由题意可知 \({a_{n + 2}} = {a_{n + 1}} + {a_n}\) ，
所以 \({a_{n + 1}} \cdot {a_{n + 2}} = {a_{n + 1}} \cdot \left( {{a_{n + 1}} + {a_n}} \right)\) ，即 \({a_{n + 1}} \cdot {a_{n + 2}} = a_{n + 1}^2 + {a_{n + 1}} \cdot {a_n}\)
所以 \({a_{2048}} \cdot {a_{2049}} = a_{2048}^2 + {a_{2048}} \cdot {a_{2047}}\) ，
\({a_{2047}} \cdot {a_{2048}} = a_{2047}^2 + {a_{2047}} \cdot {a_{2046}}\) ，
……
\({a_2} \cdot {a_3} = a_2^2 + {a_2}\cdot{a_1}\) ，
所以 \({a_{2048}} \cdot {a_{2049}} = a_{2048}^2 + a_{2047}^2 + \ldots + a_2^2 + {a_2}\cdot{a_1}\) ，
又 \({a_2} = {a_1}\)
所以 \({a_{2048}} \cdot {a_{2049}} = a_{2048}^2 + a_{2047}^2 + \ldots + a_2^2 + a_1^2\)
∴ \(\frac{{a_1^2 + a_2^2 + a_3^2 + \cdots + a_{2048}^2}}{{{a_{2048}}}} = {a_{2049}}\) ．
所以 \(\frac{{a_1^2 + a_2^2 + a_3^2 + \cdots + a_{2048}^2}}{{{a_{2048}}}}\) 是数列中的第\(2049\)项.
故答案为：\(2049\) ．
