A254404 a(n) = floor(a(n-1)^(3/2)), a(1) = 4.
4, 8, 22, 103, 1045, 33781, 6208815, 15470810960, 1924285995519778, 84412046357073810845036, 24524866000840265443624861229739287, 3840696279681724751490160062668766360564070043725605, 238020820035391513516910494940635409432084649486247633125762760898691133120625
Offset: 1
Crossrefs
Cf. A254403.
Programs
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Mathematica
RecurrenceTable[{a[1]==4,a[n]==Floor[a[n-1]^(3/2)]},a,{n,1,15}]
Formula
a(n) ~ c^((3/2)^n), where c = 2.4979223179189793375148967824709800001224474904593996103376... .