[seqfan] Help with defining a sequence

Ali Sada pemd70 at yahoo.com
Fri Oct 23 08:56:20 CEST 2020


Hi Everyone,

I hope everything is going well with you. I would really appreciate your help defining the sequence below. 

We start by pairing positive integers with themselves
(1,1), (2,2), (3,3), (4,4), (5,5)..…

The idea is to change each pair so that one element is twice the other element. We take the difference from the next pair. (We can only add to the element(s) of the designated pair.)
We start by taking 1 from the second pair and add it to one of the elements in the first pair. The first pair becomes (1,2). The second pair is also (1,2), so, it doesn’t need any change.
For the third pair, we take 3 from one element of the fourth pair and add it. The third pair becomes (3,6). 
The fourth pair is now (1,4), so we take 1 from one element of the fifth pair and add it to the 1 in the fourth pair. The fourth pair is (2,4) now, but the fifth pair is (4,5).  So, we take 3 from one element in the sixth pair and add it to the 5 in the fifth pair. Now, the fifth pair is (4,8) and the sixth pair is (3,6), which doesn’t need any change. And so on. 

The pairs produced by this process are:
(1,2), (1,2), (3,6), (2,4), (4,8), (3,6), (7,14), (4,8), (6,12), (7,14), (7,14), ….
The sequence associated with this process is the smallest element in each pair:
1, 1, 3, 2, 4, 3, 7, 4, 6, 7, 7, 9, 7, 13, 8, 10, 13, 9, 19, 10, 12, 19, 12, 18, 13, 25, 14, 16, 25, 15, 25, 16, 30, 17, 25, 21, 31, 19, 33, 20, 34, 21, 37, 22, 36, 23, 43, 24, 34, 31, 39, 26, 52, 27, 31, 49, 29, 43, 31, 57, 31, 37, 51

A second sequence could be the number added to adjust each pair: 
1, 0, 3, 1, 3, 0, 7, 3, 3, 4, 3, 6, 1, 12, 6, 4, 9, 0, 19, 9, 3, 16, 6, 12, 1, 24, 12, 4, 21, 6, 19, 3, 27, 10, 15, 6, 25, 6, 27, 7, 27, 6, 31, 9, 27, 4, 39, 15, 19, 12, 27, 1, 51, 24, 7, 42

And a third sequence could be the pairs where we didn’t need to add.  I calculated 10,000 pairs and got zeros at these points. 
2, 6, 18, 74, 198, 486

Thank you in advance. 

Best,

Ali




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