Numbers $n$ such that the digit sum of $n^2$ is a square

Let $S(n)$ be the digit sum of $n\in\mathbb N$ in the decimal system. About a month ago, a friend of mine taught me the following:

$$S\left(9\color{red}{^2}\right)=S(81)=8+1=3\color{red}{^2}$$ $$S\left(10\color{red}{^2}\right)=S(100)=1+0+0=1\color{red}{^2}$$ $$S\left(11\color{red}{^2}\right)=S(121)=1+2+1=2\color{red}{^2}$$ $$S\left(12\color{red}{^2}\right)=S(144)=1+4+4=3\color{red}{^2}$$ $$S\left(13\color{red}{^2}\right)=S(169)=1+6+9=4\color{red}{^2}$$ $$S\left(14\color{red}{^2}\right)=S(196)=1+9+6=4\color{red}{^2}$$ $$S\left(15\color{red}{^2}\right)=S(225)=2+2+5=3\color{red}{^2}$$

Then, I've got the following:

For every $m\in\mathbb N$, each of the following $7$ numbers is a square. $$S\left(\left(10^{(3m-2)^2}-1\right)^2\right),S\left(\left(10^{(3m-2)^2}\right)^2\right),\cdots,S\left(\left(10^{(3m-2)^2}+5\right)^2\right)$$

However, I'm facing difficulty in finding such $8$ consecutive numbers. So, here is my question:

Question : What is the max of $k\in\mathbb N$ such that there exists at least one $n$ which satisfies the following condition?

Condition : Each of the following $k$ numbers is a square. $$S\left((n+1)^2\right),S\left((n+2)^2\right),\cdots,S\left((n+k-1)^2\right),S\left((n+k)^2\right)$$

Note that we have $k\ge 7$. Can anyone help?

Added : A user Peter found the following example of $k=8$ : $$S\left(46045846^2\right)=8^2,S\left(46045847^2\right)=7^2,\cdots,S\left(46045852^2\right)=7^2,S\left(46045853^2\right)=8^2$$ Hence, we have $k\ge 8$.

• There are no examples of 8 up to $n=10^6$. – Robert Israel Jan 6 '15 at 17:55
• $46045846$-$46045853$ is a sequence of length $8$, so $k\ge 8$. – Peter Jan 7 '15 at 18:01
• For $k = 9$ there are no such sequences for $n < 10^{11}$. Also, for $k = 8$ the only one with $n < 10^{11}$ is $46045846 - 46045853$. – ikbuzsak Apr 30 '17 at 5:44
• ...and so the probability for this sum to be a square is $\frac{1}{\sqrt{18\left[ 2\log _{10}n\right] +16}}$ and then the probability to have the next $p-1$ integers ($n+1,n+2,\ldots ,n+p-1$) sharing the same property is close to $\left( \frac{1}{\sqrt{18\left[ 2\log _{10}n\right] +16}}\right) ^{p}$ so a good candidate is an integer $n$ for which $\left( \frac{1}{\sqrt{18\left[ 2\log _{10}n\right] +16}}\right) ^{p}n\approx 1$. We can solve for $p$ and obtain $p\approx \frac{2\ln n}{\ln \left( 18\left[ 2\log _{10}n\right] +16\right) }\approx \frac{2\ln n}{\ln \left( \ln n\right) }$ – Djalal Ounadjela Jul 8 '17 at 18:09
• ...so $p$ is unbounded. for $p=8,$ $n\approx 3\times 10^{10}$ for $p=9,$ $n\approx 10^{12}$ for $p=10,$ $n\approx 3\times 10^{13}$ for $p=11,$ $n\approx 10^{15}$ for $p=12,$ $n\approx 5\times 10^{16}$ for $p=13,$ $n\approx 3\times 10^{18}$ – Djalal Ounadjela Jul 8 '17 at 18:09

As Djalal Ounadjela outlined in the comments, there is probably no such maximal k, we can expect to find an n for any k at an order of magnitude n ~ 10^m with approximately $m/\log_{10}(m) \sim k$.