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

1,3

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.

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) = 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.

Prog

(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

Crossrefs

Cf. A317640 (7x+-1 function), A102419 (3x+1 equivalent).

Sequence in context: A140378 A085094 A010214 * A121985 A245839 A068329

Adjacent sequences: A318486 A318487 A318488 * A318490 A318491 A318492

Keyword

nonn,easy

Author

David Barina, Aug 27 2018

Status

approved