A337149 Positive integers k such that the number of steps it takes to reach 1 in the '3x+1' problem is different for all j < k.

1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 14, 16, 17, 18, 22, 24, 25, 27, 28, 31, 33, 34, 36, 39, 41, 43, 47, 48, 49, 54, 57, 62, 65, 71, 72, 73, 78, 82, 86, 91, 94, 97, 98, 103, 105, 107, 108, 111, 114, 121, 123, 124, 129, 130, 135, 137, 142, 145, 153, 155, 159

Offset

1,2

Comments

Positive integers k such that A337144(k) = 1.

Or positive integers k such that A006577(k) != A006577(j) for all j = 1..k-1.

Different from A129304.

Links

Alois P. Heinz, Table of n, a(n) for n = 1..1000

Wikipedia, Collatz Conjecture

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

Formula

A006577(a(n)) = A337150(n).

Maple

collatz:= proc(n) option remember; `if`(n=1, 0,
   1 + collatz(`if`(n::even, n/2, 3*n+1)))
end:
b:= proc() 0 end:
g:= proc(n) option remember; local t;
     `if`(n=1, 0, g(n-1));
      t:= collatz(n); b(t):= b(t)+1
    end:
a:= proc(n) option remember; local k; for k
      from 1+a(n-1) while g(k)>1 do od; k
    end: a(0):=0:
seq(a(n), n=1..100);

Mathematica

collatz[n_] := collatz[n] = If[n==1, 0,
   1+collatz[If[EvenQ[n], n/2, 3n+1]]];
b[_] = 0;
g[n_] := g[n] = Module[{t}, If[n==1, 0, g[n-1]];
   t = collatz[n]; b[t] = b[t]+1];
a[n_] := a[n] = Module[{k}, For[k = 1+a[n-1],
   g[k] > 1, k++]; k]; a[0] = 0;
Array[a, 100] (* Jean-François Alcover, Jan 30 2021, after Alois P. Heinz *)

Crossrefs

Cf. A006577, A129304, A337144, A337150.

Sequence in context: A330236 A302593 A129304 * A294358 A262358 A032962

Adjacent sequences: A337146 A337147 A337148 * A337150 A337151 A337152

Keyword

nonn,look

Author

Alois P. Heinz, Jan 27 2021

Status

approved