A254402 a(n) = floor(a(n-1)^(3/2)), a(1) = 3.
3, 5, 11, 36, 216, 3174, 178817, 75615914, 657536414322, 533187198278014959, 389331524481948767367453560, 7682089373864047165337173800294919710603, 673316152668899727226337604512171297382634328995985790889798
Offset: 1
Crossrefs
Cf. A254401.
Programs
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Mathematica
RecurrenceTable[{a[1]==3,a[n]==Floor[a[n-1]^(3/2)]},a,{n,1,15}]
Formula
a(n) ~ c^((3/2)^n), where c = 2.02960355478331314223272956490362941573262985416270194089589212743... .