A347409 - OEIS

%I #38 Oct 24 2022 21:59:33

%S 0,1,4,2,4,4,4,3,4,4,4,4,4,4,5,4,4,4,4,4,6,4,5,4,4,4,5,4,4,5,5,5,4,4,

%T 5,4,4,4,4,4,5,6,4,4,4,5,5,4,4,4,4,4,5,5,5,4,4,4,4,5,5,5,5,6,4,4,4,4,

%U 4,5,5,4,5,4,8,4,4,4,4,4,5,5,5,6,8,4,4

%N Longest run of halving steps in the trajectory from n to 1 in the Collatz map (or 3x+1 problem), or -1 if no such trajectory exists.

%C If the longest run of halving steps occurs as the final part of the trajectory, a(n) = A135282(n), otherwise a(n) > A135282(n).

%C Every nonnegative integer appears in the sequence at least once, since for any k >= 0 a(2^k) = k.

%C Conjecture: every integer >= 4 appears in the sequence infinitely many times.

%H Charles R Greathouse IV, <a href="/A347409/b347409.txt">Table of n, a(n) for n = 1..10000</a>

%H <a href="/index/3#3x1">Index entries for sequences related to 3x+1 (or Collatz) problem</a>

%e a(15) = 5 because the Collatz trajectory starting at 15 contains, as the longest halving run, a 5-step subtrajectory (namely, 160 -> 80 -> 40 -> 20 -> 10 -> 5).

%t nterms=100;Table[c=n;sm=0;While[c>1,If[OddQ[c],c=3c+1,If[(s=IntegerExponent[c,2])>sm,sm=s];c/=2^s]];sm,{n,nterms}]

%o (PARI) a(n)=my(nb=0); while (n != 1, if (n % 2, n=3*n+1, my(x = valuation(n, 2)); n /= 2^x; nb = max(nb, x));); nb; \\ _Michel Marcus_, Sep 03 2021

%o (Python)

%o def A347409(n):

%o     m, r = n, 0

%o     while m > 1:

%o         if m % 2:

%o             m = 3*m + 1

%o         else:

%o             s = bin(m)[2:]

%o             c = len(s)-len(s.rstrip('0'))

%o             m //= 2**c

%o             r = max(r,c)

%o     return r # _Chai Wah Wu_, Sep 29 2021

%Y Cf. A006370, A070165, A135282.

%Y Cf. A347668 (indices of records), A347669 (indices of first occurrences), A348007.

%K nonn,easy

%O 1,3

%A _Paolo Xausa_, Aug 30 2021