do you know of any examples where 0 is not in D , and you can prove that S is infinite? Sure. Base 4, D = {3}, starting value {11001}. The right ends of successive squarings will be xxx11[0*]1 and xxx22[0*]1. This is an infinite set, as the string of internal 0s grows by one 0 with each 4th-powering.
Rich -----Original Message----- From: math-fun-bounces+rschroe=sandia.gov@mailman.xmission.com on behalf of Michael Reid Sent: Thu 1/12/2006 9:05 AM To: math-fun@mailman.xmission.com Subject: Re: [math-fun] Digital silliness
There are many ways to vary this problem, most notably by changing the base b, the set of initial numbers A, and the set of splitting digits, which I will call D.
Case 1: b = 10, A = {2}, D = {1}. In this case S is infinite, containing an infinitude of numbers of the form 6*10^k.
Case 2: b = 10, A = {2}, D = {0,1,5}, In this case S is finite, with largest element 6347862777922.
Case 3: b = 10, A = {2}, D = {0}. Finiteness of S unknown.
i think it is reasonable to expect the behavior to be different in the case that 0 is not in D , versus the case when 0 is in D . if 0 is in D , you will never get elements of S that end with 0 (unless they're in the initial set). if 0 is not in D , then it's possible to get elements of S that end with 0 . this may not always occur, but it seems likely if the set S is reasonably large. the rightmost 0's double with each squaring and do not affect the other digits. thus, keeping the rightmost part of the splitting, we get arbitrarily large numbers, unless a splitting digit occurs as the rightmost non-zero digit. do you know of any examples where 0 is not in D , and you can prove that S is infinite? mike _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun