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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.

%I A351850 #21 Apr 10 2022 15:33:34
%S A351850 0,2,24,6,20,30,128,14,152,30,120,42,64,142,300,30,108,170,236,50,84,
%T A351850 142,284,66,300,90,40656,170,216,330,40524,62,384,142,260,206,264,274,
%U A351850 996,90,40628,126,596,186,256,330,40492,114,388,350,520,142,224,40710
%N 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.
%C A351850 If x is even, the A351849 tag system evolves from the word encoding x to the word encoding x/2 in x iterations; if x is odd, x+1 iterations are required to produce the word encoding (3x+1)/2.
%C A351850 See A351849 for additional comments, links and examples.
%H A351850 Paolo Xausa, <a href="/A351850/b351850.txt">Table of n, a(n) for n = 1..10000</a>
%H A351850 Liesbeth De Mol, <a href="https://doi.org/10.1016/j.tcs.2007.10.020">Tag systems and Collatz-like functions</a>, Theoretical Computer Science, Volume 390, Issue 1, 2008, pp. 92-101.
%H A351850 <a href="/index/3#3x1">Index entries for sequences related to 3x+1 (or Collatz) problem</a>.
%F A351850 a(n) = (Sum_{k=0..A006666(n)} 2*floor((A070168(n,k)+1)/2)) - 2.
%e A351850 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.
%t A351850 (* First program, based on the tag system definition *)
%t A351850 t[s_]:=StringDrop[s,2]<>StringReplace[StringTake[s,1],{"1"->"23","2"->"1","3"->"111"}];
%t A351850 nterms=100;Table[Length[NestWhileList[t,StringRepeat["1",n],#!="1"&]]-1,{n,nterms}]
%t A351850 (* Second program, more efficient, based on formula *)
%t A351850 c[x_]:=If[OddQ[x],(3x+1)/2,x/2];
%t A351850 nterms=100;Table[Total[Map[If[OddQ[#],#+1,#]&,NestWhileList[c,n,#>1&]]]-2,{n,nterms}]
%o A351850 (Python)
%o A351850 def A351850(n):
%o A351850     s, steps = "1" * n, 0
%o A351850     while s != "1":
%o A351850         if s[0] == "1": s += "23"
%o A351850         elif s[0] == "2": s += "1"
%o A351850         else: s += "111"
%o A351850         s = s[2:]
%o A351850         steps += 1
%o A351850     return steps
%o A351850 nterms = 100
%o A351850 print([A351850(n) for n in range(1, nterms + 1)])
%Y A351850 Cf. A006666, A070168, A285098, A351849.
%K A351850 nonn
%O A351850 1,2
%A A351850 _Paolo Xausa_, Feb 22 2022