A318489 Number of steps to reach a lower number than starting value in 7x+-1 problem, or 0 if never reached.

0, 1, 12, 1, 8, 1, 4, 1, 4, 1, 42, 1, 8, 1, 4, 1, 4, 1, 23, 1, 20, 1, 4, 1, 4, 1, 12, 1, 16, 1, 4, 1, 4, 1, 282, 1, 12, 1, 4, 1, 4, 1, 229, 1, 50, 1, 4, 1, 4, 1, 8, 1, 35, 1, 4, 1, 4, 1, 8, 1, 50, 1, 4, 1, 4, 1, 46, 1, 8, 1, 4, 1, 4, 1, 225, 1, 8, 1, 4, 1, 4, 1, 35, 1, 16, 1, 4, 1, 4, 1, 46, 1, 27, 1, 4, 1, 4, 1, 16

Offset: 1

Author

David Barina, Aug 27 2018

Keywords

• nonn
• easy

Comments

The least positive k for which the iterate A317640^k(n) < n.

Also called the dropping time, glide, or stopping time.

a(2n) = 1.

Examples

			a(5) = 8 because the trajectory is (5, 36, 18, 9, 64, 32, 16, 8, 4, 2, 1, ...) and the first lower number is 4. Thus 8 steps to reach the value 4 starting from the value 5.
		

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), A102419 (3x+1 equivalent).

Programs

C

int a(int n0) {
        if( n0 == 1 ) return 0;
        int s = 0;
        for(int n = n0; n >= n0; s++) {
                switch(n%4) {
                        case 1: n = 7*n+1; break;
                        case 3: n = 7*n-1; break;
                        default: n = n/2;
                }
        }
        return s;
}

PARI

a7(n) = {my(m=(n+2)%4-2); if(m%2, 7*n + m, n/2)};
a(n) = if (n==1, 0, my(nb=1, m=n, nm); while((nm=a7(m)) >= n, m = nm; nb++); nb); \\ Michel Marcus, Aug 28 2018