# is this time complexity subexponential? [duplicate]

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.

• "next" in what sense? – Raphael Oct 25 '17 at 9:48

$$\textsf{SUBEXP} = \bigcap_{\varepsilon > 0}\textsf{DTIME}(2^{n^\varepsilon})$$
$$\textsf{SUBEXP} = \textsf{DTIME}(2^{o(n)})$$
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.
• First definition can be rewritten as $\mathsf{DTIME}(2^{n^{\ o(1)}}\ \ )$, right? – rus9384 Oct 25 '17 at 10:56