18. There is a sequence of natural numbers, starting from the third, each natural number is the sum of the previous two; it is known that the 3rd natural number is 7, and the 2014th natural number leaves a remainder of 1 when divided by 4. Therefore, the first natural number could be ____ (please list all possibilities).
【Analysis】Consider the remainder of this sequence when divided by 4
Since the 3rd number divided by 4 leaves a remainder of 3,
I. The 1st number divided by 4 leaves a remainder of 1
Then the remainder pattern of this sequence when divided by 4 is:
$(1,2, 3, 1, 0, 1), 1,2, \ldots \ldots$
Cycles with $1, 2, 3, 1, 0, 1$
$2014 \div 6=335 \cdots 4$, the 2014th number divided by 4 leaves a remainder of 1
So the 1st number divided by 4 leaving a remainder of 1 meets the requirement
II. The 1st number divided by 4 leaves a remainder of 2
Then the remainder pattern of this sequence when divided by 4 is:
$(2, 1, 3, 0, 3, 3), 2, 1, \ldots \ldots$
Cycles with $2, 1, 3, 0, 3, 3$
$2014 \div 6=335 \cdots 4$, the 2014th number divided by 4 leaves a remainder of 0
So the 1st number divided by 4 leaving a remainder of 2 does not meet the requirement
III. The 1st number divided by 4 leaves a remainder of 3
Then the remainder pattern of this sequence when divided by 4 is:
$(3, 0, 3, 3, 2, 1), 3, 0, \ldots \ldots$

Cycles with $3, 0, 3, 3, 2, 1$
$2014 \div 6=335 \cdots 4$, the 2014th number divided by 4 leaves a remainder of 3
So the 1st number divided by 4 leaving a remainder of 3 does not meet the requirement
IV. The 1st number divided by 4 leaves a remainder of 0
Then the remainder pattern of this sequence when divided by 4 is:
$(0,3, 3, 2, 1, 3), 0, 3, \ldots \ldots$

Cycles with $0, 3, 3, 2, 1, 3$
$2014 \div 6=335 \cdots 4$, the 2014th number divided by 4 leaves a remainder of 2
So the 1st number divided by 4 leaving a remainder of 0 does not meet the requirement
In summary, the first number divided by 4 should leave a remainder of 1
And the first number should be less than or equal to 7, so it could be $1, 5$