Re: [math-fun] better posed problem?

12 Mar 2005

      even the unlikely f(6n)= 10^n *seems* to hold : up to n=24:
{1,1,2,2,4,2,10,2,16,8,34,2,100,2,130,38,256,2,1000,2,1156,134,2050,2,10000}

this is as far as I can go. the rest is for educated folks.
W.

----- Original Message -----
From: "Gareth McCaughan" <gareth.mccaughan@pobox.com>
To: <math-fun@mailman.xmission.com>; <dasimov@earthlink.net>
Sent: Saturday, March 12, 2005 3:44 PM
Subject: Re: [math-fun] better posed problem?
...
Some patterns here look intriguing.  Like
f(2^n) = 2^(2^(n-1))
for one, and
f(6n) = 10^n
for another.  Do these extend?
I'm doubtful about the second. But the first one does.

Re: [math-fun] better posed problem?

wouter meeussen