

A317753


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


1



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
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OFFSET

1,3


COMMENTS

The 7x+1 problem is as follows. Start with any natural number n. If 4 divides n1, 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*n1; break;
default: n = n/2;
}
s++;
}
return s;
}
(PARI) a(n) = my(nb=0); while(n != 1, if (!((n1)%4), n = 7*n+1, if (!((n+1)%4), n = 7*n1, 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 A124686
Adjacent sequences: A317750 A317751 A317752 * A317754 A317755 A317756


KEYWORD

nonn,easy,hear,look


AUTHOR

David Barina, Aug 06 2018


STATUS

approved



