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.

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, 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, 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

Offset: 0

Author

N. J. A. Sloane, Jun 17 2005

Keywords

• nonn
• easy
• base

Comments

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)).

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.

Examples

			n=0: s=0 -> f(s) = 101 -> f(f(s)) = 123, stop, a(0) = 2.
n=1: s=1 => f(s) = 011 -> f(f(s)) = 123, stop, f(1) = 2.
		

Links

• Reinhard Zumkeller, Table of n, a(n) for n = 0..10000

Crossrefs

A073054 gives another version. f(n) is (essentially) A171797 or A073053.

Cf. A065031, A308002.

Extensions

More terms from Zak Seidov, Jun 18 2005