# [seqfan] Re: 2 seqs with diffs distinct

William Keith wjk26 at drexel.edu
Sat Mar 27 18:00:30 CET 2010

```> Some scribbling suggests that we might end up with
> A = 0 1 7 18 35 ...
> B = 0 3 12 26 45 ...

That seems quite reasonable.  You start with 0 1... and 0 3..., and, beginning with A and alternating A and B, add (last entry in other sequence) + (lowest number not yet appearing as a difference).

Dashing this off so there may be an off-by-one error, but the argument should be clear:

At the n-th add, you already have n-2 distinct numbers.  (Here I would count 18 above as the 7th add, 26 as the 8th, 35 as the 9th, etc.)  If you are adding to A, n is odd, and you currently have ((n-1)/2)^2 distinct differences.  If adding to B, n is even, and you currently have (n/2)((n/2)-1) distinct differences.  In both cases, n-5 of those numbers (a_n - 0 and b_n - 0) are themselves numbers in the sequences.  Hence, for the n-th add when n is in A, to the previous number add ((n-1)/2)^2 + n-5+1, and for the n-th add when n is in B, add (n/2)((n/2)-1) + n-5+1.

You could make it a single sequence by applying each addition rule alternately, using floor(n/2).

William

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