A340672 Number of repeated applications of the '4x+-1' operation needed to reach 1 from n, or -1 if 1 is never reached.

0, 3, 1, 8, 6, 4, 4, 11, 2, 36, 9, 9, 34, 16, 7, 7, 32, 5, 14, 5, 5, 39, 30, 12, 12, 12, 3, 46, 37, 37, 28, 19, 10, 10, 10, 10, 44, 35, 35, 26, 26, 17, 17, 17, 8, 8, 8, 8, 42, 33, 33, 33, 24, 6, 24, 15, 15, 15, 15, 6, 6, 49, 6, 40, 40, 40, 31, 31, 31, 31, 22

Offset

1,2

Comments

Start with any positive number n. Define f(n) as 'if n mod 3 = 1, f(n) = 4n-1; if n mod 3 = 2, f(n) = 4n+1; if n is divisible by 3, f(n) = n/3'. Apply this procedure repeatedly to reach 1. This sequence gives the number of steps required to reach 1, or -1 if 1 is never reached. The procedure is a modification of the famous Collatz procedure.

We can make a conjecture similar to that of Collatz to claim that the process will always reach 1. However, this is not proven yet.

It seems that more than 65 percent of the terms in this sequence satisfy a(i) = a(i+1). For example, 6655 of the first 10000 terms satisfy this condition.

Links

Md. Towhidul Islam, Table of n, a(n) for n = 1..10001

Example

a(5)=6 because the trajectory of 5 is (5,21,7,27,9,3,1). That needs 6 repeated applications of the procedure to reach 1.

Maple

a:= proc(n) option remember; `if`(n=1, 0, 1 + (r->
      a(`if`(r=0, q, 4*n+2*r-3)))(irem(n, 3, 'q')))
    end:
seq(a(n), n=1..75);  # Alois P. Heinz, Jan 20 2021

Mathematica

a[n_] := a[n] = If[n == 1, 0, {q, r} = QuotientRemainder[n, 3]; 1 +
     a[If[r == 0, q, 4n + 2r - 3]]];
Table[a[n], {n, 1, 75}] (* Jean-François Alcover, May 29 2022, after Alois P. Heinz *)

Prog

(PARI) f(n) = my(m=n%3); if (m==1, 4*n-1, if (m==2, 4*n+1, n/3));
a(n) = my(s=n, c=0); while(s>1, s=f(s); c++); c; \\ Michel Marcus, Jan 18 2021

Crossrefs

Cf. A006577.

Sequence in context: A124025 A257488 A286416 * A005295 A077897 A308737

Adjacent sequences: A340669 A340670 A340671 * A340673 A340674 A340675

Keyword

nonn,easy,look

Author

Md. Towhidul Islam, Jan 15 2021

Status

approved