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

%I #16 Nov 13 2024 23:02:09

%S 0,1,8,27,3,34,89,174,24,127,272,465,81,298,575,918,192,565,1016,1551,

%T 375,946,1613,2382,648,1459,2384,62,1029,2122,3347,263,1536,2953,4520,

%U 566,2187,3970,5921,989,3000,5191,7568,1550,3993,6634,9479,2267,5184,8317

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

%C Is there a simple expression for a nontrivial lower bound for a(n)?

%e a(4) = 2*4^3 - floor(2^(1/3)*4)^3 = 2*64 - 5^3 = 3.

%e a(5) = 2*5^3 - floor(2^(1/3)*5)^3 = 2*125 - 6^3 = 34.

%t Table[2 n^3 - Floor[2^(1/3) n]^3, {n, 0, 60}] (* _Vincenzo Librandi_, May 29 2015 *)

%o (PARI) f(n,e=3,b=2)=n^e*b-floor(sqrtn(b,e)*n)^e

%o (Magma) [2*n^3 - Floor(2^(1/3)*n)^3: n in [0..50]]; // _Vincenzo Librandi_, May 29 2015

%Y Cf. A087056 (analog for squares), A257854, A257855 (4th & 5th power).

%K nonn,easy

%O 0,3

%A _M. F. Hasler_, May 28 2015