Eikonal Blog

2012.03.14

Pretty little tables

Filed under: mathematics, number theory, puzzles — Tags: , — sandokan65 @ 14:59

Recently I have seen in an math forum this:

and just a few days later this one, too:

Pretty little tables, aren’t they? How could they be so regular? Can they be generalized somehow?

Answers are: yes, you will see, and yes.

Wonder #1

Let’s first take a look at the first table:

 Table 1 $1 \times 8 + 1 = 9$, $12 \times 8 + 2 = 98$, $123 \times 8 + 3 = 987$, $1,234 \times 8 + 4 = 9,876$, $12,345 \times 8 + 5 = 98,765$, $123,456 \times 8 + 6 = 987,654$, $1,234,567 \times 8 + 7 = 9,876,543$, $12,345,678 \times 8 + 8 = 98,765,432$, $123,456,789 \times 8 + 9 = 987,654,321$,

In order to understand it one has to work not with the specific numbers (digits), but with their abstract representations. For this, we will work with a number system of base $B$ ($\in {\Bbb N}$), which in the orginal tables is $B=10$. Then we can rewrite several first members of the Table 1 as follows:

• $1 \cdot B^0 \times (B-2) + 1 = (B-1) B^0$,
• $(1 \cdot B^1 + 2 B^0) \times (B-2) + 2 = (B-1) B^1 + (B-2)B^0$,
• $(1 \cdot B^2 + 2 B^1 + 3 B^0) \times (B-2) + 3 = (B-1)B^2 + (B-2)B^1 + (B-3)B^0$,
• etc

Ok, we see some regularity here. To proceed further, rewrite the $n^{th}$ row of that table in the form a mathematical equation $y_n :\equiv x_n \cdot \hbox{some number} + \hbox{some other number}$, transforming the first beautifully looking number ($x_n$) into second beautifully looking number $y_n$.

Here the series $\{x_n\}$ is:

• $x_1 = 1_B = 1\times B^0$,
• $x_2 = 12_B = 1\times B^1 + 2 \times B^0$,
• $x_3 = 123_B = 1\times B^2 + 2 \times B^1 + 3 \times B^0$,
• $x_n = 123...n_B = 1\times B^{n-1} + 2 \times B^{n-2} + \cdots + n \times B^0 = \sum_{k=1}^n k B^{n-k}$.
 A side note: Note that $x_n = \sum_{s=0}^{n-1} (n-s) B^s$. It can be explicitly summarized as follows: $x_ n = (n- B \partial_B) \sum_{s=0}^{n-1} B^s = (n-B\partial_B) \frac{B^n-1}{B-1} = \frac{B(B^n-1)-n(B-1)}{(B-1)^2}$. For example, for $B=10$ that formula yields $x_n = \frac{10(10^n-1)-9n}{81}$: $x_1 = 1$, $x_2 = 12$, …, $x_5 = 12345$, etc.

Let’s go back to the main line of discussion.

Now we are interested in the following derivative series $y_n :\equiv x_n \cdot (B-2) + n$. The straightforward manipulation leads to the anticipated result:

$y_ n = B^n + \sum_{s=1}^{n-1} (s+1-n)B^s - n$
$= (B-1)B^{n-1} + B^{n-1} + \sum_{s=1}^{n-2} (s+1-n)B^s - n$
$= (B-1)B^{n-1} + (B-2) B^{n-2} + \sum_{s=1}^{n-3} (s+1-n)B^s - n$

$\cdots$

$= \sum_{r=1}^{m} (B-r)B^{n-r} + B^{n-m} + \sum_{s=1}^{n-m-1} (s+1-n)B^s - n$

$\cdots$

$= \sum_{r=1}^{n-2} (B-r)B^{n-r} + B^{2} + \sum_{s=1}^{1} (s+1-n)B^s - n$
$= \sum_{r=1}^{n-2} (B-r)B^{n-r} + B^{2} + (2-n)B^1 - n$
$= \sum_{r=1}^{n-1} (B-r)B^{n-r} + B^{1} - n$
$= \sum_{r=1}^{n} (B-r)B^{n-r}$.

i.e. $y_n = (B-1)(B-2)...(B-n+2)(B-n+1)(B-n)_B$. The initial pyramid of simple results holds for every base $B$.

Example: for $B=5$ we have $x_n = \frac{5(5^n-1)-4n}{16}$, so $x_1 = \frac{5\cdot 4 - 4 \cdot 1}{16} = 1$, $x_2 = \frac{5\cdot 24 - 4 \cdot 2}{16} = 7_{10} = 12_5$, etc. Then, for example $y_2 = x_2 \cdot 3 + 2 = 7\cdot 3 + 2 = 21 + 2 = 23_{10} = 43_5$.

