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
AUTHOR
David Barina, Aug 06 2018
STATUS
approved