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

Author

David Barina, Aug 06 2018

Keywords

• nonn
• easy
• hear
• look

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.

Examples

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

Crossrefs

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

Programs

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;
}

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 *)

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