Bonus Puzzle: Sequence Challenge – Mystery Prize

This is a puzzle that ran for three weeks in September 2015. The idea was to work out the rule for a sequence, with a new term given each day.

Although the puzzle was solved after twenty-two terms had been revealed, I have provided thirty-one terms below. I think that this gives the best chance of finding the solution for anyone else who wants to have a go.

Here is a sequence:

1 1 2 1 2 2 2 1 2 2 2 2 2 2 3 1 2 2 2 2 2 2 3 2 2 2 3 2 3 3 3 …

Each day, the next term in this sequence will be revealed.

Your challenge is to determine the rule governing the sequence.
.

UPDATE 19/09/2015:
The sequence challenge has been solved!
Click here for the solution.

Here is an Excel version of the sequence for ease of reference:
Sequencechallenge_Spreadsheet

The sequence follows a moderately straightforward mathematical rule and all the terms are non-negative integers. I don’t know how difficult it will be to work out; fairly tough I think, but who knows. Perhaps someone will get it straight away, perhaps we will still be going in a year’s time. I don’t think the answer is searchable, but if it is then I may end up looking rather silly.

New terms will be posted here and on Twitter, from my @mathistopheles account, with the hashtag #sequencechallenge.

In mathematics, there can often be several different but equivalent ways of saying the same thing. For this reason, any proposed rule that would generate the complete correct sequence will be considered a correct solution. The rule does not have to be phrased in precisely the way that I have in mind.

As an added incentive, a Mystery Science Prize* is available** for the first person to post the correct answer in the comments below. Only one guess per person per day, please; i.e. you can’t guess again until the next term is released.

UPDATE 11/09/2015: cinnamon_carter has generously pledged 15000 units of crypto-currency to the winner, alongside my prize. I don’t really know much about this, so for further details, start from this comment and this tweet.

* Don’t get too excited. It is worth under £10 and that was when it was new.

** Provided that you can later provide me with a valid postal address to a sensible location (No Antarctic survey bases or remote Pacific atolls, please. I reserve the right to duck out of ludicrously high postage costs.) and that the postal service successfully deliver it to you.

Bonus points are awarded each day for anyone who correctly guesses the following day’s term (in a reply to my latest tweet).

FINAL LEADERBOARD (19/09/2015)
Troy R – 11 pts
Jesse Thompson – 9 pts
Aaron Percival – 8 pts
John G – 7 pts
Michael Bench-Capon – 6 pts
Brad Smith – 3 pts
BurHanSolo – 3 pts
Chris Purcell – 2 pts
Alexander – 1 pt
Aoife Hunt – 1 pt
Gareth Adams – 1 pt
Henry W McLaughlin – 1 pt
H J – 1 pt
Ignacio Mancera – 1 pt
John Doe – 1 pt
Luke – 1 pt
Matthew Jackson – 1 pt
mike – 1 pt
Myria – 1 pt
Nick T – 1 pt
Simon Capstick – 1 pt
Timo – 1 pt
Tom Flowerdew – 1 pt
.

IMAGE: Creative Commons – Wikimedia Commons – Manfred Werner

Click here to return to the Puzzles page

Thomas Oléron Evans, 2015

147 thoughts on “Bonus Puzzle: Sequence Challenge – Mystery Prize”

  1. thomas Post author

    Rather than approve and respond to suggestions in an ad hoc way throughout the day (UK time), I will approve and respond to them all at the end of the day (or early next morning). If you don’t see your suggestion, that’s why.

    This will keep things simpler for me. Thanks.

    Reply ↓
  2. Troy

    Number of ones needed to write the number in binary…

    [Edited for length, since there might or might not be quite a few responses to the thread]

    Something tells me your sequence might diverge quickly to huge numbers, so I wanted to get this guess in quickly before it was clearly wrong. 😉

    Reply ↓
    1. thomas Post author

      You’ve sort of set me a puzzle here, because I don’t know what the rule is for the sequence that you have come up with. I’ll have to take a neither confirm nor deny approach.

