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a(n) = floor(a(n-1)^(3/2)), a(1) = 3.
1

%I #5 Jan 30 2015 06:35:09

%S 3,5,11,36,216,3174,178817,75615914,657536414322,533187198278014959,

%T 389331524481948767367453560,7682089373864047165337173800294919710603,

%U 673316152668899727226337604512171297382634328995985790889798

%N a(n) = floor(a(n-1)^(3/2)), a(1) = 3.

%F a(n) ~ c^((3/2)^n), where c = 2.02960355478331314223272956490362941573262985416270194089589212743... .

%t RecurrenceTable[{a[1]==3,a[n]==Floor[a[n-1]^(3/2)]},a,{n,1,15}]

%Y Cf. A254401.

%K nonn,easy

%O 1,1

%A _Vaclav Kotesovec_, Jan 30 2015