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a(n) = floor(sqrt(5*2^n)).
5

%I #12 Jun 26 2022 20:05:35

%S 2,3,4,6,8,12,17,25,35,50,71,101,143,202,286,404,572,809,1144,1619,

%T 2289,3238,4579,6476,9158,12952,18317,25905,36635,51810,73271,103621,

%U 146542,207243,293085,414486,586171,828972,1172343,1657944,2344687,3315888,4689374,6631776,9378748,13263553,18757497,26527107

%N a(n) = floor(sqrt(5*2^n)).

%C Theorem 3 of Dubickas implies that infinitely many terms of this sequence are divisible by 2 or 3 (and hence infinitely many composites). - _Charles R Greathouse IV_, Feb 04 2016

%D Artūras Dubickas, Prime and composite integers close to powers of a number, Monatsh. Math. 158:3 (2009), pp. 271-284.

%o (PARI) a(n)=sqrtint(5*2^n) \\ _Charles R Greathouse IV_, Apr 18 2013

%Y Cf. A017910, A017913, A017919, A114183, A221718, A221942-A221946.

%K nonn

%O 0,1

%A _N. J. A. Sloane_, Feb 01 2013