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%I #4 Jan 30 2015 06:35:02
%S 4,8,22,107,1116,37315,7208411,19353494040,2692396855225368,
%T 139703926313910081688758,52217120356716278672533411477879775,
%U 11932168478692303941858114447713697732545410824169841
%N a(n) = floor(b(n)), where b(n) = b(n-1)^(3/2), b(1) = 4.
%F a(n) = floor(4^((3/2)^(n-1))).
%t Floor[RecurrenceTable[{a[1]==4,a[n]==a[n-1]^(3/2)},a,{n,1,15}]]
%t Table[Floor[4^((3/2)^(n-1))], {n, 1, 15}]
%Y Cf. A254404.
%K nonn,easy
%O 1,1
%A _Vaclav Kotesovec_, Jan 30 2015
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