%I #28 Jun 18 2024 21:05:19
%S 1,1,1,2,2,3,4,5,6,8,10,12,16,20,25,32,40,50,64,80,101,128,161,203,
%T 256,322,406,512,645,812,1024,1290,1625,2048,2580,3250,4096,5160,6501,
%U 8192,10321,13003,16384,20642,26007,32768,41285,52015,65536,82570,104031
%N Powers of cube root of 2 rounded down.
%C Rounding has no effect when n is a multiple of 3, because then obviously (2^(1/3))^n = 2^(n/3). - _Alonso del Arte_, Jan 04 2014
%H Vincenzo Librandi, <a href="/A017979/b017979.txt">Table of n, a(n) for n = 0..200</a>
%e a(2) = 1 because the cube root of 2 squared is 1.5874...
%e a(3) = 2 because the cube root of 2 cubed is 2 exactly.
%e a(4) = 2 because the cube root of 2 to the fourth power is 2.519842...
%t Table[Floor[(2^(1/3))^n], {n, 0, 49}] (* _Alonso del Arte_, Jan 04 2014 *)
%o (Magma) [Floor(2^(n/3)): n in [0..50]]; // _Vincenzo Librandi_, Jan 06 2014
%o (Python)
%o from sympy import integer_nthroot
%o def A017979(n): return integer_nthroot(1<<n,3)[0] # _Chai Wah Wu_, Jun 18 2024
%Y Cf. A017981, A002580.
%Y Sequences of the type: Powers of cube root of (k) rounded down: this sequence (k=2), A017982 (k=3), A017985 (k=4), A017988 (k=5), A017991 (k=6), A017994 (k=7), A018000 (k=9), A018003 (k=10), A018006 (k=11), A018009 (k=12), A018012 (k=13), A018015 (k=14), A018018 (k=15), A018021 (k=16), A018024 (k=17), A018027 (k=18), A018030 (k=19), A018033 (k=20), A018036 (k=21), A018039 (k=22), A018042 (k=23), A018045 (k=24).
%K nonn
%O 0,4
%A _N. J. A. Sloane_
%E a(44)-a(50) from _Alex Ratushnyak_, Jan 04 2014