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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: $\sqrt[n]{x} = x^{\frac{1}{n}}$.

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2
  • $\begingroup$ Could you clarify whether we are allowed unary minus ? $\endgroup$
    – Gareth McCaughan
    Commented Jan 21, 2018 at 23:46
  • $\begingroup$ Unary minus is allowed. I edited the question to clarify this. $\endgroup$
    – A. P.
    Commented Jan 21, 2018 at 23:48

2 Answers 2

7
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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,

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

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  • $\begingroup$ In that case I think it should be possible in 18 $\endgroup$ Commented Jan 21, 2018 at 23:43
  • $\begingroup$ Ah - you saw it too :) $\endgroup$ Commented Jan 21, 2018 at 23:44
  • $\begingroup$ Yup, seconds after posting 38. $\endgroup$
    – Gareth McCaughan
    Commented Jan 21, 2018 at 23:45
  • $\begingroup$ 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 $\endgroup$ Commented Jan 21, 2018 at 23:48
  • $\begingroup$ 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.) $\endgroup$
    – Gareth McCaughan
    Commented Jan 21, 2018 at 23:58
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Using a similar method to my answer to the previous question...

 43  42

Since

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

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

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