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

0,1

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.

Links

Table of n, a(n) for n=0..69.

Example

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.

Mathematica

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

Prog

(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

Crossrefs

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

Sequence in context: A343966 A195846 A201666 * A215737 A154718 A334424

Adjacent sequences: A349247 A349248 A349249 * A349251 A349252 A349253

Keyword

nonn,base

Author

M. F. Hasler, Nov 17 2021

Status

approved