Problem 1110

Considering the last digit, if it is 1, then the number of corresponding natural numbers $N$ is $a_{n-1}$; if it is 3, then the number of corresponding natural numbers $N$ is $a_{n-3}$; if it is 4, then the number of corresponding natural numbers $N$ is $a_{n-4}$, thus $a_{n}=a_{n-4}+a_{n-3}+a_{n-1}$, and $a_{1}=1, a_{2}=1, a_{3}=2, a_{4}=4$.
Calculations yield $a_{2}=1=1^{2}, a_{4}=4=2^{2}, a_{6}=9=3^{2}, a_{8}=25=5^{2}, a_{10}=64=8^{2}, a_{12}=$ $169=13^{2}, a_{14}=441=21^{2} ; a_{1}=1, a_{3}=2=1 \times 2, a_{5}=6=2 \times 3, a_{7}=15=$ $3 \times 5, a_{9}=40=5 \times 8, a_{11}=104=8 \times 13, a_{13}=273=13 \times 21$,
Let the sequence $f_{n}$ satisfy $f_{0}=f_{1}=1, f_{n}=f_{n-2}+f_{n-1}$, then $f_{2}=2, f_{3}=3, f_{4}=5, f_{5}=8$, $f_{6}=13, f_{7}=21$. Conjecture $a_{2 n-1}=f_{n-1} f_{n}, a_{2 n}=f_{n}^{2}$.

When $n=1,2$, the conclusion holds; assuming for $n=k-1, k$, the conclusion holds, i.e., $a_{2 k-3}=f_{k-2} f_{k-1}, a_{2 k-2}=f_{k-1}^{2}, a_{2 k-1}=f_{k-1} f_{k}, a_{2 k}=f_{k}^{2}$,
then $a_{2 k+1}=a_{2 k-3}+a_{2 k-2}+a_{2 k}=f_{k-2} f_{k-1}+f_{k-1}^{2}+f_{k}^{2}=f_{k-1}\left(f_{k-2}+f_{k-1}\right)+f_{k}^{2}$ $=f_{k-1} f_{k}+f_{k}^{2}=f_{k}\left(f_{k-1}+f_{k}\right)=f_{k} f_{k+1} ;$
$a_{2 k+2}=a_{2 k-2}+a_{2 k-1}+a_{2 k+1}=f_{k-1}^{2}+f_{k-1} f_{k}+f_{k} f_{k+1}=f_{k-1}\left(f_{k-1}+f_{k}\right)+f_{k} f_{k+1}$ $=f_{k-1} f_{k+1}+f_{k} f_{k+1}=f_{k+1}\left(f_{k-1}+f_{k}\right)=f_{k+1}^{2}$, thus when $n=k+1$ the conclusion also holds. By the principle of mathematical induction, for all natural numbers $n$, $a_{2 n-1}=f_{n-1} f_{n}, a_{2 n}=f_{n}^{2}$. Therefore, $a_{2 n}$ is a perfect square.