From: no-cet1-spam@cam.ac.uk (Chris Thompson)
Newsgroups: sci.math
Subject: Re: About a family of natural numbers sequences
Date: 9 Sep 1997 17:58:32 GMT

In article <uz990x6ak66.fsf@lifl.fr>, wegrzyno@lifl.fr (Eric.Wegrzynowski) writes:
|> 
|> Consider the following transformation of natural number
|> 	n ---> f(n) = the product of the decimal digits of n
|> Example
|> 	f(1234) = 1.2.3.4 = 24
|> 
|> Now consider the following sequence
|> 	u(0) = p	 
|> 	u(n+1) = f(u(n))
[...]
|> 
|> It is clear that for all initial number p, there is a number m such that
|> for all n>=m u(n)=u(m)= an 1-digit number.
|> 
|> The problem is how big can be m.
|> 
|> The longest sequence I found is the following
|> 
|> u(0) = 277777788888899   
[...]
|> 
|> In this case m=11
|> 
|> I've tried unsuccessfully to find longer sequence 

This problem was described by Neil Sloane in the Journal of Recreational 
Mathematics back in 1973. See section F25 of Richard Guy's "Unsolved 
Problems in Number Theory" -- not that much is known about it.

|> My question is
|> 
|> Are there numbers p for which m>=12 ?

It is conjectured that there are not. UPINT says none such up to 10^50: 
I expect this may have been improved by now.

|> If yes, are there such p for all m ?

It is conjectured that, even if m=12 occurs, there is an upper limit on m.
And that the same is true in other bases.

|> If no why ?

Well, heuristically, because the number of different f(x) for n-digit x's
is a lot smaller than 10^n, and too many of them will have zero digits.
But this is not exactly a proof. :-)

Chris Thompson
Email: cet1 at cam.ac.uk (edit From field in obvious way if replying)

==============================================================================

From: Richard Pinch <rgep@cam.ac.uk>
Newsgroups: sci.math
Subject: Re: About a family of natural numbers sequences
Date: Wed, 10 Sep 1997 09:56:13 +0100

Chris Thompson wrote:
Chris Thompson wrote:
> 
> In article <uz990x6ak66.fsf@lifl.fr>, wegrzyno@lifl.fr (Eric.Wegrzynowski) writes:
> |>
> |> Consider the following transformation of natural number
> |>      n ---> f(n) = the product of the decimal digits of n
> |> Example
> |>      f(1234) = 1.2.3.4 = 24
> |>
> |> Now consider the following sequence
> |>      u(0) = p
> |>      u(n+1) = f(u(n))
> [...]
> |>
> |> It is clear that for all initial number p, there is a number m such that
> |> for all n>=m u(n)=u(m)= an 1-digit number.
> |>
> |> The problem is how big can be m.
> |>
> |> The longest sequence I found is the following
> |>
> |> u(0) = 277777788888899

We are interested in numbers which are 7-smooth (composed only
of the primes 2,3,5,7) and have no digit zero.  The number of
7-smooth k-digit numbers is approximately 0.014 k^3 and the
probability that a randomly chosen k-digit number has all its
digits non-zero is (9/10)^k.  Assuming that this remains true
for the 7-smooth numbers (which in fact it obviously doesn't)
we would have about \sum_k 0.014 k^3 (9/10)^k numbers of the
form we are interested in, and this sum is convergent (to
about 688.5).  This suggests that m is bounded.
-- 
Richard Pinch		Queens' College, Cambridge
rgep@cam.ac.uk		http://www.dpmms.cam.ac.uk/~rgep