Regarding storage size for a number column, Oracle gives the formula at
http://download-west.oracle.com/docs/cd/B14117_01/server.101/b10743/datatype.htm
#sthref3808

"the column size in bytes for a particular numeric data value NUMBER(p), where
p is the precision of a given value, can be calculated using the following
formula:

ROUND((length(p)+s)/2))+1

where s equals zero if the number is positive, and s equals 1 if the number is
negative."

I quoted 10g documentation but it's the same since at least 8i on.

There're a few mistakes in this formula and its interpretation.

(1) Parentheses don't match. We can drop the rightmost closing parenthesis:
ROUND((length(p)+s)/2)+1
Alternatively, we can drop the parenthesis to the right of scale, with a 
totally different calculation though:
ROUND((length(p)+s/2))+1
But I don't think this is what they intended. The double opening and closing 
parentheses become unnecessary.

(2) Since p is already the precision, defined as "total number of digits", 
length(p) makes no sense. E.g., for 123.1, p is 4; length(p)=length(4) would be 
1. We should replace length(p) with just p. (Length(the_number) is not right 
either since it's 1 more than precision if the number contains a decimal 
point.) Now the formula looks like:
ROUND((p+s)/2)+1

(3) How can it be that "s equals zero if the number is positive"? For 123.1, s, 
defined as "number of digits to the right of the decimal point" earlier in 
documentation, is clearly 1.

(4) How can it be that "s equals 1 if the number is negative"?

Regarding (3) and (4), there's a sentence just above the formula "Negative 
numbers include the sign in their length". I think the documentation writer 
accidentally dropped the last part in the formula, "+sign". So it really should 
look like:
ROUND((p+s)/2)+1+sign
where sign equals zero if the number is positive, and sign equals 1 if the 
number is negative.

However, the modified formula still doesn't take into account of exponent. 
vsize(110) is 3 while vsize(1100) is 2, because they're 1.1 x 10^2 and 1.1 x 
10^3, respectively. But applying the (modified) formula would yield the same 
value, ROUND((3+0)/2)+1+0=3 and ROUND((4+0)/2)+1+0=3, respectively.