      Reply ↓
  9. Jesse Thompson

    Fibonacci sequence mod 3? 🙂

    Also, it looked as though Chris was asking if the sequence is added to sequentially. You’ve given us the first five numbers, but did they get added in sequential order? will each new term always appear on the right, or will some new terms insert themselves at the beginning or in the middle? I think that’s what Chris was trying to ask, and I’m curious as well.

    Reply ↓
    1. thomas Post author

      Not Fibonacci mod 3 – the fourth term would be 0 in that case. Good idea though.

      These are terms 1 to 5. Every new term will be added in sequence. If we get really stuck (I don’t think we will), I may give other clues and later terms, but not for the moment.

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    2. Chris

      to clarify – I wrote out the sequence as I did for clarity, I was asking if the sequence would go

      1,1,2,1,2,3,1,2,3,4,1,2,3,4,5,1,2,3,4,5,6

      etc.

      Reply ↓
  10. Simon K

    You’ve said that you don’t think your answer is searchable. Presumably that means that we can eliminate anything that is listed in ?

    Reply ↓
  11. Aditya Sharad

    Here’s an idea, with the first few steps and then a general process.
    Start with 1. This 1 is now our ‘current position’.
    Move 0 steps back from the current position and add the number there (1) to the sequence, then increment it and add that number (2) to the sequence. So far: 1 1 2, current position is at the 2.
    Move 1 step back from the current position and add the number there (1) to the sequence, then increment it and add that number (2) to the sequence. So far: 1 1 2 1 2.
    Move 2 steps back from the current position and add the number there (2) to the sequence, then increment it and add that number (3) to the sequence. So far: 1 1 2 1 2 2 3.
    In general:
    Start the sequence with 1. Repeat, starting with n=0:
    On the nth iteration, move n steps back, add the number found there to the end of the sequence, then add its successor to the end of the sequence.

    Reply ↓
  12. Burhan ul Haq

    Number of odd divisors on “n” where n is a natural number.
    In that case, the odd divisors of 1,2,3,4,5,6,7,8… Should be
    1,1,2,1,2,2,2,1,3,2,2,2…
    I tried the simplest approach. Fingers crossed.

    Reply ↓
      1. Jesse Thompson

        Great news, but didn’t Burhan ul Haq above make the same next-number guess that I did?

        It looks like his sequence guess matched mine also, though when I made mine none of today’s guesses had been moderated yet. ;3

        Reply ↓
  16. Poppy

    It’s a pretty weird guess but if we took the terms in the sequence 3 at a time, we can see that for the first set of 3 terms, the difference between the nth term and n-1 is 0,1,-1. Another terms is needed to evaluate the next set of 3 terms but so far the diff between two terms of the next set is 1,0. My guess is that the the pattern of difference between the terms, 0,1,-1 stays the same but with each set of 3 terms, the 0 shifts to the right. If so, the 6th term in thr sequence would be 1.

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    1. thomas Post author

      OK, this is my new favourite suggestion. Absolutely tremendous. Wrong, but I expect you probably guessed that.

      Reply ↓
  18. Burhan ul Haq

    I have a couple of proposed sequence theories. Here’s the first one:

    If your function is described by f(n), then
    f(n)=number of divisors of “n” in binary that are also suffixes of n written in binary.

    so if n=6 (6th term) i.e. 111
    the divisors of “n” will be
    1, 2, 3, 6
    OR
    1, 10, 11, 111 in binary.
    Hence, f(6) = 2 because only “1” in binary is a suffix of “6” in binary.
    based on that, the 8th term will be 1
    Let’s see how well this goes.

    Reply ↓
    1. thomas Post author

      This is a good idea (I think you mean 6 = ‘110’ and therefore ’10’ is the only suffix). Not the answer though.

      Reply ↓
    1. thomas Post author

      Thanks for the guess. Easier for me if the next number guesses go on Twitter though (but I think you’ve already noticed that and reposted there).

      Reply ↓
  20. Simon K

    I reckon this is probably too easy to be the answer, but I’ll give it a go:

    Number of prime divisors of 2^n + 1 (counting repeated terms more than once).

