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 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 (list; graph; refs; listen; history; text; internal format)
 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[(l-1)//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

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Last modified July 25 03:53 EDT 2021. Contains 346283 sequences. (Running on oeis4.)