Find a rule that applies for the first six entries and add at least one more.






The rule is simple.

Each individual entry has multiple valid solution, this is just one such set. The fact that the next entry is longer than the previous one doesn't need to be true for all entries.

  • 6
    $\begingroup$ Next words: in lieu of pre-determined donut payment. $\endgroup$ Apr 20, 2017 at 15:47
  • $\begingroup$ @IanMacDonald why would you think that? I'm curious. $\endgroup$
    – Vepir
    Apr 20, 2017 at 15:49
  • $\begingroup$ Is the trailing space after recompensations significant? $\endgroup$
    – Scott M
    Apr 20, 2017 at 15:49
  • $\begingroup$ @ScottM Yes it is. $\endgroup$
    – Vepir
    Apr 20, 2017 at 15:50
  • 9
    $\begingroup$ I cannot see any meaningful similarities amongst the words apart from them grammatically forming the beginning of a sentence when taken in the order provided. I have simply chosen to finish the sentence as though this were a court case involving a dispute about missing donuts. $\endgroup$ Apr 20, 2017 at 15:53

1 Answer 1


How about

at every opportunity that I can.

Each line

is double the byte count of the previous one.

  • $\begingroup$ Very Nice.Short and precise.Lovely. $\endgroup$ Apr 20, 2017 at 16:03
  • 5
    $\begingroup$ This leads me to believe that in lieu of agreed donut payment. would also be an acceptable solution. If that is the case, this puzzle is not specific enough and has far too many possible correct answers. $\endgroup$ Apr 20, 2017 at 17:10
  • 3
    $\begingroup$ @IanMacDonald The question is to find the rule. Which is unique. A hidden trailing space after the last entry hints toward the rule itself. It is stated that multiple valid solutions can replace the existing entries. $\endgroup$
    – Vepir
    Apr 20, 2017 at 18:22
  • 1
    $\begingroup$ @Vepir: But the rule isn't unique. $\endgroup$
    – Deusovi
    Apr 20, 2017 at 20:38
  • 1
    $\begingroup$ @Deusovi If you consider the fact we are dealing with bytes, which leads to binary, which leads to powers of 2, which is the rule; And the fact the the rule is simple. But I guess nothing is unique if you try to fit things like "a(n)=2a(n-1), except every tenth time you multiply by 1000/512 instead of by 2" ? or something like: a(1) = 1; for n>1, a(n) is the smallest positive integer not already present which is entailed by the rules (i) k present => 2k present; (ii) 3k+1 present and k odd => k present., // which is among "similar sequences". $\endgroup$
    – Vepir
    Apr 20, 2017 at 20:45

This site is temporarily in read-only mode and not accepting new answers.

Not the answer you're looking for? Browse other questions tagged .