A056959 In repeated iterations of function m -> m/2 if m even, m -> 3m+1 if m odd, a(n) is maximum value achieved if starting from n.

4, 4, 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

Offset: 1

Author

Henry Bottomley, Jul 18 2000

Keywords

• nonn

Comments

If a(n) exists (which is the essence of the "3x+1" problem) then a(n) must be a multiple of 4, since if a(n) was odd then the next iteration 3*a(n)+1 would be greater than a(n), while if a(n) was twice an odd number then the next-but-one iteration (3/2)*a(n)+1 would be greater.

The variant A025586 considers the trajectory ending in 1, by definition. Therefore the two sequences differ just at a(1) and a(2). - M. F. Hasler, Oct 20 2019

Examples

			a(6) = 16 since iteration starts: 6, 3, 10, 5, 16, 8, 4, 2, 1, 4, 2, 1, ... and 16 is highest value.
		

Links

• Index entries for sequences related to 3x+1 (or Collatz) problem

Crossrefs

Cf. A006370, A056957, A056958.

Essentially the same as A025586.

Programs

Maple

a:= proc(n) option remember; `if`(n=1, 4,
      max(n, a(`if`(n::even, n/2, 3*n+1))))
    end:
seq(a(n), n=1..88);  # Alois P. Heinz, Oct 16 2021

Mathematica

a[n_] := Module[{r = n, m = n}, If[n <= 2, 4, While[m > 2, If[OddQ[m], m = 3*m + 1; If[m > r, r = m], m = m/2]]; r]];
Table[a[n], {n, 1, 100}] (* Jean-François Alcover, May 20 2022 *)

PARI

a(n)=my(r=max(4,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