[seqfan] Re: Binary Complement Sequences

[M. F. Hasler]

(seeing Allen's reply I notice I forgot to send off the one I wrote
yesterday... Widely agreeing with Allen.)

I think this is definitely interesting.
The first sequence that should be submitted (if not already in OEIS) is the
function itself,
a(n) = binary complement of 3n,
And then, yes,
b(n) = number of iterations of a(.) required to reach 0.

 I think a key to understanding and proving that the sequence always
terminates are the "max values" (your 1b), so these should also be listed,
but maybe (also / rather) in increasing order without repetitions.  Define
them as starting values that are the maximum of their orbit (or some other
way to say that all subsequent values are smaller).

Then, to prove that all values terminate, it is sufficient (and equivalent)
that any starting values  eventually reaches such a max number".
I think it should be possible to prove by considering the binary expansion,
and in particular the initial digits of the numbers. My hunch is that it
should be sufficient to distinguish a finite number of patterns for the
initial digits, and what might possibly come after that.

- Maximilian

On Fri, Dec 16, 2022, 15:01 Joshua Searle (larry) <jprsearle at gmail.com>
wrote:

> Hello,
>
> (In my enthusiasm, I sent this first time around before I got confirmation
> of being added to the mailing list so I don’t think anyone saw it, oops)
>
> I am looking for some help finding some more terms for a set of sequences
> I intend to add to the OEIS.
>
> It is a similar algorithm to that of the collatz algorithm, but instead of
> of multiplying by 3 and adding when odd, and dividing when even, it goes as
> follows:
>
> on any number:
> -multiply by 3
> -find the binary complement (if it is 1001010 in binary, the complement is
> 0110101). This is equivalent to subtracting from the next highest mersenne
> number.
>
> this is treated as all one step, so a seed of 2 produces the sequence
> [2,1,0]
> 3 produces the longer [3, 6, 13, 24, 55, 90, 241, 300, 123, 142, 85, 0].
>
> For lack of a better name I’ve called these binary complement sequences.
>
> While you might expect similar behaviour to the collatz algorithm (and it
> largely does), it turns out this can support sequences that are
> staggeringly long in length. The starting seed of 28 takes 7572 terms to
> terminate and I terminated my code after seed 425720 exceeded 10 billion
> terms! I do think all sequences terminate.
>
> The following sequences can be made from it:
>
> 1a) step length: (seed = term 0, natural numbers)
> 1 <= n <= 30
> 1, 2, 11, 12, 1, 10, 3, 4, 13, 2, 19, 80, 9, 2, 15, 16, 81, 14, 11, 12, 1,
> 6, 83, 8, 73, 22, 79, 7572, 5, 18…
>
> 1b) max value: (natural numbers)
> 1 <= n <= 20
> 1, 2, 300, 300, 5, 300, 10, 10, 300, 10, 300, 328536, 300, 21, 300, 300,
> 328536, 300, 300, 300…
>
> 2a) seeds with record step length:
> 1 <= n <= 25, all known terms.
> 1, 2, 3, 4, 9, 11, 12, 17, 23, 28, 33, 74, 86, 180, 227, 350, 821, 3822,
> 4187, 5561, 6380, 6398, 22174, 22246, 26494
>
> 2b) step lengths of 2a:
> 1 <= n <= 25, all known terms
> 1, 2, 11, 12, 13, 19, 80, 81, 83, 7572, 7573, 7574, 7578, 7580, 664475,
> 664882, 3180929, 3180930, 3180931, 3181981, 3181988, 3182002, 3182226,
> 120796790, 556068798
>
> 2c) max values of 2a:
> 1 <= n <= 25, al known terms, abbreviated for readability
> 1, 2, 300 (x4), 328536 (x3), ~1.23*10^53 (x5), ~3.26*10^552 (x2),
> ~2.03*10^933 (x7), ~9.38*10^8306, ~1.67*10^16667
>
> 3a) seeds with record step length and new maxima (excludes all the side
> sequences, new maxima are not necessarily larger than the previous):
> 1 <= n <= 12, all known terms
> 1, 2, 3, 12, 28, 227, 821, 22246, 26494, 103721, 204953, 425720
>
> 3b) step lengths of 3a
> 1 <= n <= 11, all known terms plus a lower bound for next one.
> 1, 2, 11, 80, 7572, 664475, 3180929, 120796790, 556068798, 572086533,
> 1246707529, 9999999999+
>
> 3c) max values of 3a
> 1 <= n <= 11, all known terms plus a lower bound for next one.
> 1, 2, 300 , 328536, ~1.23*10^53, ~3.26*10^552, ~2.03*10^933,
> ~9.38*10^8306, ~1.67*10^16667, ~2.42*10^14081, ~9.81*10^25580,
> >=2.09*10^114778
>
> Observations and questions:
> -The max value achieved by a sequence has roughly sqrt(step count) digits.
> -For how many terms can a sequence continually increase? I haven’t tracked
> it but even 3 has 6 consecutively increasing terms in its sequence.
> -The penultimate term of a sequence must be of the form [(2^3n-1)-1]/3. I
> haven’t tracked how often sequences fall into these.
> -What does a log plot look like of these sequences? They have had far too
> many data points for basic graphing software to handle!
> -And of course, does every sequence terminate? (probably unanswerable)
>
> Being able to terminate 425720 would be nice, despite several drastic
> speedups from my rickety initial coding effort, still took 67 hours to
> compute 10 billion terms of the sequence. I can provide a data file where I
> copy and pasted results from general searches if requested. For example, I
> can give you term 9,999,999,999 of seed 425720, or the step lengths/maxima
> of sequences up to 425720 that didn’t get caught by my side-sequence filter.
>
> I’m worrying that this is too long; I hope that at least someone reads
> until the end!
>
> Joshua Searle.
>
>
>