<!DOCTYPE HTML PUBLIC "-//W3C//DTD HTML 4.0 Transitional//EN">
<HTML><HEAD>
<META http-equiv=Content-Type content="text/html; charset=iso-8859-1">
<META content="MSHTML 6.00.2900.2912" name=GENERATOR>
<STYLE></STYLE>
</HEAD>
<BODY
style="WORD-WRAP: break-word; khtml-nbsp-mode: space; khtml-line-break: after-white-space"
bgColor=#ffffff>
<DIV><FONT face=Arial size=2>Let b >= 2, k >= 2. For x >= 1, Let r_b(x)
be the standard base-b representation of the integer part of x. Let d_b(x) and
s_b(x) be the number and sum, respectively, of the digits of
r_b(x).</FONT></DIV>
<DIV><FONT face=Arial size=2></FONT> </DIV>
<DIV><FONT face=Arial size=2>d_b(x) = floor(log(x)/log(b))+1. If gcd(k, b)
= 1, empirical evidence suggests that the digits of r_b(k^n) for large n are
equidistributed, which, if true, implies that s_b(k^n)/d_b(k^n) approaches
(b-1)/2, the average of the base-b digits, as n grows without bound. This in
turn implies that</FONT></DIV>
<DIV><FONT face=Arial size=2></FONT> </DIV>
<DIV><FONT face=Arial size=2>
lim s_b(k^n)/n = ((b-1)/2)(log(k)/log(b)).</FONT></DIV>
<DIV><FONT face=Arial size=2>
n->inf</FONT></DIV>
<DIV><FONT face=Arial size=2></FONT> </DIV>
<DIV><FONT face=Arial size=2>When b = 10 and k = 5, this value is
(9/2)(log(5)/log(10)) = 3.14536... Close, but no pi.</FONT></DIV>
<DIV><FONT face=Arial size=2></FONT> </DIV>
<DIV><FONT face=Arial size=2>We expect s_b(k^n) = n to have an infinite
number of solutions exactly when ((b-1)/2)(log(k)/log(b)) = 1, that is, k =
b^(2/(b-1)), and modular considerations do not preclude s_b(k^n) = n.
For all other k, we expect s_b(k^n) = n to have a finite number of solutions
(e.g, when b = 10 and k = 2, which was Tanya's original problem). For base
b = 10, we expect s_10(k^n) = n to have an infinite number of solutions only
when k = 10^(2/9) = 1.6681+.</FONT></DIV>
<DIV><FONT face=Arial size=2></FONT> </DIV>
<DIV>----- Original Message ----- </DIV>
<BLOCKQUOTE
style="PADDING-RIGHT: 0px; PADDING-LEFT: 5px; MARGIN-LEFT: 5px; BORDER-LEFT: #000000 2px solid; MARGIN-RIGHT: 0px">
<DIV
style="BACKGROUND: #e4e4e4; FONT: 10pt arial; font-color: black"><B>From:</B>
<A title=pxp@rogers.com href="mailto:pxp@rogers.com">Hans Havermann</A> </DIV>
<DIV style="FONT: 10pt arial"><B>To:</B> <A title=seqfan@ext.jussieu.fr
href="mailto:seqfan@ext.jussieu.fr">seqfan@ext.jussieu.fr</A> </DIV>
<DIV style="FONT: 10pt arial"><B>Sent:</B> Monday, July 24, 2006 7:09 AM</DIV>
<DIV style="FONT: 10pt arial"><B>Subject:</B> (sum-of-digits of 5^n)/n
approximates pi</DIV>
<DIV><BR></DIV>
<DIV>
<BLOCKQUOTE type="cite"><SPAN class=Apple-style-span
style="WORD-SPACING: 0px; FONT: 16px Monaco; TEXT-TRANSFORM: none; COLOR: rgb(0,0,0); TEXT-INDENT: 0px; WHITE-SPACE: normal; LETTER-SPACING: normal; BORDER-COLLAPSE: separate; border-spacing: 0px 0px; khtml-text-decorations-in-effect: none; apple-text-size-adjust: auto; orphans: 2; widows: 2">Each
n provides a value of (sum-of-digits of 5^n)/n that is closer to pi than the
previous value. I'm guessing there aren't many more
terms.</SPAN></BLOCKQUOTE></DIV><BR>
<DIV>I let this run overnight and did get two more terms: {1, 2, 4, 6, 8,
139, 309, 390, 819, 2868, 6751, 8045, 9414, 15008, 15375, 56839,
84383, 151286, ...}</DIV>
<DIV><BR class=khtml-block-placeholder></DIV>
<DIV>(sum-of-digits of 5^151286)/151286 is about 3.141592745.</DIV>
<P>
<HR>
<P></P>No virus found in this incoming message.<BR>Checked by AVG Free
Edition.<BR>Version: 7.1.394 / Virus Database: 268.10.4/396 - Release Date:
7/24/2006<BR></BLOCKQUOTE></BODY></HTML>