A254400 a(n) = floor(b(n)), where b(n) = b(n-1)^(3/2), b(1) = 2.
2, 2, 4, 10, 33, 193, 2684, 139116, 51888311, 373769884171, 228510656987187971, 109234465617278065859643766, 1141667222716533804555279991265973169394, 38575298818045633410275497202805726438253675072452563405216
Offset: 1
Programs
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Mathematica
Floor[RecurrenceTable[{a[1]==2,a[n]==a[n-1]^(3/2)},a,{n,1,15}]] Table[Floor[2^((3/2)^(n-1))], {n, 1, 15}]
Formula
a(n) = floor(2^((3/2)^(n-1))).