Wonder #2

Let’s look at the Table 2:

 Table 2 $1 \times 9 + 2 = 11$, $12 \times 9 + 3 = 111$, $123 \times 9 + 4 = 1,111$, $1,234 \times 9 + 5 = 11,111$, $12,345 \times 9 + 6 = 111,111$, $123,456 \times 9 + 7 = 1,111,111$, $1,234,567 \times 9 + 8 = 11,111,111$, $12,345,678 \times 9 + 9 = 111,111,111$, $123,456,789 \times 9 + 10 = 1,111,111,111$,

Here the first (i.e. the independent) variable $x_n$ is the exactly same as the one used for Table 1. The second (i.e. the dependent) variable $z_n$ is new one, determined by defining equation:

$x_n = \sum_{s=0}^{n-1}(n-s)B^s$,

Then, using steps similar to these used in analysis of the Table 1, we get:

$z_n :\equiv x_b \cdot (B-1) + (n+1) =$
$= \sum_{s=0}^{n-1} (n-s) B^{s+1} - \sum_{s=0}^{n-1} (n-s) B^s + (n+1) =$
$= \sum_{s=1}^{n} (n+1-s) B^{s} - \sum_{s=0}^{n-1} (n-s) B^s + (n+1) =$
$= B^n + \sum_{s=1}^{n-1} (n+1-s - n +s) B^{s} - n + (n+1) =$
$= B^n + \sum_{s=1}^{n-1} B^{s} +1 =$
$= \sum_{s=0}^{n} B^{s} =$
$= 1\cdots 1_B$,

where there are $(n+1)$ copies of digit 1.

Nice. Easy.

Wonder #3

 Table 3 $9 \times 9 + 7 = 88$, $98 \times 9 + 6 = 888$, $987 \times 9 + 5 = 8,888$, $9,876 \times 9 + 4 = 88,888$, $98,765 \times 9 + 3 = 888,888$, $987,654 \times 9 + 2 = 8,888,888$, $9,876,543 \times 9 + 1 = 88,888,888$, $98,765,432 \times 9 + 0 = 888,888,888$,

That series of relations has form $v_n = u_n \cdot 9 + (8-n)$ where the dependent variable is
$v_n = \underbrace{8\cdots8}_{n+1}$ in the normal decimal system ($B=10$).

For general basis $B$ this generalizes to the:
$v_n = u_n \cdot (B-1) + (B - n -2)$. Here the independent variable $u_n$ is

$u_n = \sum_{k=1}^n (B-k) B^{n-k} =$
$= \sum_{s=0}^{n-1} (B- n -s) B^s =$
$= (B - n + B \partial_B) \sum_{s=0}^{n-1}B^s =$
$= (B - n + B \partial_B)\frac{B^n-1}{B-1} =$
$= \frac{B(B-2)(B^n-1)+n (B-1)}{(B-1)^2}$.

Now

$v_ n :\equiv u_n (B-1) + (B-n-2) =$
$= (B-1) B^n - \sum_{s=1}^{n-1} B^s - 2 =$
$= (B-2) B^n + (B-1) B^{n-1} - \sum_{s=1}^{n-2} B^s - 2 =$
$= (B-2) B^n + (B-2) B^{n-1} + (B-1)B^{n-2} - \sum_{s=1}^{n-3} B^s - 2 =$

$\cdots$

$= (B-2) B^n + \cdots + (B-2) B^{n-m+1} + (B-1)B^{n-m} - \sum_{s=1}^{n-m-1} B^s - 2 =$

$\cdots$

$= (B-2) B^n + \cdots + (B-2) B^{3} + (B-1)B^{2} - B^1 - 2 =$
$= (B-2) B^n + \cdots + (B-2) B^{3} + (B-2)B^{2} + B^2 - B^1 - 2 =$
$= (B-2) B^n + \cdots + (B-2) B^{2} + (B-1)B^{1} - 2 =$
$= (B-2) B^n + \cdots + (B-2) B^{1} + B - 2 =$
$= \sum_{r=0}^{n} (B-2) B^r$.

For $B=10$ that covers all examples in the Table 3.

Note: Even more, we can add two more members to it, corresponding to $n=9$ and $n=10$:

• $u_9 = 987,654,321$ corresponds to $v_9 = x_9 \cdot 9 + (-1) = 8,888,888,888$,
• $u_{10} = 9,876,543,210$ corresponds to $v_{10} = x_{10} \cdot 9 + (-2) = 88,888,888,888$.

