Relay-Version: version B 2.10 5/3/83; site utzoo.UUCP Path: utzoo!watmath!clyde!burl!ulysses!unc!mcnc!ecsvax!pizer From: pizer@ecsvax.UUCP Newsgroups: net.math Subject: Re: Iterated sums of digits of divisors Message-ID: <2976@ecsvax.UUCP> Date: Sun, 22-Jul-84 16:58:04 EDT Article-I.D.: ecsvax.2976 Posted: Sun Jul 22 16:58:04 1984 Date-Received: Mon, 23-Jul-84 02:19:51 EDT Lines: 23 References: I don't have a definite answer to the problem, I do however have some idea. The easiest way to take this problem is to look at different ranges of numbers and make general assumptions. First of all, any incredibly high number (ie above 1000) would have to decrease as the function progressed. Take the number 5000, for example, the biggest divisor is only 2500, and since you are adding up only the digits up to get the next number, you only have the number 7 so far. Even if a large number had a divisor like 592347, the sum of the digits (30) is nothing compared to the number itself. So we can assume, at least for large numbers, the number will decrease. Below 1000, the same holds true, all the way down to the number 15. There are a few exceptions where the number will increase (for example 16 becomes 22, 24 becomes 33), however the next number will decrease (22 will become 9, 33 will become 12), or if not the next number, the one after that. Below 15, the numbers will increase (except for 10, 11 and 13, which then go down but come back up again). At 15, which seems to be the only number that does this, it will reproduce itself. I didn't study this problem for an incredibly long period of time, so there may be errors in my reasoning, but I will be happy to accept remarks and criticism. Billy Pizer (pizer@ecsvax)