According to Wolfram Alpha,
$$\sum_{j=0}^{k+1} 2^{k+1-j} \biggl(\frac{(k+1+j)!}{(k+1)!j!}\biggl) = 4^{k+1},$$
but I don't see how this sum can be simplified to this. How is it done? Thanks in advance.
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1You'll probably need to use the Binomial Theorem. – Dan Aug 25 '22 at 00:21
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@Dan could you give me an example of how that'd be done? I'm not aware of that theorem and the web pages I'm finding on it aren't very informative for this precise case. Thanks. – Baguette Boy Aug 25 '22 at 00:30
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1A good place to start is probably to use the Binomial Theorem to get that $$4^{k+1}=(2+2)^{k+1}=\sum_{j=0}^{k+1}2^{k+1-j}2^j{k+1\choose j}$$ – Lorago Aug 25 '22 at 00:38
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If only that numerator $(k + 1 + j)!$ didn't depend on the loop counter $j$. – Dan Aug 25 '22 at 00:47
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@Lorago I see that a $\frac{1}{2^j}$ needs to factored out of the $2^{k+1}$ and into the ${k+1 \choose j}$, but I don't know how to do that. Any pointers there? – Baguette Boy Aug 25 '22 at 01:13
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1Does this answer your question? How to show $\sum_{k=0}^{n}\binom{n+k}{k}\frac{1}{2^k}=2^{n}$ – Alexander Burstein Aug 25 '22 at 03:02
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It seems to me that you found the particular case of a more general problem. $$f(x)=\sum_{j=0}^{k+1} \frac{(k+1+j)!}{(k+1)!\,j!}x^{k+1-j}$$ Using the Gaussian hypergeometric function for $x>1$ $$f(x)=\frac{x^{2 k+3} } {(x-1)^{k+2} }-\frac{(2 k+3)! }{x (k+1)! (k+2)!}\,\, _2F_1\left(1,2 k+4;k+3;\frac{1}{x}\right)$$ Using the regularized form of the Gaussian hypergeometric function
$$f(x)=\frac{x^{2 k+3} } {(x-1)^{k+2} }-\frac{2^{2 k+3} \, \Gamma \left(k+\frac{5}{2}\right)}{x\sqrt{\pi } }\,\,\, _2\tilde{F}_1\left(1,2 k+4;k+3;\frac{1}{x}\right)$$ which gives
$$f(2)=2^{2 k+3}-4^{k+1}=4^{k+1}$$
Ifif not find any other case where, for integer values of $n$ $$\sqrt[k+1] {f(x)}$$ could be an integer.
Claude Leibovici
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