    Reply ↓
  21. Diederik Feilzer

    If i is the index. Then the ith number is the number of elements in the prime factorization of (2^i) + 1.
    But for example 3*3*7 is counted as three elements instead of two distinct ones.
    Next number up: 4. (because: 2^9+1+513=3*3*3*19)
    And so on: 1, 1, 2, 1, 2, 2, 2, 1, 4, 3, 2, 2,

    Reply ↓
  22. Burhan ul Haq

    Hmm…. So glad to be on the leaderboard. I didn’t fail twice, I just found two solutions that didn’t work, right?
    Okay, here’s today’s prediction
    number of prime factors of 2^n +1, where n is a natural number
    By that logic, the 9th term is the prime factors of 513, which is 4.

    Reply ↓
  23. Rene

    Does the answer depend on the reader being English, or would it work for people who speak other languages (i.e. who might use “un” instead of “one” for 1)?

    Reply ↓
    1. thomas Post author

      No. That sequence would go:

      0 0 0 0 0 0 14 0 …

      As I have told you more than once, I only eat Weetabix in large weekly binges.

      Reply ↓
  25. Troy

    Exponents of 2 = 1
    Else = 2

    I almost feel silly guessing it, but I might as well be the one.

    1 1 2 1 2 2 2 1 2 2 2 2 2 2 2 1

    Reply ↓
    1. thomas Post author

      I think it is sensible to eliminate it from your enquiries. Consider it thus eliminated.

      I will take this as a guess that today’s term is 2 (since you have explicitly written the next term) but please tweet guesses in reply to my updates for convenience. Thanks.

      Reply ↓
  27. John

    This is too convoluted to be correct, but hey…

    Let A = the nth prime, B = the nth composite number. If A > B calculate A mod B if A < B calculate B mod A. If the answer is even write down a 1. If odd write a 2. So 2 is the 9th term.

    Reply ↓
    1. thomas Post author

      Great idea. Well thought through. But no.

      Please tweet next term guesses in reply to my updates for convenience. Thanks.

      Reply ↓
    1. thomas Post author

      Good try. Your sequence must indeed repeat, because there are only two symbols in it and it refers back across four previous terms, so repetition is necessary over a maximum cycle length of 16 steps (2^4 possibilities for the 4 previous terms), I think.

      I’m afraid it’s not the answer though.

      Reply ↓
  30. Aoife Hunt

    So, there is a 1 in each power of 2 so far (2^0, 2^1, 2^2 and 2^3) and a 2 everywhere else…
    So… 1121222122222221… Etc?

    Reply ↓
    1. thomas Post author

      That’s not the rule, I’m afraid, but I will take this as a guess that today’s term is 2 (since you have explicitly written the next term).

      Please tweet guesses for the next term in reply to my updates for convenience. Thanks.

      Reply ↓
  31. Burhan ul Haq

    The sequence might indicate the number of distinctive integers when n is represented in binary:
    0, 1, 10, 11, 100, 101, 110, 111, 1000, 1001, 1010, 1011, 1100, 1101, 1110, 1111 and so on
    which makes the sequence
    1, 1, 2, 1, 2, 2, 2, 1, 2, 2, 2…..
    This solution predicts the next number to be 2.

    Reply ↓
  32. Simon Capstick

    If the position in the sequence is a power of two then represent it with a ‘1’, otherwise it is a ‘2’.

    Reply ↓
  34. John Doe

    Number of distinct numbers in binary representation of n. Where n is the index starting with 0. Tomorrow’s number 2. (Ignoring higher trailing zeros to fill the byte.)

    Reply ↓
  35. Jesse Thompson

    Now I’m curious about a few properties of our sequence, hehe. I suppose you can just decline to answer questions you feel take away from the fun, though. ?

    Will it diverge, or are the terms bounded?

    Especially in the latter case: will it repeat at some point?

    Will it contain elements beyond the natural (counting) numbers at some finite point? (zero, negatives, fractions, irrationals, complex, quaternions, infinities, infintesimals, transinfinites, supernaturals, games, 1980’s glam bands, etc?)

