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a(n) = floor(e^(n!)).
0

%I #23 Aug 07 2015 05:14:47

%S 2,7,403,26489122129,

%T 13041808783936322797338790280986488113446079415755132

%N a(n) = floor(e^(n!)).

%C The exponential growth in the number of permutations of n elements.

%C Next term is too big to be included.

%F a(n) = A000149(A000142(n)).

%F a(n) = floor(sqrt(e^A052849(n) - e^A000142(n) + sqrt(e^A052849(n) - e^A000142(n) + sqrt(e^A052849(n) - e^A000142(n) + ...)))).

%e a(1) = floor(e^(1!)) = floor(e) = 2.

%t Table[Floor[E^n!], {n, 1, 7}]

%o (PARI) default(realprecision, 100); vector(5, n, floor(exp(n!))) \\ _Michel Marcus_, Aug 06 2015

%Y Cf. A000149, A000142.

%Y Cf. A050923, A100731.

%K nonn

%O 1,1

%A _Ilya Gutkovskiy_, Jul 20 2015