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Let me get to the point. We see that 121 is a palindrome, also 12321 is a palindrome. If we were to add up the digits in each palindrome, for example, 121 we get 4. If we square root four, the root is the middle number of 121. This pattern goes on from what I worked on two years ago. Another example: 1234321 <= Palindrome 1+2+3+4+3+2+1 = 16 <= Palindrome's Digits Sum square root of 16 = 4 Also the number 11 is excluded from this pattern for obvious reason, is there any logic behind this exclusion? What is the real mathematics behind this pattern? EDIT: The palindromes are digits ranging from one to n and the same n to one. (1 + n1 + n2 + n + n2 + n1 + 1) 121 12321 1234321 123454321 This is what is meant by palindromes. EDIT: I need to clarify based on the comments. When I heard about palindromes and ran into this pattern I associated these digit sequences as palindromes. These numbers are symmetrical in terms that there is a middle digit and the digits left and right are descending being a digit till one is reached( 1, 2, 3, 2, 1). Sorry if I have mistaken these numbers as palindromes.

calculatormaster20
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    What pattern? $131$ is a palindrome and the sum of the digits is $5$. If you see a general pattern (that is, something more than a couple of isolated examples) please edit your post to include it. – lulu Oct 21 '20 at 19:34
  • @lulu I started to make the same comment that you did, until I saw the pattern in the OP's examples - which means that the answer that has been given is relevant. – user2661923 Oct 21 '20 at 19:37
  • @Gae.S. Probably the OP only looks for middle numbers of type $2^n$ – Anindya Prithvi Oct 21 '20 at 19:38
  • Not all palindromes are of the form $123.....n.....321$. So sum of those digits we get $\frac {n(n+1)}2 + \frac {(n-1)n}2 = n^2$ so .... yeah it works but take $39876167893$ where the middle digit is $1$ and the sum of the digits is...not. – fleablood Oct 21 '20 at 19:39
  • Ok, But in this special case you have simple closed formulas for the sum of consecutive digits. You are just pointing out that $\frac {n(n+1)}2+\frac {n(n-1)}2=n^2$. – lulu Oct 21 '20 at 19:42
  • "This is what is meant by palindromes." No, that is certainly NOT what is meant by palindromes. A palindrome is a number whose digits read the same from left to right as right to left. There is no condition that the numbers are ascending consecutive numbers starting from one. – fleablood Oct 21 '20 at 19:56

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Sum from $1$ to $n$ is given by

$$1 + 2 + 3 + \cdots + n = \frac{n(n+1)}{2}.$$

Sum from $1$ to $n$ and then $n$ to $1$ is

$$ 1 + 2 + \cdots + n + (n-1) + \cdots + 1 = \frac{n(n+1)}{2} + \frac{n(n-1)}{2} = \frac{n^2 + n + n^2 - n}{2} = \frac{2n^2}{2} = n^2.$$

Take the square root and you get $n$, your middle number.

Reference: Proof $1+2+3+4+\cdots+n = \frac{n\times(n+1)}2$

User203940
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  • There is good conversation going on in the comments of the question of how to actually interpret the question, I figured I'd answer the cases that the question provided. – User203940 Oct 21 '20 at 19:40
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You palindromes cited are all of the form $123.....(n-1)n(n-1)....321$.

If we add up the digits we then

$\color{red}1 + \color{green}2 + \color{blue}3 + ......... + \color{purple}{(n-1)} + $

$\color{orange}n +$

$\color{red}{(n-1)}+\color{green}{(n-2)} + \color{blue}{(n-3)} +..... + \color{purple}1$ and if we add them up in order of position (indicated by color) we get

$= [\color{red}{1+(n-1)} ]+ [\color{green}{2+(n-2)}] + [\color{blue}{2+(n-3)}] +..... + [\color{purple}{(n-1) + 1} ]+ \color{orange}n =$

$\underbrace{\underbrace{\color{red}n + \color{green}n + \color{blue}n + ......... + \color{purple}n}_{n-1\text{ terms}} +\color{orange}n}_{n\text{ terms}}=$

$n\times n = n^2$.

This should remind you of the prove that $1 + 2 + 3+ ....... + n = \frac {n(n+1)}2$.

So $(1 + 2 + 3 +....... + n)+(n-1+......+3+2+1) =\frac {n(n+1)}2 + \frac {(n-1)n}2 = n^2$.

This will only work for "ascending" palindromic numbers and not all palindromes. For example $979$ is a palindrome but $7 \ne \sqrt{9+7+9}$

BTW $123.....n.....321= {\underbrace{1111....1}_{n\text{ ones}}}^2$.

fleablood
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Saw this as a link suggested to another (more recent) question and thought I'd give you a belated graphical intuition. Let's say why, when adding digits in $123454321$ you get a square of $5$. It is because you are just counting the $5^2=25$ fields in the following square:

$$\begin{array}{ccccc}1&2&3&4&5\\2&3&4&5&4\\3&4&5&4&3\\4&5&4&3&2\\5&4&3&2&1\end{array}$$

(going from top left to bottom right you encounter "small diagonals" of length $1, 2, 3, 4, 5, 4, 3, 2, 1$, respective).

Why does $11$ not follow the pattern? Because it is not properly formed. You have to have one middle number, not two. If you have just one, this would make for the number $1$ (not $11$) and it does follow the pattern.