cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

A009101 Fixed point when iterating the function f on n, where f(x) = x + product of digits of x.

Original entry on oeis.org

0, 102, 102, 102, 102, 10, 102, 102, 102, 102, 10, 102, 102, 102, 102, 20, 102, 102, 102, 60, 20, 110, 102, 110, 102, 50, 102, 140, 60, 110, 30, 70, 102, 50, 70, 50, 102, 170, 102, 102, 40, 140, 50, 80, 60, 140, 70, 110, 80, 150, 50, 170, 102, 202, 102, 80, 170, 110, 170
Offset: 0

Views

Author

Keywords

Comments

Starting at n and iterating f, a(n) is the first number reached that contains the digit zero.
Record values of the number of steps to obtain a(n) are 10 for n = 1, 27 for n = 187, 28 for n = 3237326, 32 for n = 3515987, 33 for n = 22572473. It is conjectured that a(n) is obtained in a finite number of steps for every n.
Comment from Adam Kabela, Nov 19 2013: (Start)
The number of steps is finite for every n. The idea is for large k, 9^k < 10^(k-1). The x+f(x) is not increasing fast enough. At some point for high k, in one step the first digit is increased by at most 1. But 9+1= 10. Hence a zero appears sooner or later for every starting x. (The idea is mentioned for example in the XKCD Forum, see link below.) The argument holds not just in base 10 but in other bases too. (End)
Comment from N. J. A. Sloane, Nov 19 2013: (Start)
The XKCD Forum mentioned by Kabela says "If it is unbounded, then there is a first element in the sequence above 10^n for any n. The previous number had n digits, so the digit product that was added to get above 10^n is at most 9^n. Therefore the current number is at most 10^n+9^n. For n>21, this number begins with the digits "10", so from then on the product of the digits is zero and the sequence no longer increases." (End)

Examples

			f(5) = 10, f(10) = 10, hence a(5) = 10.
f(19) = 28, f(28) = 44, f(44) = 60, f(60) = 60, hence a(19) = 60.
		

Crossrefs

Cf. A007954 (product of digits of n), A230099.

Programs

  • Maple
    a:= n-> (m-> `if`(m=0, n, a(n+m)))(mul(i, i=convert(n, base, 10))):
    seq(a(n), n=0..58);  # Alois P. Heinz, Jun 21 2022
  • Mathematica
    Table[FixedPoint[#+Times@@IntegerDigits[#]&,n],{n,0,60}] (* Harvey P. Dale, Oct 11 2012 *)
  • Python
    from math import prod
    def f(x): return x + prod(map(int, str(x)))
    def a(n):
        x, fx = n, f(n)
        while x != fx: x, fx = fx, f(fx)
        return x
    print([a(n) for n in range(60)]) # Michael S. Branicky, Jun 21 2022

Extensions

Additional comments from Klaus Brockhaus, Mar 12 2006
Edited by N. J. A. Sloane, Aug 19 2008 at the suggestion of R. J. Mathar