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A350709
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).
2
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
OFFSET
0,1
COMMENTS
If we start with n and repeatedly apply the map i -> a(i), we eventually get the cycle {4031, 4112, 4220}
REFERENCES
M. E. Coppenbarger, Iterations of a modified Sisyphus function, Fib. Q., 56 (No. 2, 2018), 130-141.
EXAMPLE
11 has two digits, both congruent to 1 modulo 3, so a(11) = 2020.
a(20) = 2101.
a(30) = 2200.
a(1111123567) = 10262.
PROG
(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)))
print([a(n) for n in range(37)]) # Michael S. Branicky, Mar 28 2022
CROSSREFS
KEYWORD
nonn,base,easy
AUTHOR
STATUS
approved