A308190 Number of steps to reach 5 when iterating x -> A111234(x) starting at x=n.

0, 1, 3, 2, 2, 4, 4, 3, 4, 3, 3, 5, 6, 5, 5, 4, 5, 5, 5, 4, 5, 4, 4, 6, 8, 7, 7, 6, 4, 6, 4, 5, 7, 6, 6, 6, 7, 6, 6, 5, 6, 6, 6, 5, 4, 5, 5, 7, 10, 9, 6, 8, 6, 8, 8, 7, 6, 5, 5, 7, 6, 5, 7, 6, 5, 8, 8, 7, 8, 7, 7, 7, 6, 8, 8, 7, 8, 7, 7, 6, 6, 7, 7, 7, 8, 7, 5, 6, 7, 5, 5, 6, 7, 6, 6, 8, 12

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

• Chai Wah Wu, Table of n, a(n) for n = 5..10000

Crossrefs

Cf. A111234, A308191, A308192, A308193, A308194.

Programs

Mathematica

a[n_] := Module[{c = 0, x = n, y}, While[x != 5, y = Min[FactorInteger[x][[All, 1]]]; x = y + Quotient[x, y]; c++]; c];
Table[a[n], {n, 5, 100}] (* Jean-François Alcover, Jun 15 2019, from Python *)

Python

from sympy import factorint
def A308190(n):
    c, x = 0, n
    while x != 5:
        y = min(factorint(x))
        x = y + x//y
        c += 1
    return c # Chai Wah Wu, Jun 14 2019