A017979 Powers of cube root of 2 rounded down.

1, 1, 1, 2, 2, 3, 4, 5, 6, 8, 10, 12, 16, 20, 25, 32, 40, 50, 64, 80, 101, 128, 161, 203, 256, 322, 406, 512, 645, 812, 1024, 1290, 1625, 2048, 2580, 3250, 4096, 5160, 6501, 8192, 10321, 13003, 16384, 20642, 26007, 32768, 41285, 52015, 65536, 82570, 104031

Offset: 0

Author

N. J. A. Sloane

Keywords

• nonn

Comments

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

Examples

			a(2) = 1 because the cube root of 2 squared is 1.5874...
a(3) = 2 because the cube root of 2 cubed is 2 exactly.
a(4) = 2 because the cube root of 2 to the fourth power is 2.519842...
		

Links

• Vincenzo Librandi, Table of n, a(n) for n = 0..200

Crossrefs

Cf. A017981, A002580.

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).

Programs

Magma

[Floor(2^(n/3)): n in [0..50]]; // Vincenzo Librandi, Jan 06 2014

Mathematica

Table[Floor[(2^(1/3))^n], {n, 0, 49}] (* Alonso del Arte, Jan 04 2014 *)

Python

from sympy import integer_nthroot
def A017979(n): return integer_nthroot(1<Chai Wah Wu, Jun 18 2024

Extensions

a(44)-a(50) from Alex Ratushnyak, Jan 04 2014