# Types of Numbers, Part III

__PALINDROMIC NUMBERS__

A palindrome is a number that reads the same backwards as forwards, such as 121, 363, 3883, 37973, to mention a few.

Palindromic numbers can be prime, composite, odd, even, square, cube, and so on. The regular numbers, such as the examples just mentioned, are infinite in quantity. The obvious squares are the squares of 1, 11, 111, 1111, etc. which are 1, 121, 12321, and 123432. There are of course others that do not follow the simplistic pattern of the numbers consisting totally of 1’s such as the square of 22 which is 484, 264 which is 69696, or 2285 which is 5221225.

The number 26 is the smallest non palindromic number whose square happens to be palindromic, 676. Can you find a palindromic square number with an even number of digits? DON’T LOOK (836^2 = 698896 There is no smaller such number.)

Some palindromic cubes derive from the cubes of 7, 11, 101, 111, 1001, 10001, 10101, etc., which are themselves palindromic.

There are an infinite number of palindromic primes such as 101, 131, 151, 181, 313, 353, 727, 757, 787, 797, 919, 929, 79997, 91019, 93139, 93739……………….7272727, 3535353, and so on into the night.

If you have some spare time: Pick a number with any number of digits. Reverse the digits and add to the original number. Continue this process until the sum is palindromic, which might require many steps. It has been said that every number will eventually produce a palindrome. For example, the number 137 leads to the

palindrome 868 and the number 89 leads to the palindrome 8,813,200,023,188.

Here are a couple of others:

……………..1284…………….68

……………+4821…………..+86

……………..6105…………..154

……………+5016…………+451

……………11121…………..605

………….+12111…………+506

……………23232…………1111

You might like to seek out other palindromic primes or better still, how many palindromic primes can you find that are in arithmetic progression such as 13931, 14741, 15551 and 16361 with a common difference of 810?

Have you ever heard of a Pandigital Prime? A Pandigital Prime is one that contains all 10 digits 0 through 9. The number 1023456987896543201 is one. 1001^2 comes close at 12345678987654321.

There is a chapter on palindromes in Martin Gardner’s book, Mathematical Circus, published by The Mathematical Association of America, 1992, pages 242 – 252.

__PANDIGITAL NUMBERS__

A pandigital number is one that contains each of the digits from zero to nine exactly once with the leading digit never being a zero. The smallest pandigital number is 1,023,456,789. By definition, all pandigital numbers are divisible by nine.

__PARASITE NUMBERS__

A parasite number is one that retains the exact same digits when multiplied or divided by another number. For example 102,564 x 4 = 410,256 arrived at by merely moving the last 4 to the beginning of the remaining 5 digits. Similarly, 179,487 x 4 = 717,948. You might say that the digits are parasitic after being operated on.

__PASCALâ€™S TRIANGLE NUMBERS__

There is one famous arrangement of numbers in a familiar geometric shape that has received the attention and admiration of mathematicians for centuries, Pascal’s Triangle. Contrary to popular belief, it was not created by Pascal but is believed to have been created, or discovered, by both the Chinese and Persians sometime during the 11th and 12th centuries. Blaise Pascal had the distinction of having it named after him merely because of his extensive 17th century work with it in relation to probability. Surprisingly, it has connections with probability, combinations, the binomial expansion, Taxicab Geometry, powers of 2, and perhaps many others which I have not yet had the privilege of hearing about.

The first appearances of the triangle were allegedly associated with the coefficients of the binomial expansion. But lets first define the array and then show how it applies. We are going to create a triangle with an array of numbers within the triangle. The apex of our triangle is the number 1 and called row 0. The following row 1 contains 2 1’s. We start out as follows:

……………………………………………………………………….1

……………………………………………………………………1……1

The next row contains a 1, 2, and 1 as in

……………………………………………………………………….1

……………………………………………………………………1……1

…………………………………………………………………1…..2……1

The next row contains a 1, 3, 3, and 1 as in

Row

…0…………………………………………………………………….1

…1…………………………………………………………………1……1

…2………………………………………………………………1…..2……1

…3……………………………………………………………1….3……3……1

If you do not see the evolving pattern, now is probably the best time to explain how to create the rest of the triangle. In all its simplicity, each number is simply the sum of the two numbers immediately above. Looking at the 2nd row, the 1 at the beginning of each row is the sum of the 1 and the implied 0 above it. The next number is the sum of the two 1’s above it. The last one is derived the same as the first 1 in the row. The 2nd and 3rd numbers in the 3rd row are the sum of the 1’s and 2’s above them. Therefore, we can continue to construct the triangle as far as we wish as follows:

