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a(n) is the least sum n + y such that 1/n + 1/y = 1/z with gcd(n,y,z) = 1, for some integers y and z.
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%I #23 Nov 06 2022 08:38:25

%S 4,9,16,25,9,49,64,81,25,121,16,169,49,25,256,289,81,361,25,49,121,

%T 529,64,625,169,729,49,841,36,961,1024,121,289,49,81,1369,361,169,64,

%U 1681,49,1849,121,81,529,2209,256,2401,625,289,169,2809,729,121,64,361

%N a(n) is the least sum n + y such that 1/n + 1/y = 1/z with gcd(n,y,z) = 1, for some integers y and z.

%C All terms are squares.

%C Proof: Consider the general equation 1/x + 1/y = 1/z, x,y,z positive integers and gcd(x,y,z) = 1. If x,y,z is a solution, let a = x-z and b = y-z. Then a*b = (x-z)*(y-z) = z^2. The equation 1/x + 1/y = 1/z describes a projective curve in 2-dimensional projective space P^2, given by the homogeneous equation x*y - x*z - y*z = 0. So the solution is of the form x = a + sqrt(a*b), y = b + sqrt(a*b), z = sqrt(a*b), for all positive integers a and b where a*b is a square. The solutions with gcd(x,y,z) = 1 are exactly x = c^2 + c*d, y = d^2 + c*d, z = c*d for coprime positive integers c and d. Therefore the sum x+y = (c+d)^2 is a square, and x-z, y-z are also squares.

%e a(3) = 9 because 1/3 + 1/6 = 1/2 with 3 + 6 = 9.

%e Table with the integers n, y, z, n+y, c and d, for n >= 2:

%e +-----+------+-----+-------------+-----+-----+

%e | n | y | z | a(n) = n+y | c | d |

%e +-----+------+-----+-------------+-----+-----+

%e | 2 | 2 | 1 | 4 | 1 | 1 |

%e | 3 | 6 | 2 | 9 | 1 | 2 |

%e | 4 | 12 | 3 | 16 | 1 | 3 |

%e | 5 | 20 | 4 | 25 | 1 | 4 |

%e | 6 | 3 | 2 | 9 | 2 | 1 |

%e | 7 | 42 | 6 | 49 | 1 | 6 |

%e | 8 | 56 | 7 | 64 | 1 | 7 |

%e | 9 | 72 | 8 | 81 | 1 | 8 |

%e | 10 | 15 | 6 | 25 | 2 | 3 |

%e | 11 | 110 | 10 | 121 | 1 | 10 |

%e | 12 | 4 | 3 | 16 | 3 | 1 |

%e | 13 | 156 | 12 | 169 | 1 | 12 |

%e | 14 | 35 | 10 | 49 | 2 | 5 |

%e | 15 | 10 | 6 | 25 | 3 | 2 |

%e | 16 | 240 | 15 | 256 | 1 | 15 |

%e | 17 | 272 | 16 | 289 | 1 | 16 |

%e | 18 | 63 | 14 | 81 | 2 | 7 |

%e | 19 | 342 | 18 | 361 | 1 | 18 |

%e | 20 | 5 | 4 | 25 | 4 | 1 |

%p nn:=3000:T:=array(1..56):

%p for n from 2 to 57 do:

%p ii:=0:

%p for y from 1 to nn while(ii=0)do:

%p x1:=evalf(n*y/(n+y)):y1:=floor(x1):

%p g1:=gcd(n,y):g2:=gcd(g1,y1):

%p if x1=y1 and g2=1

%p then

%p printf(`%d %d %d %d\n`,n,y,y1,n+y):ii:=1:T[n-1]:=n+y:

%p else fi:

%p od:

%p od:

%p print(T):

%Y Cf. A000290 (squares), A034699.

%K nonn

%O 2,1

%A _Michel Lagneau_, Sep 26 2022