1, 11, 111, 1111, ... squared

[Function]

I don't know how to ask this question, so I'll just give something; my question: is there a formula which can put this 'phenomenon' in a general form?

 

[math]1^2 = 1[/math]

[math]11^2 = 121 \rightarrow 1+2+1=4=2^2[/math]

[math]111^2=12321\rightarrow 1+2+3+2+1=9=3^2[/math]

[math]1,111^2=1234321\rightarrow 1+2+3+4+3+2+1=16=4^2[/math]

[math]\cdots[/math]

 

How can this be put in a general, mathematically more plausible expression?

 

I think that this is a 'crucial' part of it: (I found it about a few minutes ago; sorry if it's wrong, haven't seen summations in school yet..) (To be honest, I'm pretty happy with the result that I found this ^^ )

 

[math]\left[\sum_{i=0}^{n}{10^i}\right]^2=\sum_{i=0}^{n}{\left[i\cdot 10^{i-1}\right]}+\sum_{i=0}^{n}{\left[\left(i+1\right)\cdot 10^{2n-i}\right]}[/math]

 

And I think the sum of the numbers of the square of this is:

 

[math]2\left[\sum_{i=0}^{n}{i}\right]+(n+1)[/math]

 

So, my conclusion would be:

 

[math]\forall n\in\mathbb{N}: 2\left[\sum_{i=0}^{n}{i}\right]+(n+1)=n^2[/math]

 

So, actually, the whole first step wasn't necessary after all.. But, I personally think it's a nice formule to display the series 1, 11, 111, 1111, 11111, ...

 

I'm pretty sure the last thing is correct; my main question (sorry that I have changed it) is now: is this formula correct?

 

[math]\left[\sum_{i=0}^{n}{10^i}\right]^2=\sum_{i=0}^{n}{\left[i\cdot 10^{i-1}\right]}+\sum_{i=0}^{n}{\left[\left(i+1\right)\cdot 10^{2n-i}\right]}[/math]

 

[math]\left[\sum_{i=0}^{n}{10^i}\right]^2=\sum_{i=0}^{n}{\left[10^{2n-i}\left(i+1\right)+i\cdot 10^{i-1}\right]}[/math]

Edited November 22, 2013 by Function

[Unity+]

Well, there can be a simple formula that can be used(depending what you mean by general).

 

[math]n^{2}[/math]

 

Where n is the amount of ones in the value. Do you want an equation where you can enter the actual value containing the ones so it puts it in the square form?

[Function]

Well, there can be a simple formula that can be used(depending what you mean by general).

 

[math]n^{2}[/math]

 

Where n is the amount of ones in the value. Do you want an equation where you can enter the actual value containing the ones so it puts it in the square form?

 

Lol.. Well, my main question now is if my last given formula is right.. (I edited my post a 'bit')?

Edited November 22, 2013 by Function

[studiot]

Did you find this in Trachtenberg?

[Function]

Did you find this in Trachtenberg?

 

(Note that I edited the formula - it contained a small mistake)

 

...Trachten-what?

 

I don't know who/what that is, so I'd say no.. I just wanted to make a general formula for squaring terms of the series 1, 11, 111, 1111, 11111, ... in function of n, with n = (the amount of ones in the value - 1).

...So I did.

 

However I think this formula I found, i.e. [math]\left[\sum_{i=0}^{n}{10^i}\right]^2=\sum_{i=0}^{n}{\left[10^{2n-i}\left(i+1\right)+i\cdot 10^{i-1}\right]}[/math] was probably already known (or perhaps another formula if this one's incorrect), I'd like to know if it's correct

Matter of testing my mathematical capacities (read: insight)

Edited November 22, 2013 by Function

[John Cuthber]

Someone should to check my arithmetic but

1111111111 ^2

=

1,234,567,900,987,654,321

1 2 3 4, 5 6 7 9 0 0, 9 8 7, 6 5 4 ,3 2 1

1+2+3+4+5+6+7+9+0+0+9+8+7+6+5+4+3+2+1

=82

[Function]

Someone should to check my arithmetic but

1111111111 ^2

=

1,234,567,900,987,654,321

1 2 3 4, 5 6 7 9 0 0, 9 8 7, 6 5 4 ,3 2 1

1+2+3+4+5+6+7+9+0+0+9+8+7+6+5+4+3+2+1

=82

 

Now that's unsettling.

