If you accumulate and discard as you go, you can substantially reduce the memory requirement.

Ie. (Using 10x10 as an example) After reading the first row, and duplicating the 9 elements to the lower triangle you need storage for 19 elements:

+---+---+---+---+---+---+---+---+---+---+
| a | b | c | d | e | f | g | h | i | j |
+---+---+---+---+---+---+---+---+---+---+
| b'|
+---+
| c'|
+---+
| d'|
+---+
| e'|
+---+
| f'|
+---+
| g'|
+---+
| h'|
+---+
| i'|
+---+
| j'|
+---+
[download]

But then you can print out the first row to the new file and free up that memory leaving only 9 elements in ram.

You then read the second row, duplicate the 8 elements, and print out the second row to the new file and discard it. And you now have 8x2=16 in memory.

The peak memory requirement comes when you have read the 5th row, at which point you will have 5x5 =25 elements in the lower triangle cache, and (breifly) the 10 elements of the 5 row before writing it out and discarding it. So rather than 10x10=100 you have a maximum requirement of (n/2)² + n = (10/2)²+10 = 35.

Or for your example (460000/2)² + 460000 = 52,900,460,000 rather than 460000² = 211,600,000,000 or ~60% less. Still needs a lot of memory, but it might help if your numbers are small enough to allow a C solution.