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A370350
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Number of steps to go from n to 1 in a variant of the Collatz iteration x -> (x / 5 if 5 divides x, x + (x+(5-(x mod 5)))/5 otherwise), or -1 if 1 is never reached.
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1
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0, 4, 3, 2, 1, 7, 17, 6, 16, 5, 15, 5, 5, 14, 4, 4, 13, 11, 11, 3, 12, 10, 10, 20, 2, 11, 9, 9, 19, 8, 69, 10, 8, 8, 18, 17, 18, 68, 9, 7, 7, 8, 77, 16, 17, 67, 8, 17, 8, 6, 7, 76, 15, 7, 16, 66, 7, 16, 7, 6, 75, 6, 75, 14, 6, 6, 16, 65, 6, 15, 6, 15, 24, 74
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OFFSET
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1,2
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COMMENTS
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Because ceiling(a/b) = (a+(b-(a mod b)))/b, this sequence equals the total number of steps to reach 1 in a variant of the Collatz problem using the iteration f(x) := x/k if k divides x, x+ceiling(x/k) otherwise.
Specifically, here we set k=5 because it is the next integer (after 2) that is not in A368136; i.e. the iteration used here has no loops for starting values up to 5*5=25, apart from the loop containing 1.
It is not known if there exists n such that a(n) = -1 (either by iteration reaching a non-elementary loop which implies n>5*5, or by iteration growing without bound).
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LINKS
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EXAMPLE
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For n = 11, the following trajectory is obtained:
11, 14, 17, 21, 26, 32, 39, 47, 57, 69, 83, 100, 20, 4, 5, 1
which requires 15 steps to reach 1, therefore a(11) = 15.
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MATHEMATICA
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s={}; Do[c=0; a=n; While[a>1, If[Divisible[a, 5], a=a/5, a=a+Ceiling[a/5]]; c++]; AppendTo[s, c], {n, 74}]; s (* James C. McMahon, Feb 28 2024 *)
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PROG
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(Python)
def a(n, C = 5):
s = 0
while n > 1:
d, r = divmod(n, C)
n = n + 1 + d if r else d
s += 1
return s
print([a(n) for n in range(1, 75)])
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CROSSREFS
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KEYWORD
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AUTHOR
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STATUS
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approved
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