A025586 Largest value in '3x+1' trajectory of n.

1, 2, 16, 4, 16, 16, 52, 8, 52, 16, 52, 16, 40, 52, 160, 16, 52, 52, 88, 20, 64, 52, 160, 24, 88, 40, 9232, 52, 88, 160, 9232, 32, 100, 52, 160, 52, 112, 88, 304, 40, 9232, 64, 196, 52, 136, 160, 9232, 48, 148, 88, 232, 52, 160, 9232, 9232, 56, 196, 88, 304, 160, 184, 9232

Offset: 1

Author

David W. Wilson

Keywords

• nonn
• nice
• look

Comments

Here by definition the trajectory ends when 1 is reached. Therefore this sequence differs for n = 1 and n = 2 from A056959, which considers the orbit ending in the infinite loop 1 -> 4 -> 2 -> 1.

a(n) = A220237(n,A006577(n)). - Reinhard Zumkeller, Jan 03 2013

A006885 and A006884 give record values and where they occur. - Reinhard Zumkeller, May 11 2013

For n > 2, a(n) is divisible by 4. See the explanatory comment in A056959. - Peter Munn, Oct 14 2019

In an email of Aug 06 2023, Guy Chouraqui observes that the digital root of a(n) appears to be either 7 or a multiple of 4 for all n > 2. (See also A006885.) - N. J. A. Sloane, Aug 11 2023

Examples

			The 3x + 1 trajectory of 9 is 9, 28, 14, 7, 22, 11, 34, 17, 52, 26, 13, 40, 20, 10, 5, 16, 8, 4, 2, 1 (see A033479). Since the largest number in that sequence is 52, a(9) = 52.
		

Links

• T. D. Noe, Table of n, a(n) for n = 1..10000
• Christian Hercher, There are no Collatz m-Cycles with m <= 91, J. Int. Seq. (2023) Vol. 26, Article 23.3.5.
• Philippe Picart, Algorithme de Collatz et conjecture de Syracuse
• Index entries for sequences related to 3x+1 (or Collatz) problem

Crossrefs

Essentially the same as A056959: only a(1) and a(2) differ, see Comments.

Cf. A006370, A006577, A006884, A006885, A220237.

Programs

Haskell

a025586 = last . a220237_row
-- Reinhard Zumkeller, Jan 03 2013, Aug 29 2012

Maple

a:= proc(n) option remember; `if`(n=1, 1,
      max(n, a(`if`(n::even, n/2, 3*n+1))))
    end:
seq(a(n), n=1..87);  # Alois P. Heinz, Oct 16 2021

Mathematica

collatz[a0_Integer, maxits_:1000] := NestWhileList[If[EvenQ[#], #/2, 3# + 1] &, a0, Unequal[#, 1, -1, -10, -34] &, 1, maxits]; (* collatz[n] function definition by Eric Weisstein *) Flatten[Table[Take[Sort[Collatz[n], Greater], 1], {n, 60}]] (* Alonso del Arte, Nov 14 2007 *)
collatzMax[n_] := Module[{r = m = n}, While[m > 2, If[OddQ[m], m = 3 * m + 1; If[m > r, r = m], m = m/2]]; r]; Table[ collatzMax[n], {n, 100}] (* Jean-François Alcover, Jan 28 2015, after Charles R Greathouse IV *)
(* Using Weisstein's collatz[n] definition above *) Table[Max[collatz[n]], {n, 100}] (* Alonso del Arte, May 25 2019 *)

PARI

a(n)=my(r=n);while(n>2,if(n%2,n=3*n+1;if(n>r,r=n),n/=2));r \\ Charles R Greathouse IV, Jul 19 2011

Python

def a(n):
    if n<2: return 1
    l=[n, ]
    while True:
        if n%2==0: n//=2
        else: n = 3*n + 1
        if not n in l:
            l+=[n, ]
            if n<2: break
        else: break
    return max(l)
print([a(n) for n in range(1, 101)]) # Indranil Ghosh, Apr 14 2017

Scala

def collatz(n: Int): Int = (n % 2) match {
  case 0 => n / 2
  case 1 => 3 * n + 1
}
def collatzTrajectory(start: Int): List[Int] = if (start == 1) List(1)
else {
  import scala.collection.mutable.ListBuffer
  var curr = start; var trajectory = new ListBuffer[Int]()
  while (curr > 1) { trajectory += curr; curr = collatz(curr) }
  trajectory.toList
}
for (n <- 1 to 100) yield collatzTrajectory(n).max // Alonso del Arte, Jun 02 2019