In One Sequence Or Another (counter-intuitive?)

[Leroy Quet]

[Based upon a post to sci.math] 

I came up with the following result, which must be well known, today.
If we let f and g be certain functions, we can get 2 integer sequences 
for each f (and g), where every positive integer is in one of these 
sequences or the other, but not in both.
(a possible source for sequences...)


Let f(x) be ANY real -> real function which is defined for all x >= 1, is 
continuous, is strictly monotonically increasing, and where floor(f(1)) = 
1.

Let g(x) the inverse of f(x) -- ie. g(x) is subject to the same 
conditions as f(x), ceiling(g(1)) = 1, 
and f(g(x)) = x for all x >= 1.


Then:

EVERY positive integer m is representable as either
(but not as both):


m = floor(f(k)) + k

or  

m = ceiling(g(k)) + k - 1


for some k.

(One k for each integer.
So each m is represented once and only once by either of the above.)


(And if g(k) is not an integer, we can replace the latter with:

m = floor(g(k)) +  k,

obviously.)  
 

(I find the general result counter-intuitive and {although simple and not 
too advanced} amazing.)
 

-

For example, this result is a generalization of that relating to Beatty 
sequences, which can be found at:

http://mathworld.wolfram.com/BeattySequence.html

(A Beatty sequence has f(k) = (y-1)*k for an irrational y {y>1}.)

-

Another example:

f(k) = sqrt(k) *(sqrt(5)+1)/2,
g(k) = k^2 *(3 -sqrt(5))/2.

floor(f(k)) +k : 2, 4, 5, 7, 8, 9,...

floor(g(k)) +k : 1, 3, 6, 10, 14,...

Are either of these in the EIS?

I wonder which sequences can be represented as resulting from a specific 
f or g.

We can not find a function for any positive integer sequence of distinct 
terms, because 1 is automatically in the g-sequence,
and 2 is automatically in the f-sequence.
But WHICH sequences can be represented by SOME f and g?


Thanks,
Leroy Quet