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a(1) = 1; a(n+1) = floor((sum{k=1 to n} a(k)^3)^(1/3)).
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%I #10 Apr 09 2014 10:13:00

%S 1,1,1,1,1,1,1,1,2,2,2,3,3,4,5,6,7,9,11,14,17,21,27,34,42,53,67,84,

%T 106,133,167,211,265,334,421,530,668,841,1060,1335,1682,2119,2670,

%U 3364,4238,5339,6727,8475,10678,13453,16950,21355,26906,33899,42710,53811

%N a(1) = 1; a(n+1) = floor((sum{k=1 to n} a(k)^3)^(1/3)).

%e a(10) = floor((1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 2^3)^(1/3)) = floor(16^(1/3)) = 2.

%t a[1] = 1; a[n_] := a[n] = Floor[ Sum[ a[k]^3, {k, 1, n - 1}]^(1/3)]; Table[ a[n], {n, 1, 56} ]

%K easy,nonn

%O 1,9

%A _Leroy Quet_, Feb 15 2002

%E Edited by _Robert G. Wilson v_, Feb 18 2002