[seqfan] Re: Newbie discovery

[Neil Sloane]

Ian,  Very interesting sequence!  It is yet another variation on the Van
Eck sequence, of course (A181391).

Please go ahead and submit it to the OEIS. I confirmed your terms.  The
initial term should have index 1, I think (i.e. the offset should be 1),
that is, start with a(1)=0.

There are typos in your definition, I think you want to say
not
where f(x,y) can be any function that receives two integers
and outputs a single value, and y represents the number of terms before
f(n-1) that are equal to f(n-1).
but
where f(x,y) can be any function that receives two integers
and outputs a single value, and y represents the number of terms before
a(n-1) that are equal to a(n-1).

What about the sequence that gives the index when k first appears, for k >=
0 ?
It starts 1, 3, 6, 8, 16, I guess.

Best regards
Neil

Neil J. A. Sloane, President, OEIS Foundation.
11 South Adelaide Avenue, Highland Park, NJ 08904, USA.
Also Visiting Scientist, Math. Dept., Rutgers University, Piscataway, NJ.
Phone: 732 828 6098; home page: http://NeilSloane.com
Email: njasloane at gmail.com



On Fri, Jan 8, 2021 at 10:26 PM Ian Hutchinson <ianrh125 at gmail.com> wrote:

> Dear Seqfans,
>
> I recently came up with a sequence that seemed to generate
> emergent patterns, which I was fascinated by. I decided to search the OEIS
> with the first ~10 terms to see if it was already known, and to my surprise
> it returned zero results. This is either because I made a brand new
> discovery or because I don't know much about existing sequences, so I
> thought I'd get a more knowledgeable opinion on this.
>
> I came up with this while looking for a pattern that would avoid repetition
> using simple rules. I came up with a formula in the form of  a(n) =
> f(a(n-1), y), where f(x,y) can be any function that receives two integers
> and outputs a single value, and y represents the number of terms before
> f(n-1) that are equal to f(n-1). This ensures that for any integer k, the
> first occurrence of k will be followed by f(k,0), the second will be
> followed by f(k,1), the third by f(k,2), and so on. After trying a version
> of this with bitxor(x, y) as the main function, I found some intriguing
> results.
>
> The first 15 terms are 0, 0, 1, 1, 0, 2, 2, 3, 3, 2, 0, 3, 1, 3, 0, and
> already start to reveal an interesting pattern. Terms 1-5 switch between 0
> and 1 until the third time 0 shows up, which causes the next term to be 2.
> Terms 6-10 switch between 2 and 3 in the exact same way, since there were
> no previous occurrences of 2 or 3. Terms 11-15 display a new behavior,
> moving all over the range [0, 3] in a less orderly manner, ending after the
> 5th occurrence of 0. This pattern of repeating everything in a new range
> and then creating a more complicated interaction in the combined range
> continues at a larger and larger scale, at which point it starts to
> generate some striking graphs:
> [image: XOR sequence.png]
> (This is a plot of the first 3,416 terms, stopping right after the 65th
> zero)
>
> I also noticed a few other interesting properties about the resulting
> sequence:
> - The graph of the first n terms will resemble the graph of the first n/4
> terms at sufficiently large values (n>100).
> - The behavior of bitxor such that bitxor(n, 0) = n for any integer n
> causes an upward sloping line to appear in the top right quadrant of the
> above graph, as the series regularly returns to zero and other small
> numbers.
> - If computed to infinite terms, this sequence may contain every possible
> pair of positive integers exactly once. This is based on a few specific
> observations that are a little too long to list out here, but I'd be happy
> to elaborate on them.
>
> Have any of you seen a sequence like this before? I'm curious to see if the
> overall structure of it has been used anywhere else.
>
> Thanks,
> Ian
>
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