    Reply ↓
    1. thomas Post author

      These are very good questions, and I have hummed and hahhed somewhat about whether to answer them… but I have decided against for the time being. I am going to stick to the ‘no clues’ line, for now.

      Oh… all right then, I will throw you this very small crumb. The terms are all non-negative integers. Not much help, I realise.

      I am still pretty confident that someone will get the solution before long.

      Reply ↓
  36. Brad Smith

    Obviously not the correct answer, but prime numbers mod 3 can eventually generate the sequence so far. The earliest match I could find started at prime 1421:
    1 1 2 1 2 2 2 1 2 2 2 1 1 1 2 1 2 2…

    So that will make my guess for tomorrow: 2

    Reply ↓
    1. thomas Post author

      Interesting. I would suspect that any finite sequence of 1s and 2s could be found somewhere in the sequence of primes mod 3. More broadly, I would have thought that any finite sequence of digits mod n (excluding 0) could be found somewhere in the sequence of primes mod n. I wonder if this is true/provable.

      I’m afraid you’re right that this is not the answer though.

      Reply ↓
      1. Brad Smith

        After thinking about it, I realized that this primes mod 3 basically just a random binary sequence, so with an arbitrary sequence of length 10, I think we should expect to find it once per ~10^2 (1024) bits. It was basically right on cue. The sequence appeared again at 2334, and 4025… With another bit added today, it becomes half as frequent.

        Reply ↓
  37. Michael Bench-Capon

    My guess today is a bit like the popular “distinct digits in binary starting with 0” and “1 for powers of 2 and 2 otherwise”, (which are equivalent, right?). 1s followed by triangular numbers of 2s?
    1
    12
    1222
    1222222
    12222222222
    …

    Reply ↓
    2. thomas Post author

      And yes, I hadn’t considered it before, but those two sequences are indeed equivalent. Although to be strictly mathematically consistent, I suppose zero should really have no digits at all.

      Reply ↓
  38. Chris

    Has anyone guessed that you’re flipping a coin to generate this sequence?

    I have a silly half-guess, which is that the sequence is generated in such a way that each string of length 5 appears exactly once before the whole thing repeats. This would determine the next term: 2. I will tweet it as well (I’m ccppurcell for future reference)

    Reply ↓
    1. thomas Post author

      Diederik makes an equivalent suggestion to your coin flip idea a few posts above. Suffice to say, it isn’t that.

      The complete set of length five binary strings (+1) is an interesting idea, but not the solution, I’m afraid. Do you have an idea of what that entire sequence would look like (for curiosity’s sake)?

      Reply ↓
      1. Chris

        Aha I knew that there was a reason that it popped into my head – years ago I covered linear feedback shift registers in a course on cipher systems. The sequence generated by an LFSR has the desired property. So a full guess might be a description of an LFSR whose output is consistent with the current sequence. Could you remind me (us all) how many elements the sequence started with?

        Reply ↓
        1. thomas Post author

          The sequence started with five terms. I will save you the trouble on the LFSR idea though, because it isn’t that. Thanks though for introducing me to yet another interesting mathematical object. This puzzle is throwing up loads of fascinating ideas.

          Reply ↓
          1. Chris

            I love it! Was that your intention?

            It also illustrates how silly it is when you get a logical puzzle of the form “what’s the next number in the following sequence?” and there is supposed to be precisely one right answer.

          2. thomas Post author

            I’m not sure quite what I expected. I think I thought it would chug along in the background for ages with very little involvement (and it still might!). Very pleased that it has attracted a fair amount of interest (for which I have to thank Hannah, for tweeting it to her many followers), and great to see the variety of ideas.

            After the flurry on the first day, I thought it would be solved very quickly, but given that it is (for the moment) identical to the ‘1s for powers of 2, 2s otherwise‘ sequence, I should have realised that it would take a bit longer. I remain fairly confident that it will be solved before too long though…

  39. Nick T

    All I’ve got is a notion that this is linked to the Stern-Brocot sequence.

    Where the nth term of the S-B sequence is 1, your sequence takes the value of 1, else your sequence takes the value of 2.

