<?xml version="1.0" encoding="utf-8"?><feed xmlns="http://www.w3.org/2005/Atom" ><generator uri="https://jekyllrb.com/" version="3.10.0">Jekyll</generator><link href="http://sam.stromwall.org/feed.xml" rel="self" type="application/atom+xml" /><link href="http://sam.stromwall.org/" rel="alternate" type="text/html" /><updated>2026-09-19T21:07:06+00:00</updated><id>http://sam.stromwall.org/feed.xml</id><title type="html">Sam Stromwall’s Blog</title><author><name>Sam Stromwall</name></author><entry><title type="html">The Only Post</title><link href="http://sam.stromwall.org/2026/09/18/initial-post.html" rel="alternate" type="text/html" title="The Only Post" /><published>2026-09-18T00:00:00+00:00</published><updated>2026-09-18T00:00:00+00:00</updated><id>http://sam.stromwall.org/2026/09/18/initial-post</id><content type="html" xml:base="http://sam.stromwall.org/2026/09/18/initial-post.html"><![CDATA[<p>This blog is under construction. I have no content to give you, so here’s a numbers fact.</p>

<p>A typical $20$-sided die is numbered from $1$ to $20$, so the sum of all its faces is $210$. If I gave you a blank die and asked you to write a non-negative integer on each face so that your die also sums to $210$, then there are $43 674 337 807 412 863 662 664 548$ distinct dice you could produce. That would be the coefficient on $z^{210}$ in the Taylor expansion of this rational function.</p>

\[\frac1{60(1-z)^{20}} + \frac1{3(1-z)^2(1-z^3)^6} + \frac1{4(1-z^2)^{10}} + \frac2{5(1-z^5)^4}\]]]></content><author><name>[&quot;Sam Stromwall&quot;]</name></author><summary type="html"><![CDATA[This blog is under construction. I have no content to give you, so here’s a numbers fact.]]></summary></entry></feed>