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<p>
Next: <a href="Random-Number-Algorithms.html#Random-Number-Algorithms" accesskey="n" rel="next">Random Number Algorithms</a>, Previous: <a href="Fibonacci-Numbers-Algorithm.html#Fibonacci-Numbers-Algorithm" accesskey="p" rel="prev">Fibonacci Numbers Algorithm</a>, Up: <a href="Other-Algorithms.html#Other-Algorithms" accesskey="u" rel="up">Other Algorithms</a> &nbsp; [<a href="Concept-Index.html#Concept-Index" title="Index" rel="index">Index</a>]</p>
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<h4 class="subsection">15.7.5 Lucas Numbers</h4>
<a name="index-Lucas-number-algorithm"></a>
<p><code>mpz_lucnum2_ui</code> derives a pair of Lucas numbers from a pair of Fibonacci
numbers with the following simple formulas.
</p>
<div class="example">
<pre class="example">L[k] = F[k] + 2*F[k-1]
L[k-1] = 2*F[k] - F[k-1]
</pre></div>
<p><code>mpz_lucnum_ui</code> is only interested in <em>L[n]</em>, and some work can be
saved. Trailing zero bits on <em>n</em> can be handled with a single square
each.
</p>
<div class="example">
<pre class="example">L[2k] = L[k]^2 - 2*(-1)^k
</pre></div>
<p>And the lowest 1 bit can be handled with one multiply of a pair of Fibonacci
numbers, similar to what <code>mpz_fib_ui</code> does.
</p>
<div class="example">
<pre class="example">L[2k+1] = 5*F[k-1]*(2*F[k]+F[k-1]) - 4*(-1)^k
</pre></div>
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