Row

0……………………………………………………………………….1

1……………………………………………………………………1……1

2…………………………………………………………………1…..2…..1

3………………………………………………………………1….3……3…..1

4……………………………………………………………1….4…..6……4…..1

5…………………………………………………………1….5…10….10…..5…..1

6………………………………………………………1….6…15….20….15….6…..1

7……………………………………………………1….7…21…35…35….21…..7…..1

8…………………………………………………1….8…28…56…70….56…28….8…..1

9………………………………………………1….9…36..84..126..126..84….36…..9…..1

10………………………………………….1…10..45.120..210.252..210..120…45…10…..1

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How can you determine the numbers in any row without knowing the numbers in the previous row?

Let the row number start from 0 at the top. Call that number n. Let the number of the number in the row be r, starting from r=1 at the left.

The value of the number that goes in the rth place of that row is

n!/[r! (n-r)!]]

Consider, for example, the fifth row of the triangle that has elements 1 4 6 4 1

n=4 in this case, since we started with n=0 at the top. For the second element, r=2

The value of that number is

4!/(2! 2!) = 3*4/1*2 = 6

What part do these numbers play in mathematics? Lets first look at the binomial expansion of (a + b)^n. Lets expand a few with increasing values of n.

(a + b)^0 = 1

(a + b)^1 = 1a + 1b

(a + b)^2 = 1a^2 + 2ab + 1b^2

(a + b)^3 = 1a^3 + 3a^2b + 3ab^2 + 1b^2

(a + b)^4 = 1a^4 + 4a^3b + 6a^2b^2 + 4ab^3 + 1b^4

By now you must have recognized that the coefficients of the terms in each expansion are the values from Pascal’s triangle. How elegant. We can write the expansion of (a + b)^n for any “n” by simply referring to Pascal’s triangle. The coefficients of (a + b)^n are the numbers from the nth row of Pascal’s triangle.

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Before getting into other applications of the triangle to various areas of mathematics, might you by now have noticed anything else about the array of numbers? Lets explore. The left edge of our triangle is all 1’s. The line of numbers parallel to this left edge are the counting numbers, 1, 2, 3, 4, 5, etc. The next row of numbers parallel to the left edge are the triangular numbers, 1, 3, 6, 10, 15, 21, etc. The next row of numbers parallel to the left edge are the tetrahedral numbers, 1, 4, 10, 20, 35, etc. The next row of numbers are the pentatope numbers (the sum of the tetrahedral numbers), 1, 5, 15, 35, 70, etc. The next row of numbers are the sums of the pentatope numbers, and so on. Also notice that each number is the cumulative sum of the numbers from the adjacent row. The first 3 in row 3 is the sum of the three 1’s in the adjacent sloping row above it. The 35 in row 7 is the sum of the 15, 10, 6, 3, and 1 in the adjacent sloping row above it.

Row

0………………………………..1…../_——————–Counting numbers

1…………………………….1……1…./_—————–Triangular numbers

2………………………….1…..2…..1…./_————–Pentatope numbers

3………………………1….3……3…..1…./_———-Sum of Pentatope numbers

4……………………1….4…..6……4…..1…./_——Sum of previous sums

5…………………1….5…10….10…..5…..1…./_—Sum of previous sums

6………………1….6…15….20….15….6…..1

7……………1….7…21…35…35….21…..7…..1

8…………1….8…28…56…70….56…28….8…..1

9………1….9…36..84..126..126..84….36…..9…..1

Another interesting aspect of Pascal’sTriangle is the hidden values of Fibonacci’s secuence.

Draw a line through the 3rd 1 from the top on the left and the 2nd 1 from the top on the right.

Draw a line from the 2nd 1 from the top on the left parallel to the first line.

Draw a line from the top 1 parallel to the first line.

Draw a line from the 4th 1 from the top on the left through the 2 in the 3rd row and parallel to the first line.

Draw a line from the 5th 1 from the top on the left side through the first 3 in the 4th row and through the 3rd 1 from the top on the right and parallel to the first line.

The geometry of the figure must be accommodating of these instructions but I think you will see the point of it all.

