A351850 a(n) is the number of iterations of the computation of the A351849 tag system when started from the word encoding n, or -1 if the number of iterations is infinite.
0, 2, 24, 6, 20, 30, 128, 14, 152, 30, 120, 42, 64, 142, 300, 30, 108, 170, 236, 50, 84, 142, 284, 66, 300, 90, 40656, 170, 216, 330, 40524, 62, 384, 142, 260, 206, 264, 274, 996, 90, 40628, 126, 596, 186, 256, 330, 40492, 114, 388, 350, 520, 142, 224, 40710
Offset: 1
Keywords
Examples
When started from 1111 (the word encoding the number 4), the system evolves as 1111 -> 1123 -> 2323 -> 231 -> 11 -> 23 -> 1, reaching the word 1 after 6 steps. a(4) is therefore 6.
Links
- Paolo Xausa, Table of n, a(n) for n = 1..10000
- Liesbeth De Mol, Tag systems and Collatz-like functions, Theoretical Computer Science, Volume 390, Issue 1, 2008, pp. 92-101.
- Index entries for sequences related to 3x+1 (or Collatz) problem.
Programs
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Mathematica
(* First program, based on the tag system definition *) t[s_]:=StringDrop[s,2]<>StringReplace[StringTake[s,1],{"1"->"23","2"->"1","3"->"111"}]; nterms=100;Table[Length[NestWhileList[t,StringRepeat["1",n],#!="1"&]]-1,{n,nterms}] (* Second program, more efficient, based on formula *) c[x_]:=If[OddQ[x],(3x+1)/2,x/2]; nterms=100;Table[Total[Map[If[OddQ[#],#+1,#]&,NestWhileList[c,n,#>1&]]]-2,{n,nterms}] -
Python
def A351850(n): s, steps = "1" * n, 0 while s != "1": if s[0] == "1": s += "23" elif s[0] == "2": s += "1" else: s += "111" s = s[2:] steps += 1 return steps nterms = 100 print([A351850(n) for n in range(1, nterms + 1)])
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