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).

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

Author

Matthew E. Coppenbarger, Mar 28 2022

Keywords

• nonn
• base
• easy

Comments

If we start with n and repeatedly apply the map i -> a(i), we eventually get the cycle {4031, 4112, 4220}

Examples

			11 has two digits, both congruent to 1 modulo 3, so a(11) = 2020.
a(20) = 2101.
a(30) = 2200.
a(1111123567) = 10262.
		

References

• M. E. Coppenbarger, Iterations of a modified Sisyphus function, Fib. Q., 56 (No. 2, 2018), 130-141.

Crossrefs

Cf. A073053 (Sisyphus), A171797, A171798, A171813, A055642, A196563, A196564, A308002.

Programs

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