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a(n) = floor(b(n)), where b(n) = b(n-1)^(3/2), b(1) = 3.
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%I #6 Aug 11 2021 17:54:47

%S 3,5,11,40,260,4198,272087,141926645,1690814280107,

%T 2198588037458589676,3259986420078467245024559414,

%U 186133132558073401497122370072981441415771,80303695148132102394972761166781400472084090796880334009212493

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

%F a(n) = floor(3^((3/2)^(n-1))).

%t Floor[RecurrenceTable[{a[1]==3,a[n]==a[n-1]^(3/2)},a,{n,1,15}]]

%t Table[Floor[3^((3/2)^(n-1))], {n, 1, 15}

%t Floor[NestList[#^(3/2)&,3,15]] (* _Harvey P. Dale_, Aug 11 2021 *)]

%Y Cf. A254402.

%K nonn,easy

%O 1,1

%A _Vaclav Kotesovec_, Jan 30 2015