The sum of the numbers on each of these lines is what is important. If your figure is anywhere near accurate enough, you will see that the first line sum is 1, the second line sum 1, the third line sum 2, the fourth line sum 3, the fifth line sum 5, the sixth line sum 8, the seventh line sum 13 and so on.

Row

0………………………………..1…..

1…………………………….1……1…

2………………………….1…..2…..1….

3………………………1….3……3…..1….

4……………………1….4…..6……4…..1….

5…………………1….5…10….10…..5…..1…

6………………1….6…15….20….15….6…..1

7……………1….7…21…35…35….21…..7…..1

8…………1….8…28…56…70….56…28….8…..1

9………1….9…36..84..126..126..84….36…..9…..1

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In looking over the numbers in the triangle, you might notice that each row follows a consistent pattern. Reading the rows as the numbers, 1, 11, 121, 1331, 14641, etc., is there anything unique that you notie about these numbers.

1 = 11^0, 11 = 11^1, 121 = 11^2, 1331 = 11^3, 14641 = 11^4, and so on. At this point you undoubtedly wonder why 11^5 = 161,051 and not 15101051. Consider the numbers in the row as the place value digits in the base 10 system which then results in 1(100,000) + 5(10,000) + 10(1000) + 10(100) + 5(10) + 1(1) = 161,051. This not only applies to all the subsequent rows but to the first 5 rows also.

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You have probably heard of permutaions and combinations, the numbers of arrangements and choices, or combinations, that n things can be arranged or chosen. We designate the permutations of n things taken n at a time as nPn and the permutations of n things taken r at a time as nPr where P stands for permutations, n stands for the number of things involved, and r is less than n. To find the number of permutations of n dissimilar things taken n at a time, the formula is nPn = n! which is n factorial which means n(n-1)(n-2)(n-3)…….3x2x1. To find the number of permutations of n dissimilar things taken r at a time, the formula is nPr = n(n-1)(n-2)(n-3)……….(n-r+1).

We designate the combinations of n things taken n at a time as nCn and the combinations of n things taken r at a time by nCr. To find the number of combinations of n dissimilar things taken r at a time, the formula is nCr = n!/[r!(n-r)!] which can be stated as n factorial divided by the product of r factorial times (n-r) factorial. Example: In how many ways can a committee of three people be selected from a group of 12 people? We have 12C3 = (12!)/[3!(9!) = 220. How many different ways can you combine A, B, C, and D in sets of three? Clearly, 4C3 = (4x3x2x1)/(3x2x1)(1) = 4. How many handshakes will take place between six people in a room when they each shakes hands with all the other people in the room one time? Here, 6C2 = (6x5x4x3x2x1)/(2×1)(4x3x2x1) =15.

Lets look at our expression for combinations, nCr = n!/[r!(n-r)!], assuming 0! = 1. What is the number of possible combinations of 4 things taken 2 at a time? 4C2 = 4!/[2!(2!)] = 6. What about 4 things taken 1 at a time? 4C1 = 4!/[1!(3!)] = 4. What about 4 things taken 3 at a time? 4C3 = 4!/[3!(1!)] = 4. What about 4 things taken none at a time? 4C0 = 4!/[0!(4!)] = 1. What about 4 things taken 4 at a time? 4C4 = 4!/[4!(0!)] = 1. Lets try 3 things taken 2 at a time? 3C2 = 3!/[2!(1!)] = 3. How about 3 things taken 1 at a time? 3C1 = 3!/[1!(2!)] = 3. How about 3 things taking none at a time? 3C0 = 3!/[3!(0!)] = 1. How about 3 things taken 3 at a time? 3C3 = 3!/[3!(0!)] = 1. Do you recognize anything familiar about the resulting combinations of 3 things taken 0, 1, 2, and 3 at a time and 4 things taken 0, 1, 2, 3, and 4 at a time? Yes, the numbers of combinations for each series of conditions match the numbers in the 3rd and 4th rows of Pascals triangle, 1, 3, 3, 1 and 1, 4, 6, 4, 1. How about that? We can use Pascal’s triangle to find any combination of n things taken r at a time by simply referring to the nth row of the triangle. In general, the sequence of numbers in the nth row represent the possible combinations of “n” things taken 0, 1, 2, 3, …n at a time, in that order across the row..

Based on this information, it can now be shown that the coefficient C(i) of any term in a binomial expansion may be derived from nCi = n!/[i!(n-i)!] where n = the exponent to which (a + b) is being raised and i = the term of the expansion, where the 1st coefficient C(0) = 1, i = 1 for the 2nd coefficient C(1), i = 2 for the 3rd coefficient C(2), etc. For example, the coefficient of the 3rd term in the expansion of (a + b)^4 is 4C2 = 4!/[2!(2!)] = 6.

