A056959 - OEIS

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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

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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]];