Also, the $u_0=0$ provides one more line, which is prepended to this more complete table 3:

 Table 3* $0 \times 9 + 8 = 8$, $9 \times 9 + 7 = 88$, $98 \times 9 + 6 = 888$, $987 \times 9 + 5 = 8,888$, $9,876 \times 9 + 4 = 88,888$, $98,765 \times 9 + 3 = 888,888$, $987,654 \times 9 + 2 = 8,888,888$, $9,876,543 \times 9 + 1 = 88,888,888$, $98,765,432 \times 9 + 0 = 888,888,888$, $987,654,321 \times 9 - 1 = 8,888,888,888$, $9,876,543,210 \times 9 - 2 = 88,888,888,888$.

Wonder #4

 Table 4 $1^2 = 1$, $11^2 = 121$, $111^2 = 12,321$, $1,111^2 = 1,234,321$, $11,111^2 = 123,454,321$, $111,111^2 = 1,2345,654,321$, $1,111,111^2 = 1,234,567,654,321$, $11,111,111^2 = 123,456,787,654,321$, $111,111,111^2 = 12,345,678,987,654,321$.

Here the independent variable is

$p_n :\equiv 1\cdot B^{n} + 1\cdot B^{n-1} + \cdots + 1\cdot B^0 = \sum_{i=0}^n B^i = \frac{B^{n+1}-1}{B-1}$.

The resulting variable is

$r_n :\equiv p_n^2 = \sum_{i=0}^n \sum_{j=0}^n B^{i+j} =$
$= \sum_{k=0}^{2n} (\sum_{i=0}^n \sum_{j=0}^n \delta_{i+j,k}) B^k$.

Now the sum in brackets can be transformed as follows:

$\sum_{i=0}^n \sum_{j=0}^n \delta_{i+j,k} = \sum_{i=0}^n \Theta(0 \le k-i \le n) =$
$= \sum_{i=0}^n \Theta(k-n \le i \le k) = \sum_{i=\hbox{max}(0,k-n)}^{\hbox{min}(n,k)} 1 =$
$= \hbox{min}(n,k) - \hbox{max}(0,k-n) + 1 = *$

which has three posible simplifications:

• $* = k - 0 + 1 = k + 1$ for $k < n$,
• $* = n + 1$ for $k = n$,
• $* = n - (k-n) + 1 = 2n - k + 1$ for $k > n$.

So, now we can write the final form for $r_n$ as following:

$r_n = \sum_{k=0}^{n-1} (k+1) B^k + (n+1) B^n + \sum_{k=n+1}^{2n} (2n-k+1) B^k =$
$= 1\cdot B^0 + 2\cdot B^1 + 3\cdot B^2 + \cdots + (n+1) B^n + \cdots 2\cdot B^{2n-1} + 1\cdot B^{2n} =$
$= 123\cdots(n+1)\cdots321_B$.

That is it.

Note that one can get some regularities for degrees higher than 2. For example, for degree 3 one has:

$1^3 = 1$
$11^3 = 1,331$
$111^3 = 135,531$

$1,111^3 = 13,577,531$
$11,111^3 = 1,357,997,531$
$111,111^3 = 135,79b,b97,531$
$1,111,111^3 = 13,579,bdd,b97,531$

$\cdots$

${\underbrace{1\cdots1}_{[B'/2]+1}}^3 = 1357\cdots B'B'\cdots 7531$.

up to the last member (of that series) where the two central digits are the highest single-digit number $B'$ allowed in the number system of base $B$ (i.e. $B'=B-2$ if $B$ is odd, and $B'=B-1$ if $B$ is even).

Note: Here $a = 10_{10}$, $b = 11_{10}$, $c = 12_{10}$, $d = 13_{10}$, $e = 14_{10}$, $f = 15_{10}$, $g = 16_{10}$, $h = 17_{10}$, $i = 18_{10}$, etc.

For the degree 4 the similar pyramid/table is:

$1^4 = 1$
$11^4 = 14,641$
$111^4 = 1,48a,841$

$1,111^4 = 1,48a,cec,841$

$\cdots$

Here one can also work some more (and I did not do that work yet) to establish which is the last member of that table (as a function of the base $B$), and what is the innermost digit in that last member.
This could be a homework for you. đź™‚

For the degree 5:

$1^5 = 1$
$11^5 = 15885$
$111^5 = 15ciic51$

$\cdots$

Similar here: More simple math wonders – https://eikonal.wordpress.com/2012/03/14/more-simple-math-wonders/ | Mental calculation of cube root of a six-digit number – https://eikonal.wordpress.com/2010/01/14/mental-calculation-of-cube-root-of-a-two-digit-number/ | Squares with just two different decimal digits – https://eikonal.wordpress.com/2010/01/05/squares-with-just-two-different-decimal-digits/ | Number theory finite concidental sums – https://eikonal.wordpress.com/2010/01/05/number-theory-finite-considental-sums/

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