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A350709
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Modified Sisyphus function of order 3: a(n) is the concatenation of (number of digits of n)(number digits of n congruent to 0 modulo 3)(number of digits of n congruent to 1 modulo 3)(number of digits of n congruent to 2 modulo 3).
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2
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1100, 1010, 1001, 1100, 1010, 1001, 1100, 1010, 1001, 1100, 2110, 2020, 2011, 2110, 2020, 2011, 2110, 2020, 2011, 2110, 2101, 2011, 2002, 2101, 2011, 2002, 2101, 2011, 2002, 2101, 2200, 2110, 2101, 2200, 2110, 2101, 2200
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OFFSET
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0,1
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COMMENTS
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If we start with n and repeatedly apply the map i -> a(i), we eventually get the cycle {4031, 4112, 4220}
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REFERENCES
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M. E. Coppenbarger, Iterations of a modified Sisyphus function, Fib. Q., 56 (No. 2, 2018), 130-141.
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LINKS
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EXAMPLE
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11 has two digits, both congruent to 1 modulo 3, so a(11) = 2020.
a(20) = 2101.
a(30) = 2200.
a(1111123567) = 10262.
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PROG
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(Python)
def a(n):
d, m = list(map(int, str(n))), [0, 0, 0]
for di in d: m[di%3] += 1
return int(str(len(d)) + "".join(map(str, m)))
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CROSSREFS
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KEYWORD
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nonn,base,easy
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AUTHOR
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STATUS
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approved
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