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I want to write a Christmas message to leave as a comment thanking the people who in the next 24th December will solve some of my problems:

I wish you Math Christmas and a Happy New Year ...

Where the dots … will be 2016 written only using particular values of Riemann zeta function.

Example. For the integer $2$, we have $$2=\frac{\zeta^2(2)}{\zeta(4)}-\zeta(0).$$

Question. Can you do the same with the integer $2016$ using sums, addition, subtraction, products, quotients and powers if it is necessary, only with particular values of Riemann zeta function. Thanks in advance.

References:

[1] https://en.wikipedia.org/wiki/Particular_values_of_Riemann_zeta_function

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  • $\begingroup$ I am looking for a simplest expression, and if it possible that likes cool (caution, I say cool, not cold!) I will choose one answer, as soon as possible. Thanks. $\endgroup$ – user243301 Dec 12 '15 at 19:35
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    $\begingroup$ First $\frac{\zeta(2)^2}{\zeta(4)}-\zeta(0) = 3 \ne 2$. Second, $2016 = \frac{\left(\zeta(2)^2-\zeta(0)\zeta(4)\right)^3}{\zeta(0)^6\zeta(4)\zeta(8)}$ $\endgroup$ – achille hui Dec 12 '15 at 20:15
  • $\begingroup$ Thanks @achillehui , today was a bad day! Very thanks much. $\endgroup$ – user243301 Dec 12 '15 at 22:18
  • $\begingroup$ May I ask how you came up with that? @achillehui $\endgroup$ – Will Fisher Dec 13 '15 at 0:50
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Well we can say that $2016 = 2^{11}-2^{5}$ which means one possibility, of probably many, is
$$2016=\left(\frac{\zeta^2(2)}{\zeta(4)}+\zeta(0)\right)^{11}-\left(\frac{\zeta^2(2)}{\zeta(4)}+\zeta(0)\right)^{5}$$
Also one thing, which I don't know if it fits your criteria, but I think it looks pretty slick, is
$$2016=\sum\limits_{i=1}^{1008}\left(\frac{\zeta^2(2)}{\zeta(4)}+\zeta(0)\right)$$
(It is a pretty point less statement because you could just multiply it by 1008 but hey, it looks cool). Finally
$$2016=\left(\frac{\zeta^2(2)}{\zeta(4)}+\zeta(0)\right)^{5}\left(\frac{\zeta^2(2)}{\zeta(4)}-\zeta(0)\right)^{2}\left(\frac{3\zeta^2(2)}{\zeta(4)}+\zeta(0)\right)$$

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    $\begingroup$ You can ever rewrite 11 as 2^(2^2)+2+1 and 5 as 2^2+1 and use the formula for 2 in each instance :-) $\endgroup$ – Mariano Suárez-Álvarez Dec 12 '15 at 20:01
  • $\begingroup$ Very thanks much @WillFisher $\endgroup$ – user243301 Dec 12 '15 at 22:22
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What about $$\left(\frac{\zeta^2(2)}{\zeta(4)}+\zeta(0)\right)^{4}-\frac{\zeta^3(-1)}{\zeta^2(-3)\zeta(-7)}$$

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    $\begingroup$ Very thanks much @Sfarla, I choose previous answer, but now I don't know what I will to put in my message. Your is so cool and short as achillehui's answer $\endgroup$ – user243301 Dec 12 '15 at 22:22

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