A317753 Number of steps to reach 1 in 7x+-1 problem, or -1 if 1 is never reached.

0, 1, 13, 2, 10, 14, 18, 3, 7, 11, 53, 15, 19, 19, 23, 4, 27, 8, 50, 12, 73, 54, 16, 16, 58, 20, 20, 20, 43, 24, 24, 5, 47, 28, 325, 9, 70, 51, 32, 13, 13, 74, 272, 55, 55, 17, 17, 17, 276, 59, 40, 21, 40, 21, 21, 21, 63, 44, 63

Offset

1,3

Comments

The 7x+-1 problem is as follows. Start with any natural number n. If 4 divides n-1, multiply it by 7 and add 1; if 4 divides n+1, multiply it by 7 and subtract 1; otherwise divide it by 2. The 7x+-1 problem concerns the question whether we always reach 1.

The number of steps to reach 1 is also called the total stopping time.

Also the least positive k for which the iterate A317640^k(n) = 1.

Links

David Barina, Table of n, a(n) for n = 1..10000

D. Barina, 7x+-1: Close Relative of Collatz Problem, arXiv:1807.00908 [math.NT], 2018.

K. Matthews, David Barina's 7x+1 conjecture.

Example

a(5)=10 because the trajectory of 5 is (5, 36, 18, 9, 64, 32, 16, 8, 4, 2, 1).

Mathematica

f[n_] := Switch[Mod[n, 4], 0, n/2, 1, 7 n + 1, 2, n/2, 3, 7 n - 1]; a[n_] := Length@NestWhileList[f, n, # > 1 &] - 1; Array[a, 70] (* Robert G. Wilson v, Aug 07 2018 *)

Prog

(C)
int a(int n) {
        int s = 0;
        while( n != 1 ) {
                switch(n%4) {
                        case 1: n = 7*n+1; break;
                        case 3: n = 7*n-1; break;
                        default: n = n/2;
                }
                s++;
        }
        return s;
}
(PARI) a(n) = my(nb=0); while(n != 1, if (!((n-1)%4), n = 7*n+1, if (!((n+1)%4), n = 7*n-1, n = n/2)); nb++); nb; \\ Michel Marcus, Aug 06 2018

Crossrefs

Cf. A317640 (7x+-1 function), A006577 (3x+1 equivalent).

Sequence in context: A100081 A078197 A070704 * A298059 A298708 A367936

Adjacent sequences: A317750 A317751 A317752 * A317754 A317755 A317756

Keyword

nonn,easy,hear,look

Author

David Barina, Aug 06 2018

Status

approved