[seqfan] Re: Binary Complement Sequences

[Tom Duff]

And it just finished! 425720 takes 87,037,147,316 steps to converge to 0.
(Or my  computer glitched, or I have a bug. I seriously doubt the latter,
because all my other results match what others have reported.)

On Mon, Dec 26, 2022 at 8:23 AM Tom Duff <eigenvectors at gmail.com> wrote:

> My run of 425720 has been going for almost 83 billion iterations. The
> length of the current iterate is down to under 167000 bits (from a maximum
> of roughly 595000 bits). Excitement reigns!
>
> On Fri, Dec 16, 2022 at 11:00 AM 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.
>>
>> Email: jprsearle at gmail.com <mailto:jprsearle at gmail.com> (if you want to
>> request files)
>>
>> --
>> Seqfan Mailing list - http://list.seqfan.eu/
>>
>