[seqfan] Binary Complement Sequences

[Joshua Searle (larry)]

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)