A254403 a(n) = floor(b(n)), where b(n) = b(n-1)^(3/2), b(1) = 4.
4, 8, 22, 107, 1116, 37315, 7208411, 19353494040, 2692396855225368, 139703926313910081688758, 52217120356716278672533411477879775, 11932168478692303941858114447713697732545410824169841
Offset: 1
Crossrefs
Cf. A254404.
Programs
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Mathematica
Floor[RecurrenceTable[{a[1]==4,a[n]==a[n-1]^(3/2)},a,{n,1,15}]] Table[Floor[4^((3/2)^(n-1))], {n, 1, 15}]
Formula
a(n) = floor(4^((3/2)^(n-1))).