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%I #13 Aug 20 2012 11:57:25
%S 1,1,11,256,10541,672136,61001951,7445096296,1172998147241,
%T 231333927486736,55747918840676411,16100914826854576456,
%U 5485174647349481371661,2175023364205612725532456,992565740287786208277022391,516241740263310751317668520736
%N 2n-th derivative of sec(x)^sec(x) at x=0.
%C sec(x) = 1/cos(x).
%C The sequence gives only 2n-th derivatives because (2n+1)-th derivatives are 0.
%t f[x_] := Sec[x]^Sec[x]; Table[Derivative[2*n] [f][0],{n,0,18}]
%Y Cf. A186250.
%K nonn
%O 0,3
%A _Michel Lagneau_, Aug 18 2012