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I’m certain that you have probably talked about the probability of tossing a head or tail with a coin. Obviously, the probability of getting either a head or a tail is 1/2, there being only 1 successful result out of two possibilities. What if we were to ask what is the probability of two people tossing 2 heads at the same time? The two tosses can result in any one of four possibilitites, HH, HT, TH, or TT. With each of these individual results being equally likely, there is only one chance out of the four possible outcomes of getting 2 heads, thus making the probability of 2 people tossing 2 heads at the same time 1 in 4 or 1/4th. Clearly, the probability of simultaneously tossing 1 head and 1 tail is 2 out of 4 or 1/2.

What about 3 people all tossing a head at the same time? The tosses can result in eight outcomes, HHH, HHT, HTH, THH, HTT, THT, TTH, and TTT for a total of 8 possible outcomes, all equally likely. The probability of tossing 3 heads is therefore 1 in 8 or 1/8. The probability of tossing 2 heads and 1 tail is 3 in 8 or 3/8. Logically, the probability of tossing 2 tails and a head or 3 tails is 3/8 and 1/8.

Might you see something in the numbers of Pascal’s triangle that reflects the probabilitites we just derived? Looking at row 2 of the triangle we see the numbers 1, 2, 1. The sum of these three numbers is 4. If we now divide the numbers 1, 2, and 1 by 4 we obtain 1/4, 2/4, and 1/4, the probabilities of two people tossing the various outcomes. Looking at row 3 of the triangle we see the numbers 1, 3, 3, 1. These sum to 8 and if we divide each of the numbers in the row by 8 we get 1/8, 3/8, 3/8, and 1/8, the probabilities of three people tossing the various outcomes. Generally speaking, each number in the row, divided by the sum of the numbers in the row, represents the probability of the possible equally likely outcomes occurring n times. What do you know? Somewhat amazing wouldn’t you say?

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You may, or may not, have heard of taxicab geometry. The whole subject can be viewed in the context of a taxicab maneuvering through a city where the streets are totally straight and cross one another in the form of multiple squares as shown in the figure below. The sides of the squares represent the streets. The intersections of the squares form a lattice of points representing the numerous junctions through which a taxicab traveling from point A to point E would have to travel. In Euclidian geometry, the shortest distance between points A and E would be the straight line joining A to E. In taxicab geometry, one must follow the lines as defined by the squares, always moving in a direction toward the target point, resulting in many paths, all equally minimum. Without going through the sequential steps of deriving the data associated with such squares, I’ll summarize the results obtained. The derivation of the data appears in another article.

A __ __ __ __

I__I__I__I__I

I__I__I__I__I

I__I__I__I__I

I__I__I__I__I

E

__No. of squares No. of Intersections External paths Internal paths Total__

……..1…………………………4………………………..2…………………..0………………2

……..2…………………………9………………………..2…………………..4………………6

……..3………………………..16……………………….2………………….18…………….20

……..4………………………..25……………………….2………………….68…………….70

……..5………………………..36……………………….2…………………250…………..252

Examination of these results might just look familiar to you. Do any numbers in Pascal’s triangle look familiar? Of course, right down the middle are our results derived above.

0……………………………………..1

1………………………………….1…….1

2……………………………..1…….2…….1

3…………………………1…….3…….3……..1

4…………………….1……..4…….6…….4…….1

5………………..1……..5…….10…..10…….5……1

6…………….1…….6…….15……20…..15……6…….1

7………..1…….7…….21……35……35…..21……7…….1

8…….1…….8……28……56……70……56….28…….8…….1

9……………………………….126….126

10………………………………….252

The significance of Pascal’s triangle in taxicab geometry is that you can draw a square or rectangle of whatever size you wish on the diagram and the lower, inside, corner of the figure indicates the number of possible paths from one corner to the diagonally opposite corner. For example, draw a square of 4 unit squares from the apex 1 down to the 2nd 1’s on either side of the triangle and down through the 3’s to the 6. The 6 indicates the number of taxicab geometry paths. Do the same thing with a 3×3 square, and the lowest corner of the square will fall on the 20 in the 7th row. Draw a 4×4 square on the triangle and the lowest corner of the square will fall on the 70 in the 9th row. Draw a 2×3 rectangle on the triangle and the lowest corner will fall on the 10 on the 6th row, either way you draw the rectangle. A 3×5 rectangle results in 56 paths. Clever?

