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a(n) = floor(b(n)), where b(n) = b(n-1)^(3/2), b(1) = 2.
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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