A122987 Possible last three digits of n^3 (leading zeros omitted).

0, 1, 3, 7, 8, 9, 11, 13, 16, 17, 19, 21, 23, 24, 27, 29, 31, 32, 33, 37, 39, 41, 43, 47, 48, 49, 51, 53, 56, 57, 59, 61, 63, 64, 67, 69, 71, 72, 73, 77, 79, 81, 83, 87, 88, 89, 91, 93, 96, 97, 99, 101, 103, 104, 107, 109, 111, 112, 113, 117, 119

Offset: 1

Author

Sergio Pimentel, Sep 22 2006

Keywords

• base
• fini
• full
• easy
• nonn

Comments

This is a finite sequence. a(505) = 999 is the last term.

Examples

			The last three digits of n^3 can be 111, 112, 113, etc. but not 114, 115, 116, etc.
		

Links

• Nathaniel Johnston, Table of n, a(n) for n = 1..505 (full sequence)
• Eric Weisstein's World of Mathematics, Cubic Numbers.

Crossrefs

Cf. A000578, A122986, A122988.

Programs

Maple

s:={}: for n from 0 to 999 do s:=s union {n^3 mod 1000}: od: op(s); # Nathaniel Johnston, Jun 22 2011

Mathematica

Union[Mod[Range[1000]^3,1000]] (* James C. McMahon, Nov 14 2024 *)