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A308190
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Number of steps to reach 5 when iterating x -> A111234(x) starting at x=n.
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6
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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
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OFFSET
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5,3
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COMMENTS
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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.
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REFERENCES
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LINKS
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MATHEMATICA
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a[n_] := Module[{c = 0, x = n, y}, While[x != 5, y = Min[FactorInteger[x][[All, 1]]]; x = y + Quotient[x, y]; c++]; c];
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PROG
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(Python)
from sympy import factorint
c, x = 0, n
while x != 5:
y = min(factorint(x))
x = y + x//y
c += 1
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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