A349249 Number of iterations of the digit-census function A348783 until an element of a cycle is reached.

6, 5, 3, 4, 6, 8, 10, 12, 14, 16, 4, 3, 1, 2, 4, 6, 8, 10, 12, 14, 2, 0, 0, 2, 4, 6, 8, 10, 12, 14, 0, 2, 2, 6, 4, 6, 8, 10, 12, 14, 4, 4, 4, 4, 6, 6, 8, 10, 12, 14, 6, 6, 6, 6, 6, 10, 8, 10, 12, 14, 8, 8, 8, 8, 8, 8, 8, 10, 12, 14, 10, 10, 10, 10, 10, 10, 10

Offset: 0

Author

M. F. Hasler, Nov 17 2021

Keywords

• nonn

Comments

The length of the cycle in which the trajectory of n will end, is equal to the size of the full orbit of n under A348783, A349250(n), minus a(n).

Examples

			Under iterations of A348783, 0 -> 1 -> 10 -> 11 -> 20 -> 101 -> 21 -> 110 -> 21 -> ..., so the cycle (21, 110) is reached after a(0) = 6 iterations, and also a(1) = 5, a(10) = 4, a(11) = 3, a(20) = 2, a(101) = 1 and a(21) = a(110) = 0 because these two are elements of this cycle.
Similarly, 2 -> 100 -> 12 -> 110 -> 21 -> ..., so the same cycle is reached in a(2) = 3 iterations, and also a(100) = 2, a(12) = 1.
Then, 3 -> 1000 -> 13 -> 1010 -> 22 -> 200 -> 102 -> 111 -> 30 -> 1001 -> 22 -> 200 -> ..., here the cycle (22, 200, 102, 111, 30, 1001) is reached after a(3) = 4 iterations, and a(1000) = 3, a(13) = 2, a(1010) = 1 and a(n) = 0 for the elements of that cycle.
		

Crossrefs

Cf. A348783, A349250 (length of the full orbit).

Programs

PARI

apply( {A349249(n, S=[n], T)=while(!setsearch(S, n=A348783(n)), S=setunion(S, [n])); if(T,T-#S,A349249(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.append(fn)
        n, fn = fn, f(fn)
    return orbit.index(fn)
print([a(n) for n in range(77)]) # Michael S. Branicky, Nov 18 2021

Formula

a(n) = 0 iff n is element of a cycle (or "loop") under iterations of A348783,

otherwise a(n) = 1 + a(A348783(n)).