 

Nevertheless; is 'my' formula in #5 correct?

Following that formula, and keeping in mind that my calculator (TI-84 Plus) may have rounding inaccuracies, it gives for both left part and right part of the formula in message #5:

1.234567901E20

Edited November 22, 2013 by Function

[studiot]

Professor Trachtenberg was imprisoned by the Nazis and came up with schemes of mental arithmetic to keep him sane.

 

It is known as the Trachtenberg system (you should) look it up.

 

The instruction to multiply any number by 11 for instance is

 

Write down the last digit.

Add successive pairs of neighbours to successively obtain the next digit to the left and so on

write doen the last digit.

 

So 11 x 11

 

write down last digit =1

 

1

 

next leftmost digit is

1+1 = 2

 

21

 

write down rightmost digit =1

 

11 x 11 = 121

 

you can see the pattern extended to 111 etc in your listing.

[Function]

Seems like the formula I have found is correct - well, according to Wolfram Alpha; for both left and right part, it gives the alternative formula:

 

[math]\frac{1}{81}\left(10^{n-1}-1\right)^2[/math]

 

(I secretly wished this wasn't known yet However, I'm glad 'my' formula is right.)

[md65536]

Now that's unsettling.

Don't worry too much about it, because

1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10 + 9 + 8 + 7 + 6 + 5 + 4 + 3 + 2 + 1

= 100 = 10^2

 

It's easy to see why... if you just shift everything past the 10 over, ie reordered to...

(1+9) + (2+8) + ...

 

 

I haven't followed all the math in the thread, but there are 2 things here, the squares of numbers with all 1s, and the sequences that add up to squares.

If you imagine multiplying eg. 11111 "the long way", you get

 11111
x11111
=
     11111
    11111
   11111
  11111
+11111
=123454321

And that might give you a clue about why it works, but if you have more than 9 rows you're going to get digits carrying and it unsettlingly stops working so easily.

 

 

 

Now as to why the digits add up to squares, what you're doing is...

Suppose you call the sequences S_n, where

S₁ = 1 = 1^2

S₂ = 1 + 2 + 1 = 2^2

S₃ = 1 + 2 + 3 + 2 + 1 = 3^2

 

Then S_n+1 = S_n + n + n+1 ---- You're adding 2 numbers into the middle of the previous sequence.

 

I'm sure there's a better way, but one way to see that this is always a square is to think of it geometrically. Imagine a square grid of n by n blocks. Then add a row of n blocks, and add a column of n+1 blocks, and you end up with a square again, with sides 1 larger than the previous square.

 

So, while the squares of 1111111111111111... doesn't keep working, the "ascending and then descending" sequences keep adding up to squares.

Edited November 22, 2013 by md65536

[John Cuthber]

I'm fairly sure that for any given number base, n, it will fail for repunits longer than n.

So, for most commonly used number systems, I can show that it fails.

 

It's like the "observation" that 11^n gives you Pascal's triangle.

[Function]

Can someone proove btw that every symmetrical number (although, that's what I suspect) is dividable by 11?

 

(1881, 3649229463, 635536, ...)

[John Cuthber]

Any number where the sum of the odd digits is the same as the sum of the even digits, is divisible by 11 so, yes.

 

http://www.wikihow.com/Check-Divisibility-of-11

 

But I don't think that citing a wikihow page counts as mathematical proof.

[Function]

Very well, then. Thank you for your help so far!

 

Now, I've found a 'less vague' formula (n is defined more clear) for squaring the series 1, 11, 111, 1 111, 11 111, 111 111, ...

 

[math]\left[\sum_{i=0}^{n-1}{\left(10^i\right)}\right]^2 = \sum_{i=0}^{n-1}{\left(i\cdot 10^{i-1}\right)}+\sum_{i=0}^{n}{\left[\left(n-1\right)\cdot 10^{n-1+i}\right]}[/math] with [math]n[/math] the number of times "1" appears in the term on place "n" of the series, represented by [math]\sum_{i=0}^{n-1}{\left(10^i\right)}[/math].

 

My final question: (how) can this be proven? (Or is this something which can be accepted solely by assumptions and axioma's?)

Edited November 23, 2013 by Function

[John Cuthber]

It might be easier to work in binary where they are all of the form 2^n -1

The base 10 repunits are of the form (10^n -1) /9

[Function]

It might be easier to work in binary where they are all of the form 2^n -1

The base 10 repunits are of the form (10^n -1) /9

 

I'm afraid I don't understand exactly what you're saying..