    I don’t think it’s that, but wanted to put it out there, partly because the Stern-Brocot sequence is just so damn cool, and I found the link interesting.

    This would give the 12th term of your sequence as 2.

    Reply ↓
  41. Aaron P

    Here is my guess for the first 100 terms:

    1 1 2 1 2 2 2 1 2 2 2 2 2 2 2 1 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 1 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 1 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2

    The last 1 I have posted in my sequence is in the 64th term. The next 1 appears at term 128, then 256, and so on. Otherwise, all else are 2’s. I stumbled on this pattern after expanding my original formula, which I’d be happy to share if the sequence is correct.

    Reply ↓
    1. Aaron P

      To further clarify, the 1’s appear at terms which are powers of 2, otherwise the term is a 2.

      As I said, I found this neat sequence through a recursive formula, which I’d be happy to share.

      Reply ↓
      1. thomas Post author

        I’m afraid that this is not the rule. My sequence is not identical to the sequence you describe (though the terms that have been released so far are). Thanks for the guess though, and I will take your previous post as a guess that the next term is a 2. If you want to make further ‘next term’ guesses, please tweet them in response to my updates since this is more convenient for me. Thanks.

        Reply ↓
  42. The one and only Dan ever

    A three-digit binary number incrementing by the number of “1”‘s in the previous number?
    But with 1’s and 2’s instead of 0’s and 1’s?

    Original:
    1 1 2
    1 2 2
    2 1 2
    2 2 2
    (Original in 0’s and 1’s)
    0 0 1 (There is one 1, so add 1)
    0 1 1 (There are two 1’s, so add 2)
    1 0 1 (There are two 1’s, so add 2)
    1 1 1 (There are three 1’s, so add 3)

    So the next numbers would be 2121 (1010) or 121 (010) depending on if we’re ditching the overflow.

    Reply ↓
  43. Tim Howard

    The sequence is based on a progression of the triangular numbers of the count of the 1s, determining the number of 2s to follow. Ie:

    position 0 1 2 3 4 5
    triangle 0 1 3 6 10 15

    result 1 12 1222 1222222 12222222222 1222222222222222

    Reply ↓
  44. Alexander

    Considering the fact that no one has found a solution, I think my idea seems too simple. But since it still fits your sequence (I looked in the comments but didn´t see this idea yet. If I overlooked it: Sorry!):

    Sequence if one follows my rule-idea: 112122212222221222222222212222222222222221…
    Rule:
    “1”
    (0)x”2″
    “1”
    (0+1)x”2″
    “1”
    (0+1+2)x”2″
    “1”
    (0+1+2+3)x”2″
    “1”
    ….

    Reply ↓
    1. thomas Post author

      Nice idea. Well spotted. But this is not the rule.

      You are in good company though, with Tim Howard (above) also suggesting the same rule today.

      Reply ↓
  45. adam kugler

    Here is my guess at the first 100 terms

    1121222122222122222221222222222122222222222122222222222221222222222222222122222222222222222122222222

    I generated it using more of a mathematical concept than an equation. Not sure if that counts 🙂

    Reply ↓
  46. Jack

    Very simple answer I came up with…although I haven’t read through all the other answers to see if this was already suggested.

    Sequence positions 1, 2, 4, & 8 are all ones….the others are all twos…

    Therefore I suggest positions 16, 32, 64 etc would be ones also. i.e. the positions where a one exists is doubled each time.

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  47. Myria

    If someone determined a rule different to your own that produced the same sequence, would you consider that a solution?

    Melissa

    Reply ↓
    1. thomas Post author

      Any rule that would generate the complete correct sequence would be a correct solution. I have clarified this in the rules. Thanks for spotting the ambiguity.

      Reply ↓
  48. Myria

    My guess is 1, 1, f(n=2…) where f(x) = 1 if x is a Mersenne prime, and 2 otherwise.

    This guess could probably be represented in closed form using the σ_1 function. Perhaps the floor of sigma(x)/x or similar would be involved.

