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%I #4 Jan 30 2015 06:35:25
%S 2,2,4,10,33,193,2684,139116,51888311,373769884171,228510656987187971,
%T 109234465617278065859643766,1141667222716533804555279991265973169394,
%U 38575298818045633410275497202805726438253675072452563405216
%N a(n) = floor(b(n)), where b(n) = b(n-1)^(3/2), b(1) = 2.
%F a(n) = floor(2^((3/2)^(n-1))).
%t Floor[RecurrenceTable[{a[1]==2,a[n]==a[n-1]^(3/2)},a,{n,1,15}]]
%t Table[Floor[2^((3/2)^(n-1))], {n, 1, 15}]
%Y Cf. A254401, A254403, A254405.
%K nonn,easy
%O 1,1
%A _Vaclav Kotesovec_, Jan 30 2015
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