1, 2, 3, 4, 5, 6, 8, 7, 9, 10, 12, 16, 11, 17, 13, 32, 14, 18, 20, 19, 33, 15, 34, 22, 64, 21, 24, 36, 28, 65, 23, 66, 25, 40, 35, 72, 42, 48, 37, 68, 26, 128, 27, 129, 29, 130, 30, 132, 31, 256, 38, 80, 49, 73, 56, 136, 41, 96, 69, 144, 67, 84, 97, 137, 112, 145, 134, 160, 50, 133, 76, 161, 100, 257, 39, 258, 43, 260, 44
Offset: 1
For n=11, we first note that a(10) = 10, and the least unused number after a(1) .. a(10) is 11. Trying to multiply 10 (= 1010_2) and 11 (= 1011_2), in the binary system results in
1011
* 1010
-------
c1011
1011
-------
1101110 = 110,
and we see that there's a carry-bit (marked c) affecting the result, thus A048720(10,11) < 10*11 and A061858(10,10) > 0, thus we cannot select 11 for a(11).
The next unused number is 12, and indeed, for numbers 10 and 12 (= 1100_2), the binary multiplication results in
1100
* 1010
-------
1100
1100
-------
1111000 = 120,
which is a clean product without carries (i.e., A061858(10,12) = 0), thus 12 is selected to be a match for 10, and we set a(11) = 12.
For a(49) = 31 (= 11111_2) and a(50) = 256 (= 100000000_2) the multiplication results in
100000000
* 11111
-------------
100000000
100000000
100000000
100000000
100000000
-------------
1111100000000 = 7936,
and we see that the carryless product is this time obtained almost trivially, as the other number is so much larger and more spacious than the other that they can easily avoid any clashing bits that would produce carries.
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