    Reply ↓
  49. cinnamon_carter

    As an additional incentive for those who are participating

    (ALSO since I am a member of the Blake Coin development team and have a lot of coins mining it for years)

    I have pledged 15,000 Blake Coin to whoever Thomas declares the winner. Once he states someone has solved the problem all you need to do is provide a blake coin address for me to send them to. Blake coin is a cryptocurrency similar to Bitcoin but with several advantages most of which are from using a lighter algorithm that allows more ‘speed’ ….

    If you ever wished you got into bitcoin early but missed it and don’t want to pay $200-$250 for one bitcoin this can be your lucky day. Learn more at the website and get a wallet if you win.

    As for the contest I find it challenging although at first I thought it would be easy to solve !!!

    To show my respect I pledged this supplement to the prize and also to help attract some potential cryptography people to help solve it !! I guess if i solve it I get to keep my coins !!!

    Reply ↓
        1. mike

          Thomas

          So the winner will be the one who can psychoanalyse which of the correct solutions you happened to pick?

          For the record my predictions are implied by the sequence

          1, 1, 2, 1, 2, 2, 2, 1, 2, 2, 2, 2, 2, 2, 3, 1, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 3, 2, 2, 3, 2, 1, 2, 2, 2, 2, 2, 2, 3, 2, 2, 2, 2, 2, 3, 2, 2, 2, 2, 2, 3, 2, 2, 3, 2, 2, 2, 2, 2, 3…

          Reply ↓
          1. thomas Post author

            Err… yes. Something like that. There is a particular rule generating the sequence and the idea is to guess that rule. No matter how many terms have been provided, there will always be multiple possible rules that could fit (e.g. an n-1 th order polynomial can always describe n terms).

            I have duly noted your guess for today’s term. I can’t take more than one guess in advance though, because that would mean trawling back through all the comments before each update, rather than just that day’s, which takes too long. Sorry about that.

  50. Tom

    so far its 1 1 2 1 2 2 2 1 2 2 2 2 2 2

    Next will be a 2 followed by a 1 followed by 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 followed by 1

    Going on the theory that the number of 2’s you see after a 1 is equal to the number of 2’s +1

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  51. Jamil Fabia

    1 (2^0) , 1 (copy previous digits) + 2^1 , 1 2 (copy previous digits) + 2 2 (2^2 expanded) , 1 2 2 2 (copy previous digits) + 2 2 2 (2^3 expanded)

    Thanks!

    Reply ↓
    1. thomas Post author

      This is a great idea. It fits. But it’s not the answer.

      I’m guessing that you came up with this by trying to find a property that makes 15 unique (in this case, it is the first number with two distinct odd prime factors) and worked back from there. This seems a good way to proceed, even if it has not been successful on this occasion.

      Reply ↓
  53. Myria

    Since so far the sequence has produced powers of 2 -> 1, I’m going to apply a heuristic and guess that the 16th term, on 2015/09/12, will be a 1.

    With 15 -> 3, it’s finally diverged from the direct powers-of-2 sequence; I have no idea anymore what it is.

    Reply ↓
  54. H J

    reminds me of filling beaker of increasing size. we start with beaker 1 and fill it to 1 unit. then we go to beaker 2, fill it to 1 unit. then add 1 and we are now filled to 2 units. then we pour off 1 unit and are back to beaker 2 at 1 unit. then we fill back to 2. and then to be fancy we fill beaker 2 directly to 2 units. then we move on to beaker 3. the question is, do we fill beaker 3 to 1 unit or to 3-1, ie 2 units. I will guess we start at 1 unit so today’s integer will be 1. More (but not everything) will be revealed by today’s integer. I can write this as a formula but for now I like thinking about beakers.
    if i am right the sequence could be:
    11 21 22 21 22 22 31 32 33 31 32 33 33 . . . but it is too soon to know several things. when we get the 3 sequence completed then, if my general approach is correct, we can extrapolate the whole pattern.

