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Problem 932

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$2025$

For the year $2025$:

$$2025 = (20 + 25)^2$$

Given positive integers $a$ and $b$, the concatenation $ab$ we call a $2025$-number if $ab = (a+b)^2$.

Other examples are $3025$ and $81$.

Note $9801$ is not a $2025$-number because the concatenation of $98$ and $1$ is $981$.

Let $T(n)$ be the sum of all $2025$-numbers with $n$ digits or less. You are given $T(4) = 5131$.

Find $T(16)$.

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$2025$

今年的年份$2025$满足

$$2025 = (20 + 25)^2$$

给定正整数$a$和$b$，如果将其连接后满足等式$ab = (a+b)^2$，则称$ab$为$2025$数。

其它的例子有$3025$和$81$。

注意$9801$不是$2025$数，因为$98$和$1$连接后是$981$。

令$T(n)$为所有不超过$n$位的$2025$数之和。已知$T(4) = 5131$。

求$T(16)$。

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