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2026-06-10 18:42:18 -04:00
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<hgroup> <hgroup>
<h1>Sums of Powers</h1> <h1>Sums of Powers</h1>
<p>Posted <time datetime="2026-04-15">April 15, 2026</time></p> <p>Posted <time datetime="2026-04-15">April 15, 2026</time></p>
<p>Updated <time datetime="2026-06-10">June 10, 2026</time></p>
</hgroup> </hgroup>
<p>This page is all about how we can efficiently compute large sums of <p>This page is all about how we can efficiently compute large sums of
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<p>This algorithm is quadratic in the value of <p>This algorithm is quadratic in the value of
<math><mi>p</mi></math>, which isn't great, but it's simple to <math><mi>p</mi></math>, which isn't great, but it's simple to
implement (and I don't know of any faster way).</p> implement.<sup>1</sup></p>
<section> <section>
<h4>Stirling numbers calculator</h4> <h4>Stirling numbers calculator</h4>
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<p>It turns out there is a very elegant identity relating <p>It turns out there is a very elegant identity relating
exponents, Stirling numbers, and falling exponents, Stirling numbers, and falling
factorials:<sup>1</sup></p> factorials:<sup>2</sup></p>
<div class="math-block"> <div class="math-block">
<math display="block"> <math display="block">
<mrow> <mrow>
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</math> </math>
</div> </div>
<p>(For <math><mi>p</mi><mo>=</mo><mn>0</mn></math>, the sum is <p>(For <math><mi>p</mi><mo>=</mo><mn>0</mn></math>, the sum is
just <math><mi>n</mi></math>.)<sup>2</sup> This lets us just <math><mi>n</mi></math>.)<sup>3</sup> This lets us
simplify.<sup>3</sup></p> simplify.<sup>4</sup></p>
<div class="math-block"> <div class="math-block">
<math display="block"> <math display="block">
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<section> <section>
<h2>Footnotes</h2> <h2>Footnotes</h2>
<ol id="footnote-list"> <ol id="footnote-list">
<li><sup>1</sup> For proofs of this identity, see chapter 1.9 of <li><sup>1</sup> I think we can technically compute all the Stirling
numbers we need in linearithmic time by applying the convolution
theorem to the numbers' explicit sum formula, but that approach
makes it difficult to get exact values. If you just want the
Stirling numbers modulo a prime, you could maybe use a
<a href="https://en.wikipedia.org/wiki/Discrete_Fourier_transform_over_a_ring">number-theoretic transform</a>,
but that's well outside my wheelhouse so I'm not going to touch
on that here.</li>
<li><sup>2</sup> For proofs of this identity, see chapter 1.9 of
<cite>Enumerative Combinatorics</cite> by Stanley and chapter 6.1 <cite>Enumerative Combinatorics</cite> by Stanley and chapter 6.1
of <cite>Concrete Mathematics</cite> by Graham, Knuth, and of <cite>Concrete Mathematics</cite> by Graham, Knuth, and
Patashnik.</li> Patashnik.</li>
<li><sup>2</sup> See chapter 2.6 of <cite>Concrete <li><sup>3</sup> See chapter 2.6 of <cite>Concrete
Mathematics</cite> for an explanation of this identity.</li> Mathematics</cite> for an explanation of this identity.</li>
<li><sup>3</sup> A version of this formula is also mentioned in <li><sup>4</sup> A version of this formula is also mentioned in
chapter 6.5 of <cite>Concrete Mathematics</cite>.</li> chapter 6.5 of <cite>Concrete Mathematics</cite>.</li>
</ol> </ol>
</section> </section>