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What is a mathematical expression for the sequence $$2,0,1,7,8,8,2,0,1,7,8,8,2,0,1,7,8,8,...$$

Like this Following it's not I want to answer $$a_{n}=\begin{cases} 2&n=6k\\ 0&n=6k+1\\ 1&n=6k+2\\ 7&n=6k+3\\ 8&n=6k+4\\ 8&n=6k+5 \end{cases}$$

such this:simple expression

math110
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    Egreg's solution to the linked problem shows a general approach you can use; you just take a combination of powers of sixth roots of unity. – John Hughes Mar 02 '17 at 14:07

4 Answers4

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It's also the decimal digits of $\dfrac{201788}{999999}$ after the radix point.

Joffan
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A generating function could be $g(x)=\dfrac{2+x^2+7x^3+8x^4+8x^5}{1-x^6}$

so the $n$th term of the sequence is the coefficient of $x^n$ in the expansion of $g(x)$, which can be found by dividing the $n$th derivative of $g(x)$ at $x=0$ by $n!$, i.e. $\dfrac{g^{(n)}(0)}{n!}$

Henry
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A DFT (discrete fourier transform) gives $$a_k={13\over3}-{2\over3}(-1)^k-{11\over6}\cos{k\pi\over3}-{5\sqrt{3}\over2}\sin{k\pi\over3}+{1\over6}\cos{2k\pi\over3}-{\sqrt{3}\over6}\sin{2k\pi\over3}\ .$$

2

You can use $$-\frac13\left(13\cos\left(\frac\pi3n\right) + 15\cos\left(\frac\pi3\left(n-2\right)\right) + 2\cos\left(\pi n\right) + \cos\left(\frac{2\pi}3\left(n-1\right)\right) - 13\right).$$

Théophile
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