# Make 439204 from Φ (Golden Ratio)

In my last question I somehow tricked myself, because I wanted to make a puzzle about a number that looks like a year. But the complicated solution I thought of turned out to be outperformed by a much simpler ansatz. Here is a puzzle that (hopefully) has my intended solution as optimum:

Make the number $439204$ by using an arbitrary number of $\Phi$s and

• the operators $+$, $−$, $\cdot$, $/$ (including unary "$-$")
• exponentiation
• brackets $($ $)$.

The aim is to use as few $\Phi$s as possible.

You may not use operators or functions other than in this list, so don't even ask for rounding ($\lfloor$ $\rfloor$, $\lceil$ $\rceil$) or logarithm ($\log_a (x)$). If you want to use roots, this is ok as long as you express them as exponent: $\sqrt[n]{x} = x^{\frac{1}{n}}$.

• Could you clarify whether we are allowed unary minus ? – Gareth McCaughan Jan 21 '18 at 23:46
• Unary minus is allowed. I edited the question to clarify this. – A. P. Jan 21 '18 at 23:48

I can do

38 ... no, 18,

as follows.

Write $L_n$ for the $n$th Lucas number, so $L_0=2$, $L_1=1$, and $L_{n+2}=L_{n+1}+L_n$. It happens that $L_{27}=439204$. And we have $L_n=\phi^n+(-\phi)^{-n}$. And $27=3^3$.

Therefore,

$$\begin{eqnarray}439204 &=& \phi^{27}+(-\phi)^{-27} \\ &=& \phi^{27}-\phi^{-27} \\ &=& \phi^{\left(\left(\frac{\phi+\phi+\phi}{\phi}\right)^{\left(\frac{\phi+\phi+\phi}{\phi}\right)}\right)} - \phi^{-\left(\left(\frac{\phi+\phi+\phi}{\phi}\right)^{\left(\frac{\phi+\phi+\phi}{\phi}\right)}\right)}. \end{eqnarray}$$

• In that case I think it should be possible in 18 – Jonathan Allan Jan 21 '18 at 23:43
• Ah - you saw it too :) – Jonathan Allan Jan 21 '18 at 23:44
• Yup, seconds after posting 38. – Gareth McCaughan Jan 21 '18 at 23:45
• I really should've done what I was going to do when I first saw the number change... an oeis search oeis.org/search?q=439204&language=english&go=Search – Jonathan Allan Jan 21 '18 at 23:48
• Yep :-). (What I actually did when I saw the new number was to look up "Lucas numbers". But I already knew about the nice formula for those.) – Gareth McCaughan Jan 21 '18 at 23:58

Using a similar method to my answer to the previous question...

43  42

Since

$N = \frac{\phi + \phi + ...}{\phi}$
and
$439204 = 2^{(2^3)}\cdot 5\cdot 7^3+2\cdot (3^4+1)$

$\frac{\phi + \phi}{\phi}^{(\frac{\phi + \phi}{\phi}^\frac{\phi + \phi + \phi}{\phi})}\cdot \frac{\phi + \phi + \phi + \phi + \phi}{\phi} \cdot \frac{\phi + \phi + \phi + \phi + \phi + \phi + \phi}{\phi} ^\frac{\phi + \phi + \phi}{\phi} + \frac{\phi + \phi}{\phi}\cdot (\frac{\phi + \phi + \phi}{\phi} ^\frac{\phi + \phi + \phi + \phi}{\phi} + \frac{\phi}{\phi})$