Is next time complexity sub-exponential?
$O(2^{N^{LOG2(1.5)}}/8)$
unformatted: O((2^N)^LOG2(1.5))/8) just in case I didn't format it properly.
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$\begingroup$ "next" in what sense? $\endgroup$ – Raphael♦ Oct 25 '17 at 9:48
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It depends on how you define subexponential. The definition I generally use is:
$$\textsf{SUBEXP} = \bigcap_{\varepsilon > 0}\textsf{DTIME}(2^{n^\varepsilon})$$
However some authors define:
$$\textsf{SUBEXP} = \textsf{DTIME}(2^{o(n)})$$
Since your function is $2^{n^c}$ for some constant $c<1$, it would be subexponential according to the second but not to the first.
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$\begingroup$ First definition can be rewritten as $\mathsf{DTIME}(2^{n^{\ o(1)}}\ \ )$, right? $\endgroup$ – rus9384 Oct 25 '17 at 10:56
