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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 $k\ge 8$.

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    $\begingroup$ There are no examples of 8 up to $n=10^6$. $\endgroup$
    – Robert Israel
    Commented Jan 6, 2015 at 17:55
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    $\begingroup$ $46045846$-$46045853$ is a sequence of length $8$, so $k\ge 8$. $\endgroup$
    – Peter
    Commented Jan 7, 2015 at 18:01
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    $\begingroup$ For $k = 9$ there are no such sequences for $n < 10^{11}$. Also, for $k = 8$ the only one with $n < 10^{11}$ is $46045846 - 46045853$. $\endgroup$
    – ikbuzsak
    Commented Apr 30, 2017 at 5:44
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    $\begingroup$ The first run of 9 consecutive squares starts at 302260461719025^2. $\endgroup$
    – Giovanni Resta
    Commented Jun 26, 2017 at 9:06
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    $\begingroup$ ...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 $p=13,$ $n\approx 3\times 10^{18}$ $\endgroup$
    – Djalal Ounadjela
    Commented Jul 8, 2017 at 18:09

1 Answer 1

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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 $.

PS: http://oeis.org/A061910 lists numbers n for which the sum of digits of n^2 is a square.

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