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Is there any way that adjusted $R^2$ would be greater than $R^2$? Including cases of extreme values of n and p and negative values of $R^2$.

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No it can't. Check out this summary.

Edit: Just to flush out a bit more

The formula is $R^2_{adj} = 1 - \frac{(N-1)}{N-p-1}(1-R^2) $ where N = sample size, p = number of predictors, and $R^2$ is, well, $R^2$. So at best with an enormous number of samples and a small number of predictors, it can approach the original $R^2$ as $\frac{(N-1)}{N-p-1}$ approaches 1.

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    $\begingroup$ The argument is correct--but it's not quite general enough to address the situation implied by the question. The suggestion that $R^2$ could be negative hints that a regression through the origin is contemplated, in which case the correct formula for adjusted $R^2$ is slightly different than the one given here. The conclusion remains true--in fact, the adjustment is even greater in that case--but it might be worth pointing out that the absolute value of the adjusted $R^2$ easily could exceed that of $R^2$ (fwiw). In the most extreme cases $R^2=0$ whereas the adjusted value is $-1$. $\endgroup$
    – whuber
    Mar 25, 2016 at 20:30

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