A308194 Number of steps to reach 5 when iterating x -> A063655(x) starting at x=n.

0, 1, 3, 2, 2, 4, 5, 4, 4, 3, 3, 3, 4, 3, 4, 3, 5, 5, 6, 5, 5, 4, 5, 6, 7, 6, 6, 5, 4, 5, 5, 5, 7, 6, 4, 5, 6, 5, 5, 4, 4, 6, 5, 4, 4, 4, 4, 5, 5, 4, 4, 4, 6, 7, 5, 4, 6, 5, 4, 4, 4, 5, 7, 6, 5, 5, 6, 5, 6, 5, 4, 7, 4, 5, 5, 4, 4, 6, 6, 5, 6, 5, 6, 5, 6, 5, 4, 6, 6, 5, 6, 4, 7, 6, 4, 4, 8, 7, 7, 6, 6, 5

Offset: 5

Author

N. J. A. Sloane, Jun 14 2019

Keywords

• nonn

Comments

It is easy to show that every number n >= 5 eventually reaches 5. This was conjectured by Ali Sada. For A111234 sends a composite n > 5 to a smaller number, and sends a prime > 5 to a smaller number in two steps. Furthermore no number >= 5 can reach a number less than 5. So all numbers >= 5 eventually reach 5.

References

• Ali Sada, Email to N. J. A. Sloane, Jun 13 2019.

Links

• Rémy Sigrist, Table of n, a(n) for n = 5..10000

Crossrefs

Cf. A063655, A308190, A308195.

Programs

PARI

b(n) = { my(c=1); fordiv(n, d, if((d*d)>=n, if((d*d)==n, return(2*d), return(c+d))); c=d); (0); } \\ after A063655
a(n) = for (k=0, oo, if (n==5, return (k), n=b(n))) \\ Rémy Sigrist, Jun 14 2019

Python

from sympy import divisors
def A308194(n):
    c, x = 0, n
    while x != 5:
        d = divisors(x)
        l = len(d)
        x = d[(l-1)//2] + d[l//2]
        c += 1
    return c # Chai Wah Wu, Jun 14 2019