cp's OEIS Frontend

A100961 For a decimal string s, let f(s) = decimal string ijk, where i = number of even digits in s, j = number of odd digits in s, k=i+j (see A171797). Start with s = decimal expansion of n; a(n) = number of applications of f needed to reach the string 123.

%I A100961 #15 May 12 2019 10:25:47
%S A100961 2,2,2,2,2,2,2,2,2,2,2,3,2,3,2,3,2,3,2,3,3,2,3,2,3,2,3,2,3,2,2,3,2,3,
%T A100961 2,3,2,3,2,3,3,2,3,2,3,2,3,2,3,2,2,3,2,3,2,3,2,3,2,3,3,2,3,2,3,2,3,2,
%U A100961 3,2,2,3,2,3,2,3,2,3,2,3,3,2,3,2,3,2,3,2,3,2,2,3,2,3,2,3,2,3,2,3,2,1,2,1,2
%N A100961 For a decimal string s, let f(s) = decimal string ijk, where i = number of even digits in s, j = number of odd digits in s, k=i+j (see A171797). Start with s = decimal expansion of n; a(n) = number of applications of f needed to reach the string 123.
%C A100961 Obviously if the digits of m and n have the same parity then a(m) = a(n). E.g. a(334) = a(110). In other words, a(n) = a(A065031(n)).
%C A100961 It is easy to show that (i) the trajectory of every number under f eventually reaches 123 (if s has more than three digits then f(s) has fewer digits than s) and (ii) since each string ijk has only finitely many preimages, a(n) is unbounded.
%H A100961 Reinhard Zumkeller, <a href="/A100961/b100961.txt">Table of n, a(n) for n = 0..10000</a>
%e A100961 n=0: s=0 -> f(s) = 101 -> f(f(s)) = 123, stop, a(0) = 2.
%e A100961 n=1: s=1 => f(s) = 011 -> f(f(s)) = 123, stop, f(1) = 2.
%Y A100961 A073054 gives another version. f(n) is (essentially) A171797 or A073053.
%Y A100961 Cf. A065031, A308002.
%K A100961 nonn,easy,base
%O A100961 0,1
%A A100961 _N. J. A. Sloane_, Jun 17 2005
%E A100961 More terms from Zak Seidov, Jun 18 2005