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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 $k\ge 8$.

mathlove
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    There are no examples of 8 up to $n=10^6$. – Robert Israel Jan 06 '15 at 17:55
  • @RobertIsrael: Thanks. By the way, I've just found [OEIS(A061910)](http://oeis.org/A061910). – mathlove Jan 06 '15 at 18:04
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    The next $7$-sequence starts at $9999$ – Peter Jan 07 '15 at 14:20
  • For $n\le 40\ 000\ 000$, there is no more sequence with length $\ge$ 7. – Peter Jan 07 '15 at 14:36
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    $46045846$-$46045853$ is a sequence of length $8$, so $k\ge 8$. – Peter Jan 07 '15 at 18:01
  • May be you should first find relationship between $S(n)$ and $S(n^2)$. – Tahir Imanov Jan 23 '15 at 16:20
  • In your expression for 7 consecutive squares, you can replace each instance of $(3m-2)^2$ with simply $m^2$. This is because $S[(10^p-1)^2] = 9p$, so if $p = m^2$, then $S[(10^p-1)^2] = 9(m^2) = (3m)^2$. The change doesn't upset the following terms either. – Archr Jan 20 '17 at 22:30
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    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 05:44
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    The first run of 9 consecutive squares starts at 302260461719025^2. – Giovanni Resta Jun 26 '17 at 09:06
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    One thing to consider, is what the properties of a square are, and when sums of digits could do that ( e.g , mod 8, squares are 0,1 or 4. The normal digits of decimal mod 8 are 0,1,2,3,4,5,6,7,0,1 . What combinations of sums of these, are 0,1 or 4 mod 8). –  Jun 29 '17 at 20:02
  • You can have some estimations using probabilities. The idea is the following The problem is so find $p$ successive positive integers such that the digit sum of their squares is a square. It is known that the probability for a big positive integer $i$ to be the square of some other positive integer is $\frac{1}{2\sqrt{i}}$. Let $n$ a very big integer so we know that the number of digits of $n^{2}$ is $\left[ 2\log _{10}n\right] +1$ and they have the average sum $\frac{9}{2}\left[ 2\log _{10}n\right] +4$ ... – Djalal Ounadjela Jul 08 '17 at 18:07
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    ...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 08 '17 at 18:09
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    ...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 08 '17 at 18:09
  • $$S((3m)^2) = 3^2 : 1 \le m \le 7$$ $$S((10^m + 7)^2) = 19 : m\in \mathbb{N}$$ But what I found most astonishing was when doing $1^2, 11^2, 111^2, \ldots$etc I discovered that depending on how many consecutive $1$'s there were (which we will denote as $R$), this means that: $$S(1^2, 11^2, 111^2, \ldots) = R_{1}^2, R_{2}^2, R_{3}^2, \ldots : R < 10$$ $$\implies n^2 = n + 2(n - 1) + \ldots + 2(n - (n - 2)) + 2(n - (n - 1))$$. $$\text{i.e}\ \ \ \ 3^2 = 3 + 2(3 - 1) + 2(3 - 2)$$ $$\ = 1 + 2 + 3 + 2 + 1$$ $$\therefore \ = S(111^2)$$ – George N. Missailidis Aug 13 '17 at 14:52

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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.

Max
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