cp's OEIS Frontend

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.

Showing 1-3 of 3 results.

A188545 Fusible numbers.

Original entry on oeis.org

1, 3, 10
Offset: 0

Views

Author

Ed Pegg Jr, Apr 03 2011

Keywords

Comments

Given irregular one minute fuses, the shortest amount of time that can be measured over n minutes is time(n) = n + 1/2^(fuse(n)) minutes. For example, fuse(1)=3, for 9/8 = 1 + 1/(2^fuse(1)). Over 2 minutes, time(2) = 2 + 1/(2^fuse(2)) = 2049/1024 minutes. The value for fuse(3) is larger than 2↑↑↑↑↑↑↑↑↑16, in Knuth's up-arrow notation. - Ed Pegg Jr, Apr 03 2011; edited by Junyan Xu, Jan 04 2012

Crossrefs

A287012 Number of time intervals that can be measured off with n ropes and a lighter.

Original entry on oeis.org

2, 6, 15, 34, 78, 174, 386, 844, 1837, 3960, 8513, 18238
Offset: 1

Views

Author

Mark Rickert, May 17 2017

Keywords

Comments

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.)
In this sequence, the time intervals begin when any rope (or safety fuse, or match cord) begins or stops burning.

Examples

			a(2)=6: (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 met, so the 2nd rope's flames finish at t=3/4. (vi) Generate 1/4 using the technique for 3/4 and measuring the time between t=1/2 and t=3/4.
For n = 3 the a(3) = 15 solutions are 1/8, 1/4, 3/8, 1/2, 5/8, 3/4, 7/8, 1, 9/8, 5/4, 3/2, 7/4, 2, 5/2, 3.
		

Crossrefs

A343264 Cardinalities of the sets of fusible numbers obtained at the consecutive steps of their construction as follows. We set S(0) = {0}. S(n+1) is obtained by adding to S(n) the sums (x+y+1)/2 for all x,y from S(n) with the property |x-y| < 1. Then, a(n) is the number of elements in S(n).

Original entry on oeis.org

1, 2, 4, 9, 21, 50, 119, 281, 656, 1513, 3449, 7777, 17363, 38422, 84355, 183915, 398526, 858901
Offset: 0

Views

Author

Mamuka Jibladze, Apr 09 2021

Keywords

Examples

			a(1) = 2 because S(1) = {0, 1/2};
a(2) = 4 because S(2) = {0, 1/2, 3/4, 1};
a(3) = 9 because S(3) = {0, 1/2, 3/4, 7/8, 1, 9/8, 5/4, 11/8, 3/2}.
		

Crossrefs

Cf. A283075.

Programs

  • Maple
    s:= proc(n) option remember; `if`(n=0, {0}, (l-> (m-> {seq([2*x, seq(
         `if`(abs(x-y) nops(s(n)):
    seq(a(n), n=0..10);  # Alois P. Heinz, Apr 09 2021
  • Mathematica
    S[n_]:=S[n]=If[n==0,{0},S[n-1]\[Union]Map[(#[[1]]+#[[2]]+1)/2&,Select[Tuples[S[n-1],{2}],Abs[#[[1]]-#[[2]]]<1&]]]; Table[Length[S[n]],{n,0,12}]
  • PARI
    \\ See Corneth link. David A. Corneth, Apr 09 2021

Extensions

a(13) from Alois P. Heinz, Apr 09 2021
a(14)-a(17) from David A. Corneth, Apr 10 2021
Showing 1-3 of 3 results.