<rss version="2.0" xmlns:dc="http://purl.org/dc/elements/1.1/" xmlns:atom="http://www.w3.org/2005/Atom"><channel><title>Hacker News: bustermellotron</title><link>https://news.ycombinator.com/user?id=bustermellotron</link><description>Hacker News RSS</description><docs>https://hnrss.org/</docs><generator>hnrss v2.1.1</generator><lastBuildDate>Fri, 15 May 2026 08:48:38 +0000</lastBuildDate><atom:link href="https://hnrss.org/user?id=bustermellotron" rel="self" type="application/rss+xml"></atom:link><item><title><![CDATA[New comment by bustermellotron in "I let AI build a tool to help me figure out what was waking me up at night"]]></title><description><![CDATA[
<p>You could use air scrubbers <a href="https://en.wikipedia.org/wiki/Soda_lime" rel="nofollow">https://en.wikipedia.org/wiki/Soda_lime</a></p>
]]></description><pubDate>Tue, 12 May 2026 05:10:57 +0000</pubDate><link>https://news.ycombinator.com/item?id=48104427</link><dc:creator>bustermellotron</dc:creator><comments>https://news.ycombinator.com/item?id=48104427</comments><guid isPermaLink="false">https://news.ycombinator.com/item?id=48104427</guid></item><item><title><![CDATA[New comment by bustermellotron in "A recent experience with ChatGPT 5.5 Pro"]]></title><description><![CDATA[
<p>I saw Tim Gowers give a talk at the AMS-MAA joint meeting in Seattle about ten years ago where he predicted that in 100 years humans would no longer be doing research mathematics. I wonder if he’s adjusted his timeline.<p>At the time I thought the key missing tool was a natural language search that acted like mathoverflow, where you could explain your problem or ideas as you understood them and get references to relevant literature (possibly outside your experience or vocabulary).</p>
]]></description><pubDate>Sat, 09 May 2026 05:12:28 +0000</pubDate><link>https://news.ycombinator.com/item?id=48072005</link><dc:creator>bustermellotron</dc:creator><comments>https://news.ycombinator.com/item?id=48072005</comments><guid isPermaLink="false">https://news.ycombinator.com/item?id=48072005</guid></item><item><title><![CDATA[New comment by bustermellotron in "After a 40-year wait, technology enables three-sided zipper design"]]></title><description><![CDATA[
<p>Go go Gadget arms!</p>
]]></description><pubDate>Wed, 06 May 2026 05:39:52 +0000</pubDate><link>https://news.ycombinator.com/item?id=48032649</link><dc:creator>bustermellotron</dc:creator><comments>https://news.ycombinator.com/item?id=48032649</comments><guid isPermaLink="false">https://news.ycombinator.com/item?id=48032649</guid></item><item><title><![CDATA[New comment by bustermellotron in "A beginner's guide to split keyboards"]]></title><description><![CDATA[
<p>I found it easy to adapt to the x-bows keyboard (column staggered and splayed). The thumb buttons and large ctrl, alt, space are great for emacs. My only complaint is that the braces are a bit far away.</p>
]]></description><pubDate>Fri, 20 Feb 2026 06:25:22 +0000</pubDate><link>https://news.ycombinator.com/item?id=47084455</link><dc:creator>bustermellotron</dc:creator><comments>https://news.ycombinator.com/item?id=47084455</comments><guid isPermaLink="false">https://news.ycombinator.com/item?id=47084455</guid></item><item><title><![CDATA[New comment by bustermellotron in "The manager's unbearable lack of endorphins"]]></title><description><![CDATA[
<p>I think the book "The Now Habit" discusses this well. (Could be another book though...)<p>I once completed a 3000 mile cycling tour across the US (and didn't take ADHD meds while on the trip) and was I was basically disappointed to reach the west coast.</p>
]]></description><pubDate>Sun, 23 Jun 2024 09:23:22 +0000</pubDate><link>https://news.ycombinator.com/item?id=40765981</link><dc:creator>bustermellotron</dc:creator><comments>https://news.ycombinator.com/item?id=40765981</comments><guid isPermaLink="false">https://news.ycombinator.com/item?id=40765981</guid></item><item><title><![CDATA[New comment by bustermellotron in "The Case for Bash (2021)"]]></title><description><![CDATA[
<p>This covers most of what you need for "housekeeping" scripts in Python: <a href="https://automatetheboringstuff.com" rel="nofollow">https://automatetheboringstuff.com</a></p>
]]></description><pubDate>Fri, 19 May 2023 09:18:53 +0000</pubDate><link>https://news.ycombinator.com/item?id=35999549</link><dc:creator>bustermellotron</dc:creator><comments>https://news.ycombinator.com/item?id=35999549</comments><guid isPermaLink="false">https://news.ycombinator.com/item?id=35999549</guid></item><item><title><![CDATA[New comment by bustermellotron in "Only one pair of distinct positive integers satisfy the equation m^n = n^m"]]></title><description><![CDATA[
<p>On the other hand, your proof really only needs the binomial theorem and geometric series.</p>
]]></description><pubDate>Thu, 20 Apr 2023 19:53:30 +0000</pubDate><link>https://news.ycombinator.com/item?id=35645366</link><dc:creator>bustermellotron</dc:creator><comments>https://news.ycombinator.com/item?id=35645366</comments><guid isPermaLink="false">https://news.ycombinator.com/item?id=35645366</guid></item><item><title><![CDATA[New comment by bustermellotron in "Only one pair of distinct positive integers satisfy the equation m^n = n^m"]]></title><description><![CDATA[
<p>The “claim” more or less proves that k-1 th root of k is less than 2 if k is larger than 2. So I think your argument is equivalent.</p>
]]></description><pubDate>Thu, 20 Apr 2023 19:45:22 +0000</pubDate><link>https://news.ycombinator.com/item?id=35645256</link><dc:creator>bustermellotron</dc:creator><comments>https://news.ycombinator.com/item?id=35645256</comments><guid isPermaLink="false">https://news.ycombinator.com/item?id=35645256</guid></item><item><title><![CDATA[New comment by bustermellotron in "Only one pair of distinct positive integers satisfy the equation m^n = n^m"]]></title><description><![CDATA[
<p>Here is an elementary proof:<p>Since m and n are distinct, we may assume that m > n >= 2. From the equation and unique factorization, we know that n divides m, so write m = nd.<p>Then (nd)^n = n^(nd).
Hence d^n = n^{n(d-1)}, which yields d = n^{d-1} >= 2^{d-1}.<p>Claim: if k is an integer greater than or equal to 3, then k < 2^{k - 1}.<p>Proof: the base case is clear: 3 < 4. Suppose k > 3 and k - 1 < 2^{k - 2}.
Then k < 2^{k-2} + 1 <= 2^{k-1}, where the last inequality holds because 2^{k-1} - 2^{k-2} = 2^{k-2} >= 1. QED<p>So, d must be less than 3. Since m = nd and m and n are distinct, d is not 1, so d = 2.
Since d = n^{d-1} and d - 1 = 1, we have n = d, so m = 4.</p>
]]></description><pubDate>Thu, 20 Apr 2023 10:06:22 +0000</pubDate><link>https://news.ycombinator.com/item?id=35638064</link><dc:creator>bustermellotron</dc:creator><comments>https://news.ycombinator.com/item?id=35638064</comments><guid isPermaLink="false">https://news.ycombinator.com/item?id=35638064</guid></item></channel></rss>