|
Search: id:A104758
|
|
|
| A104758 |
|
Number of cubes m = k^3 such that n <= m <= (n+1)^2. |
|
+0
4
|
|
| 2, 1, 1, 1, 1, 2, 2, 3, 3, 2, 2, 3, 3, 3, 4, 4, 4, 4, 5, 5, 5, 5, 6, 6, 6, 6, 7, 7, 6, 6, 6, 7, 7, 7, 7, 7, 8, 8, 8, 8, 8, 9, 9, 9, 9, 9, 10, 10, 10, 10, 10, 10, 11, 11, 11, 11, 11, 11, 12, 12, 12, 12, 12, 13, 13, 12, 12, 12, 12, 12, 13, 13, 13, 13, 13, 13, 14, 14, 14, 14, 14, 14, 15, 15, 15
(list; graph; listen)
|
|
|
OFFSET
|
0,1
|
|
|
COMMENT
|
a(n)>=1 because between n and (n+1)^2 there is always at least one cubic number.
a(6)=2 because between 5 and 6^2 there are two cubic numbers: 8 and 27
|
|
MATHEMATICA
|
f[n_] := Floor[(n + 1)^(2/3)] - Floor[n^(1/3)] + If[ IntegerQ[n^(1/3)], 1, 0]; Table[ f[n], {n, 0, 84}] (from Robert G. Wilson v (rgwv(AT)rgwv.com), Apr 24 2005)
|
|
CROSSREFS
|
Sequence in context: A109035 A064823 A140225 this_sequence A143227 A026791 A080576
Adjacent sequences: A104755 A104756 A104757 this_sequence A104759 A104760 A104761
|
|
KEYWORD
|
easy,nonn
|
|
AUTHOR
|
Giovanni Teofilatto (g.teofilatto(AT)tiscalinet.it), Apr 23 2005
|
|
EXTENSIONS
|
More terms from Robert G. Wilson v (rgwv(AT)rgwv.com), Apr 24 2005
|
|
|
Search completed in 0.002 seconds
|
