06 2022
EXTENSIONS
a(6) from Chai Wah Wu, Apr 10 2022
a(7)-a(22) from Max Alekseyev, Mar 01 2023
STATUS
approved