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a(n) = [ a(n-1)/(sqrt(6) - 2) ], where a(0) = 1.
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%I #7 Aug 03 2014 14:27:15

%S 1,2,4,8,17,37,82,182,404,898,1997,4442,9882,21984,48908,108807,

%T 242067,538537,1198107,2665482,5930017,13192774,29350556,65297498,

%U 145270273,323189294,719013724,1599622094,3558751049,7917313144,17614001812,39186660195

%N a(n) = [ a(n-1)/(sqrt(6) - 2) ], where a(0) = 1.

%H Clark Kimberling, <a href="/A024557/b024557.txt">Table of n, a(n) for n = 0..250</a>

%t a[0] = 1;

%t a[n_] := Floor[a[n - 1]/FractionalPart[Sqrt[6]]]

%t Table[a[n], {n, 0, 60}]

%t (* _Clark Kimberling_, Aug 16 2012 *)

%K nonn

%O 0,2

%A _Clark Kimberling_