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  55. H J

    I see that I wrote the original sequence incorrectly. without going into a revised rationale, I will just say that today’s integer should still be 1 or 2 and I am going to guess 1.

    okay the revised sequence
    11 21 22 21 22 22 22 31 32 33 31 32 33 33 33 33 41 . . . the same caveats apply and the same revelation will be possible when we see what happens with 3

    Reply ↓
    1. thomas Post author

      I can’t really follow how your sequence works, I’m afraid. It certainly looks clever, but could you explain it again?

      Reply ↓
    1. thomas Post author

      Ahhrgh. It’s not that, I’m sorry!

      I really thought you had it when I saw your tweet, but no. We will diverge from omega(2n) (the number of distinct prime factors of 2*n) at some point.

      Sorry…

      Reply ↓
    1. thomas Post author

      Thanks. It’s easier for me if you tweet your guesses in response to my daily updates (if you have a Twitter account), but your guess for the next term is duly noted.

      Reply ↓
  59. Troy

    Since 1 isn’t prime, how about a slight adjustment:

    The count of non-composite, odd factors.

    1,1,2,1,2,2,2,1,2,2,2,2,2,2,3,1,2,2,2,2,3,2,2

    Reply ↓
  60. Jesse Thompson

    Hmm. Long ago I took a stab at “the number of odd integers that can divide evenly into n”, but I still feel like the factors of N are somehow involved. So let’s try something different.

    How about, the number of factors F of N (including 1, and including N) that satisfy “(N-F) divides evenly into 2 ^ (floor(log₂F)+1)”?

    This will render terms 17-26 = 2, while term 27 = 3 again because 27’s factors are 1, 3, 9, 27.
    27-1 = 26, does divide evenly into 2 ^ (floor(log₂1)+1)=2 (counts).
    27-3 = 24, does divide evenly into 2 ^ (floor(log₂3)+1)=4 (counts).
    27-9 = 18, does NOT divide evenly into 2 ^ (floor(log₂9)+1)=16 (no count).
    27-27 = 0, divides evenly into everything including 2 ^ (floor(log₂27)+1)=32 (counts).

    All of the Mersenne terms in this sequence like 15 are interesting, but the Mersenne Primes like 31 and 127 just render “2” at us. The non-primes are the fun ones, 63 should give us our first 4. Out of it’s factors 1, 3, 7, 9, 21, 63 we’d kick out the 9 and the 21 as not qualifying, leaving 4 remaining.

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  61. cinnamon_carter

    Looking the big picture over the reciprocals of primes and variations of produce divergent series that diverge ‘slowly’…. I think the solution could lie here…. but one never knows !

    Reply ↓
  62. Michael Bench-Capon

    This is what it looks like to me at the moment: f(n) = 1 + number of distinct odd prime factors of n. It seems an odd choice, but perhaps you’re using a more elegant rule that generates the same sequence.

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  63. Jesse Thompson

    Hmm. It looks like:
    f(N) = ceil(hamming_weight(N) * k) for any 0.5 < k <= (2/3) will generate this exact sequence.

    In fact, no k within that range is capable of diverging any sequence you put it into until term 31 when hamming distance finally hits 5 (k=0.5 + Ɛ predicts 3 while k=(2/3) predicts 4).

    Is it legal for me to guess whether or not this generator with an unknown, yet bounded k is capable of describing your sequence, or are you going to force me to choose a k before you can let us dismiss this slice of potentials? hehe :3

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    1. thomas Post author

      Good thought. However, there is no value of k that would make this equal to the correct sequence.

      (For those who don’t know, ‘hamming_weight’ here is referring to the number of 1s in the binary representation of N – the term number.)

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  64. hj

    Just got back to my email account. You mentioned that you didn’t understand my rationale but recent numbers render it incorrect anyway. It is an operation according to rules of addition and subtraction. I have a revised idea but for now will simply offer the next number and see what happens.
    1121222122222231222 . . and the next number would be 1.

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  66. Tom Flowerdew

    Is it the number of factors of the sum of the binary representations of the numbers from 1-20?

    e.g. 12 is 1100. 1+1+0+0 = 2 which has two factors
    15 is 1111. 1+1+1+1 = 4 which has three factors, etc…

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