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Can someone please verify and maybe simplify the closed formula for A119016 :

$ a(n)=\frac{1}{4} \left(1+\sqrt{2}\right)^{n/2} \left((-1)^n+1\right)+\frac{1}{2} \left(\sqrt{2}-1\right)^{n/2} \cos \left(\frac{\pi n}{2}\right) +\frac{1}{4 \sqrt{8119+5741\sqrt{2}}} \left(58+41 \sqrt{2}\right) \left(1+\sqrt{2}\right)^{n/2} \left(1-(-1)^n\right)-2 \left(140+99 \sqrt{2}\right) \left(\sqrt{2}-1\right)^{n/2} \sin \left(\frac{\pi n}{2}\right) $

Considering even and odd functions I could simplify the formula to : $ a(n)=\frac{1}{4} \left(\sqrt{2}-2\right) \left(1+\sqrt{2}\right) \left(1-(-1)^n\right) \left(\left(1-\sqrt{2}\right)^{\frac{n-1}{2}}-\left(1+\sqrt{2}\right)^{\frac{n-1}{2}}\right)+\frac{1 }{4} \left((-1)^n+1\right) \left(\left(1-\sqrt{2}\right)^{n/2}+\left(1+\sqrt{2}\right)^{n/2}\right) $

Thanks

1 Answers 1

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One observation is that $\cos\left(\frac{\pi n}2\right)=\left\{\begin{array}{cc}(-1)^{n/2}& n\equiv0,2(\mod4)\\0&n\equiv1,3(\mod4) \end{array}\right.$ and $\sin\left(\frac{\pi n}2\right)=\left\{\begin{array}{cc}(-1)^{n/2}& n\equiv1,3(\mod4)\\0&n\equiv0,2(\mod4) \end{array}\right.$

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    yes, also only taking the formula from $a(n)=1/4$ .. to .. $cos( \pi n / 2)$ you will notice that this represents the sequence A001333 for $n$ odd.2012-11-04