

A308194


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


3



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
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OFFSET

5,3


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


PROG

(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[(l1)//2] + d[l//2]
c += 1
return c # Chai Wah Wu, Jun 14 2019


CROSSREFS

Cf. A063655, A308190, A308195.
Sequence in context: A058743 A117970 A298904 * A245572 A344985 A325596
Adjacent sequences: A308191 A308192 A308193 * A308195 A308196 A308197


KEYWORD

nonn


AUTHOR

N. J. A. Sloane, Jun 14 2019


STATUS

approved



