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EXAMPLE
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Array begins:
n\k| 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24
---+-------------------------------------------------------------------------
1 | 0, 0, 0, 1, 1, 2, 0, 1, 1, 2, 0, 1, 2, 3, 1, 2, 2, 3, 1, 2, 3, 4, 2, 3,
2 | 0, 0, 0, 2, 1, 1, 0, 1, 0, 1, 0, 3, 2, 3, 2, 3, 2, 0, 1, 3, 2, 2, 1, 2,
3 | 0, 0, 0, 1, 2, 0, 0, 3, 0, 2, 0, 0, 1, 2, 1, 2, 2, 0, 2, 2, 3, 1, 0, 5,
4 | 0, 0, 0, 0, 1, 0, 0, 0, 0, 1, 0, 0, 2, 3, 0, 5, 1, 0, 1, 3, 2, 1, 0, 4,
5 | 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 2, 0, 4, 2, 0, 0, 2, 5, 1, 0, 3,
6 | 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 5, 0, 3, 1, 0, 0, 2, 4, 2, 0, 3,
7 | 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 4, 0, 3, 0, 0, 0, 1, 3, 1, 0, 4,
8 | 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 3, 0, 4, 0, 0, 0, 2, 3, 0, 0, 3,
9 | 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 3, 0, 3, 0, 0, 0, 1, 4, 0, 0, 4,
10 | 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 4, 0, 4, 0, 0, 0, 0, 3, 0, 0, 4,
11 | 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 3, 0, 4, 0, 0, 0, 0, 4, 0, 0, 5,
12 | 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 4, 0, 5, 0, 0, 0, 0, 4, 0, 0, 3,
13 | 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 4, 0, 3, 0, 0, 0, 0, 5, 0, 0, 6,
14 | 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 5, 0, 6, 0, 0, 0, 0, 3, 0, 0, 2,
15 | 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 3, 0, 2, 0, 0, 0, 0, 6, 0, 0, 4,
16 | 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 6, 0, 4, 0, 0, 0, 0, 2, 0, 0, 4,
17 | 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 2, 0, 4, 0, 0, 0, 0, 4, 0, 0, 4,
18 | 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 4, 0, 4, 0, 0, 0, 0, 4, 0, 0, 3,
19 | 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 4, 0, 3, 0, 0, 0, 0, 4, 0, 0, 4,
20 | 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 4, 0, 4, 0, 0, 0, 0, 3, 0, 0, 6,
21 | 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 3, 0, 6, 0, 0, 0, 0, 4, 0, 0, 4,
We have A372283(5, 14) = 71, and when we compare the binary expansion of 71 = 1000111_2 with the term of A086893 that has a binary expansion of the same length, which in this case is 85 = 1010101_2, we see that only the bits at positions 1 and 4 (indexed from the right hand end, with 0 being the least significant bit position at right) need to be toggled to obtain the 71 from 85 or vice versa, therefore A(5, 14) = 2.
We have A372283(6, 14) = 107 = 1101011_2, and when xored with A086893(7) = 85 = 1010101_2, we obtain A372361(6, 14) = 62 = 111110_2, with five 1-bits, therefore A(6, 14) = 5. I.e., five bits (all except the least and the most significant bit) need to be flipped to change 85 to 107 or vice versa.
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