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PostPosted: Thu, 5 Nov 2009 17:44:00 UTC 
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My friend Eric told me this one, and I thought it interesting enough to post.

Consider the formula

$S(n)={\sum_p{n\choose p}\over 2^n} (p ranges over the primes)

Find a function f(z) of an analytic variable, z, such that

$\lim_{z\to\infty}{f(z)\over S(z)}=1.

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PostPosted: Wed, 11 Nov 2009 15:00:16 UTC 
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Shadow wrote:
My friend Eric told me this one, and I thought it interesting enough to post.

Consider the formula

$S(n)={\sum_p{n\choose p}\over 2^n} (p ranges over the primes)

Find a function f(z) of an analytic variable, z, such that

$\lim_{z\to\infty}{f(z)\over S(z)}=1.


Presumably you can take a branch cut?

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\begin{aligned}
Spin(1)&=O(1)=\mathbb{Z}/2&\quad&\text{and}\\
Spin(2)&=U(1)=SO(2)&&\text{are obvious}\\
Spin(3)&=Sp(1)=SU(2)&&\text{by }q\mapsto(\mathop{\mathrm{Im}}\mathbb{H}\ni p\mapsto qp\bar{q})\\
Spin(4)&=Sp(1)\times Sp(1)&&\text{by }(q_1,q_2)\mapsto(\mathbb{H}\ni p\mapsto q_1p\bar{q_2})\\
Spin(5)&=Sp(2)&&\text{by }\mathbb{HP}^1\cong S^4_{round}\hookrightarrow\mathbb{R}^5\\
Spin(6)&=SU(4)&&\text{by the irrep }\Lambda_+\mathbb{C}^4
\end{aligned}


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PostPosted: Thu, 12 Nov 2009 02:28:26 UTC 
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outermeasure wrote:
Shadow wrote:
My friend Eric told me this one, and I thought it interesting enough to post.

Consider the formula

$S(n)={\sum_p{n\choose p}\over 2^n} (p ranges over the primes)

Find a function f(z) of an analytic variable, z, such that

$\lim_{z\to\infty}{f(z)\over S(z)}=1.


Presumably you can take a branch cut?


Yes. And the right one is fairly obvious I would say.

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