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

1,1

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

Links

Table of n, a(n) for n=1..61.

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

Example

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

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 *)

Prog

(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

Crossrefs

Cf. A006370, A056957, A056958.

Essentially the same as A025586.

Sequence in context: A177241 A076821 A165825 * A255300 A255298 A255302

Adjacent sequences: A056956 A056957 A056958 * A056960 A056961 A056962

Keyword

nonn

Author

Henry Bottomley, Jul 18 2000

Status

approved