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
Keywords
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.
Programs
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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
Comments