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%I #14 Jun 03 2019 03:21:49
%S 2,5,17,44,95,188,377,479,887,1361,1451,1811,2060,3056,3992,5843,5876,
%T 6008,6536,8648,10592,14585,16868,20597,23339,27500,29000,35000,41696,
%U 48872,55520,57464,65240,68960,69077,69545,71417,78905,93356,100049
%N a(1) = 2, a(n) = a(n-1) + 3*(a(n-1)-floor(a(n-1)^(1/3))^3).
%C A cubic version of the Weintraub recursion.
%H Steven H. Weintraub, <a href="https://www.jstor.org/stable/4145074">An Interesting Recursion</a>, Amer. Math. Monthly, v 111, no. 6, 2004, page 528.
%t digits=200
%t a[n_Integer?Positive] := a[n] = a[n-1] + 3*(a[n-1] - Floor[a[n-1]^(1/3)]^3)
%t a[1] = 2
%t a0=Table[a[n], {n, 1, digits}]
%K nonn,easy
%O 1,1
%A _Roger L. Bagula_, Jun 20 2004
%E Name edited by _Michel Marcus_, Jun 03 2019
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