A349250 Size of the orbit of n under iterations of the digit-census function A348783.

8, 7, 5, 10, 12, 14, 16, 18, 20, 22, 6, 5, 3, 8, 10, 12, 14, 16, 18, 20, 4, 2, 6, 8, 10, 12, 14, 16, 18, 20, 6, 8, 8, 12, 10, 12, 14, 16, 18, 20, 10, 10, 10, 10, 12, 12, 14, 16, 18, 20, 12, 12, 12, 12, 12, 16, 14, 16, 18, 20, 14, 14, 14, 14, 14, 14, 14, 16, 18, 20

Offset: 0

Author

M. F. Hasler, Nov 17 2021

Keywords

• nonn
• base

Comments

Equivalently: the number of times the function A348783 must be iterated, starting with n, before a value occurs for the second time.

It can be shown that the trajectory of all numbers end in one of the two limit cycles (21, 110) or (22, 200, 102, 111, 30, 1001). Therefore a(n) >= 2 with equality iff n = 21 or n = 110.

Examples

			Under iterations of A348783, 0 -> 1 -> 10 -> 11 -> 20 -> 101 -> 21 -> 110 -> 21 -> ...: here, the cycle (21, 110) is reached, so the orbit of 0 is O(0) = {0, 1, 10, 11, 20, 101, 21, 110}, of size a(0) = 8. We may also deduce that a(1) = 7, a(10) = 6, a(11) = 5, a(20) = 4, a(101) = 3 and a(21) = a(110) = 2 which is the length of the cycle these elements belong to.
Similarly, 2 -> 100 -> 12 -> 110 -> 21 -> ..., so the orbit of 2 is O(2) = {2, 100, 12, 110, 21}, of size a(2) = 5. We also deduce a(100) = 4 and a(12) = 3.
Then, 3 -> 1000 -> 13 -> 1010 -> 22 -> 200 -> 102 -> 111 -> 30 -> 1001 -> 22 -> 200 -> ..., here the cycle (22, 200, 102, 111, 30, 1001) is reached, and we have the orbit O(3) = {3, 1000, 13, 1010, 22, 200, 102, 111, 30, 1001}, of size a(3) = 10, and also a(1000) = 9, a(13) = 8, a(1010) = 7 and a(n) = 6 for the elements of that cycle.
		

Crossrefs

Cf. A348783 ("digit census function"), A349249 (number of iterations to reach a limiting cycle).

Programs

Mathematica

Array[-1 + Length@ NestWhileList[FromDigits@ RotateLeft@ Reverse@ DigitCount[#] &, #, UnsameQ[##] &, All] &, 60, 0] (* Michael De Vlieger, Nov 23 2021 *)

PARI

apply( {A349250(n,S=[n])=while(!setsearch(S,n=A348783(n)),S=setunion(S,[n]));#S}, [0..99])

Python

def f(n):
    s = str(n)
    return int("".join(str(s.count(d)) for d in "9876543210").lstrip("0"))
def a(n):
    orbit, fn = {n}, f(n)
    while fn not in orbit:
        orbit.add(fn)
        n, fn = fn, f(fn)
    return len(orbit)
print([a(n) for n in range(70)]) # Michael S. Branicky, Nov 18 2021