A254401 a(n) = floor(b(n)), where b(n) = b(n-1)^(3/2), b(1) = 3.
3, 5, 11, 40, 260, 4198, 272087, 141926645, 1690814280107, 2198588037458589676, 3259986420078467245024559414, 186133132558073401497122370072981441415771, 80303695148132102394972761166781400472084090796880334009212493
Offset: 1
Crossrefs
Cf. A254402.
Programs
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Mathematica
Floor[RecurrenceTable[{a[1]==3,a[n]==a[n-1]^(3/2)},a,{n,1,15}]] Table[Floor[3^((3/2)^(n-1))], {n, 1, 15} Floor[NestList[#^(3/2)&,3,15]] (* Harvey P. Dale, Aug 11 2021 *)]
Formula
a(n) = floor(3^((3/2)^(n-1))).