<story><title>Golden powers are nearly integers</title><url>https://www.johndcook.com/blog/2017/03/22/golden-powers-are-nearly-integers/</url></story><parent_chain><item><author>Sniffnoy</author><text>Specifically, they&amp;#x27;re nearly Lucas numbers.  For those unfamiliar, the Lucas numbers are defined by the recursion L_0 = 2, L_1 = 1, and L_n = L_{n-1} + L_n.  (I.e., the Fibonacci recursion, but with different starting values.)  They can also be given explicitly by L_n = phi^n + (-1&amp;#x2F;phi)^n.  Since (-1&amp;#x2F;phi)^n goes rapidly to zero, phi^n will be very close to L_n.  More specifically, for n&amp;gt;=2, (1&amp;#x2F;phi)^n &amp;lt; 1&amp;#x2F;2, so L_n will be the nearest integer to phi^n.&lt;p&gt;The Fibonacci numbers, meanwhile, can be given by the formula F_n = (phi^n - (-1&amp;#x2F;phi)^n) &amp;#x2F; sqrt(5).  So a similar thing holds for them.&lt;p&gt;More generally, phi is a Pisot number: &lt;a href=&quot;https:&amp;#x2F;&amp;#x2F;en.wikipedia.org&amp;#x2F;wiki&amp;#x2F;Pisot%E2%80%93Vijayaraghavan_number&quot; rel=&quot;nofollow&quot;&gt;https:&amp;#x2F;&amp;#x2F;en.wikipedia.org&amp;#x2F;wiki&amp;#x2F;Pisot%E2%80%93Vijayaraghavan_n...&lt;/a&gt;  Any Pisot number will have its powers approach integers at an exponential rate.  (In that, the distance to the nearest integer decreases exponentially.)</text></item></parent_chain><comment><author>jacobolus</author><text>Another way to get at this:&lt;p&gt;&lt;pre&gt;&lt;code&gt;  φ^1 = φ
  φ^2 = φ + 1 [this is the definition of φ]
  φ^3 = φ(φ + 1) = 2φ + 1
  φ^4 = φ(2φ + 1) = 3φ + 2
  φ^5 = φ(3φ + 2) = 5φ + 3
  φ^6 = φ(5φ + 3) = 8φ + 5
  ...
  φ^n = F(n)φ + F(n-1)
&lt;/code&gt;&lt;/pre&gt;
Note that F(n+1)&amp;#x2F;F(n) → φ as n → ∞. Or flipped around, F(n)φ → F(n+1).&lt;p&gt;So φ^n = F(n)φ + F(n-1) → F(n+1) + F(n-1) = L(n) as n → ∞. This gets close to being an integer because F(n+1) and F(n-1) are both integers.</text></comment>
<story><title>Golden powers are nearly integers</title><url>https://www.johndcook.com/blog/2017/03/22/golden-powers-are-nearly-integers/</url></story><parent_chain><item><author>Sniffnoy</author><text>Specifically, they&amp;#x27;re nearly Lucas numbers.  For those unfamiliar, the Lucas numbers are defined by the recursion L_0 = 2, L_1 = 1, and L_n = L_{n-1} + L_n.  (I.e., the Fibonacci recursion, but with different starting values.)  They can also be given explicitly by L_n = phi^n + (-1&amp;#x2F;phi)^n.  Since (-1&amp;#x2F;phi)^n goes rapidly to zero, phi^n will be very close to L_n.  More specifically, for n&amp;gt;=2, (1&amp;#x2F;phi)^n &amp;lt; 1&amp;#x2F;2, so L_n will be the nearest integer to phi^n.&lt;p&gt;The Fibonacci numbers, meanwhile, can be given by the formula F_n = (phi^n - (-1&amp;#x2F;phi)^n) &amp;#x2F; sqrt(5).  So a similar thing holds for them.&lt;p&gt;More generally, phi is a Pisot number: &lt;a href=&quot;https:&amp;#x2F;&amp;#x2F;en.wikipedia.org&amp;#x2F;wiki&amp;#x2F;Pisot%E2%80%93Vijayaraghavan_number&quot; rel=&quot;nofollow&quot;&gt;https:&amp;#x2F;&amp;#x2F;en.wikipedia.org&amp;#x2F;wiki&amp;#x2F;Pisot%E2%80%93Vijayaraghavan_n...&lt;/a&gt;  Any Pisot number will have its powers approach integers at an exponential rate.  (In that, the distance to the nearest integer decreases exponentially.)</text></item></parent_chain><comment><author>tgb</author><text>I&amp;#x27;ve always enjoyed that formula for the Fibonacci numbers, primarily because it&amp;#x27;s so unintuitive that the right hand side would actually even be an integer. It&amp;#x27;s a fun exercise to prove that (without resorting to the fact that the Fibonacci numbers are obviously integers). Hint: start with -1&amp;#x2F;phi = 1-phi and use phi=(sqrt(5)+1)&amp;#x2F;2.</text></comment>