[Olinguito]

So, my conclusion would be:

 

[math]\forall n\in\mathbb{N}: 2\left[\sum_{i=0}^{n}{i}\right]+(n+1)=n^2[/math]

[latex]2\left[\sum_{i=0}^{n}{i}\right]+(n+1)\ =\ 2\left[\frac{n(n+1)}2\right]+(n+1)\ =\ (n+1)^2[/latex]

 

Also [latex]\left(\sum_{i=0}^n10^i\right)^2\ =\ \sum_{i=0}^n10^{2i}+2\sum_{0\leqslant i<j\leqslant n}^n10^{i+j}[/latex] using the formula

 

[latex]\left(a_n+a_{n-1}+\cdots+a_1+a_0\right)^2\ =\ a_n^2+\cdots+a_0^2+2\sum_{0\leqslant i<j\leqslant n}^na_ia_j[/latex]

Given [latex]1\leqslant r\leqslant n[/latex], how many pairs [latex]i,j[/latex] are there with [latex]0\leqslant i<j\leqslant n[/latex] such that [latex]i+j=r[/latex]?

 

Suppose [latex]r=2k[/latex] is even. Then [latex]r=0+2k=1+(2k-1)=\cdots=(k-1)+(k+1)[/latex] so there are [latex]k[/latex] such [latex]i,j[/latex] pairs. Thus the coefficent of [latex]10^r[/latex] is [latex]2k+1=r+1[/latex].

 

Suppose [latex]r=2k-1[/latex] is odd. Then [latex]r=0+(2k-1)=1+(2k-2)=\cdots=(k-1)+k[/latex] so again there are [latex]k[/latex] such [latex]i,j[/latex] pairs. Then the coefficent of [latex]10^r[/latex] is [latex]2k=r+1[/latex] again.

 

For [latex]n+1\leqslant r\leqslant2n-1[/latex], we let [latex]s=r-n[/latex]; then [latex]r=s+n=(s+1)+(n-1)=\cdots=(s+k)+(n-k)[/latex] giving [latex]k+1[/latex] pairs. Since [latex]s-k<n-k[/latex] we have [latex]n-s>2k[/latex]; therefore [latex]n-s=2k+1[/latex] or [latex]n-s=2k+2[/latex]. If [latex]r[/latex] is even, then so is [latex]n-s=r-2s[/latex]; then [latex]n-s=2k+2[/latex] and the coefficient of [latex]10^r[/latex] is [latex]2(k+1)+1=n-s+1=2n-r+1[/latex]. If [latex]r[/latex] is odd, then so is [latex]n-s[/latex]; so [latex]n-s=2k+1[/latex] and the coefficient of [latex]10^r[/latex] is [latex]2(k+1)=n-s+1=2n-r+1[/latex] again.

 

Hence [latex]\left(\sum_{i=0}^n10^i\right)^2\ =\ 10^{2n}+1+\sum_{r=1}^n(r+1)10^r+\sum_{r=n+1}^{2n-1}(2n-r+1)10^r\ =\ \sum_{r=0}^n(r+1)10^r+\sum_{r=n+1}^{2n}(2n-r+1)10^r[/latex].

Edited November 24, 2013 by Olinguito

[John]

Can someone proove btw that every symmetrical number (although, that's what I suspect) is dividable by 11?

 

(1881, 3649229463, 635536, ...)

This is true if the number has an even number of digits. For odd numbers of digits, it's not necessarily the case, e.g. in the case of palindromic primes.

[John Cuthber]

This is true if the number has an even number of digits. For odd numbers of digits, it's not necessarily the case, e.g. in the case of palindromic primes.

Good point, I had assumed an even number of digits.

 

 

Function,

Say n=4

10^n = 10000

10^n -1 = 9999

(10^n-1)/9 =1111

 

It's just another way of expressing a string of 1s in terms of the number of 1s.

[Xzyt_Now]

8 + 2 = 10

1 + 0 = 1

1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 = 10

1 + 0 = 1

So the pattern does continue.

[Function]

8 + 2 = 10

 

1 + 0 = 1

 

1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 = 10

 

1 + 0 = 1

 

So the pattern does continue.

 

Yes, well... I already believed that in primary school..