This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.
%I A283075 #33 Jul 22 2017 21:36:30 %S A283075 2,5,11,23,48,101,218,473,1044,2284,5011,10979,23954 %N A283075 Number of time intervals that can be measured with n ropes and a lighter from the moment when the first rope begins burning. %C A283075 Suppose you have n ropes and a lighter. Each rope burns at a nonconstant rate but takes exactly one hour to burn completely from one end to the other. You can only light the ropes at either of their ends but can decide when to light each end as you see fit. If you're strategic in how you burn the ropes, how many specific lengths of time can you measure? (For example, if you had one rope, you could measure two lengths of time: one hour, by simply burning the entire rope from one end, and half an hour, by burning the rope from both ends and marking when the flames meet.) %C A283075 In this sequence, the time intervals begin when the first rope (or safety fuse, or match cord) begins burning. Without this restriction, if we are allowed to burn some ropes beforehand, more time intervals are available for measuring off. - _Andrey Zabolotskiy_, Feb 28 2017 %H A283075 FiveThirtyEight, <a href="https://fivethirtyeight.com/features/a-riddle-how-many-crooked-politicians-are-there/">The Riddler</a> (see section Riddler Classic), see also <a href="https://fivethirtyeight.com/features/two-mysteries-of-the-gold-coins/">the solution</a> (for a(4)) %H A283075 Samuel Cormier-Iijima, <a href="https://github.com/sciyoshi/riddler/blob/master/ropes.py">Python implementation</a> %e A283075 a(2)=5: (i) Generate 1 by burning one rope from one end. (ii) Generate 2 by burning one rope from one end at t=0 and the other afterwards at t=1 from one end. (iii) Generate 1/2 by burning 1 rope from both ends. (iv) Generate 3/2 by burning 1 rope from one end at t=0 then the other from both ends at t=1 (or swapped order). (v) Generate 3/4 by burning one rope at t=0 from both ends, starting the other also at t=0 at one end, and lighting the other's second end at t=1/2 when the first rope's flames have reached the middle, so the 2nd rope's flames finish at t=3/4. %e A283075 For n = 3 the a(3) = 11 solutions are 1/2, 3/4, 7/8, 1, 9/8, 5/4, 3/2, 7/4, 2, 5/2, 3. %Y A283075 Cf. A287012. %K A283075 nonn,hard,more %O A283075 1,1 %A A283075 _Samuel Cormier-Iijima_, Feb 28 2017 %E A283075 a(9) from _Andrey Zabolotskiy_, Mar 02 2017 %E A283075 a(10)-a(13) from _Mark Rickert_, Apr 10 2017 %E A283075 Name clarified by _Mark Rickert_, Jun 08 2017