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Another view of Pascal’s triangle leads us to another interesting find. Not being able to draw what is required, we will have to rely on the following description. Draw a straight line between the 1st 1 on row 1 and the 1st 1 in row 2 and the 2nd 1 in row 1. Now draw a line parallel to this line below the 1 in row 1. Now draw parallel lines between successive rows down the remainder of the triangle. You should now have a 1 above the 1st sloping line, a 1 above the 2nd sloping line, two 1’s above the 3rd sloping line, a 1 and a 2 above the 4th sloping line, a 1, 3, and 1 above the 5th sloping line, a 1, 4, and 3 above the 6th sloping line, a 1, 5, 6, and 1 above the 7th sloping line, and so on. If you sum the numbers within each sloping row, you arrive at 1, 1, 2, 3, 5, 8, 13, 21, etc., Fibonacci’s famous series of numbers.

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A few more subtle fallouts of Pascal’s triangle are the following:

The product of the six numbers surrounding any number in the triangle, i.e., the two above, the two below, and the two on both sides, produces a perfect square.

From the 2 in row 2, 1x1x1x3x3x1 = 9 = 3^2

From the 3 in row 3, 1x2x3x6x4x1 = 144 = 12^2

From the 4 in row 4, 1x3x6x10x5x1 = 900 = 30^2

From the 5 in row 5, 1x4x10x15x6x1 = 3600 = 60^2

From the 6 in row 4, 3x3x4x10x10x4 = 14,400 = 120^2

From the 35 in row 7, 15x20x35x70x56x21 = 864,360,000 = 29,400^2

..0………………………………1……………………………= 2^0

..1…………………………..1..+..1………………………..= 2^1

..2………………………..1.+..2..+..1……………………..= 2^2

..3……………………..1.+.3..+..3..+..1………………….= 2^3

..4…………………..1.+.4.+..6..+..4..+.1……………….= 2^4

..5………………..1.+.5.+10.+.10.+..5.+..1…………….= 2^5

..6…………….1….6…15….20…15…..6…..1…………..= 2^6

..7………….1….7…21…35…35….21…..7…..1………..= 2^7

..8……….1….8…28…56…70….56…28….8…..1……..= 2^8

..9……1…..9…36..84..126..126..84….36…..9…….1..= 2^9

………………………………and

..0……………………………..1……………………………

..1…………………………..1..-..1………………………= 0

..2………………………..1.-..2..+..1…………………..= 0

..3……………………..1.-.3..+..3..-..1………………..= 0

..4…………………..1.-.4.+..6..-..4..+.1………………= 0

..5………………..1.-.5.+10..-.10.+..5.-..1……………= 0

..6…………….1.-..6..15.-.20.+.15.-..6.+…1………..= 0

..7………….1….7…21…35…35….21….7…..1……..= 0

..8……….1….8…28…56…70….56…28….8…..1…..= 0

..9……1….9…36..84..126..126..84….36…..9…..1..= 0

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You might have heard at one time or another the problem of determining the number of regions created by joining n randomly located points by straight lines within a circle. The number of regions created within the circle is really the number of regions defined by joining all the vertices within an n-gon plus the n arc/chord regions created when inscribing the n-gon within the circle.

One point on the circle allows for no lines leaving one region. Drawing one line between two points creates 2 regions. Drawing 3 lines between 3 points creates 4 regions, the triangle and the 3 regions surrounding the triangle. Drawing 6 lines between 4 points creates 8 regions. Drawing 10 lines between 5 points creates 16 regions. Contnuing this further, we end up with the partial table

nth term…1…..2…..3…..4…..5……6…..7…..8…..9….10…………n

Points……1…..2…..3…..4…..5…..6……7…..8…..9….10…………n

Lines……..0…..1…..3…..6….10…15….21…28….36…45………..N

Regions….1…..2…..4…..8….16…31….57…99…163..256

1st diff………1…..2…..4……8…15….26…42….64….93

2nd diff…………1…..2…..4…..7….11…16….22….29

3rd diff…………….1…..2…..3….4……5…..6……7

4th diff……………….1…..1…..1…..1……1…..1

The resulting numbers of regions form a finite difference series. An expression can be derived enabling the definition of the nth term of the finite difference series. This expression is N = (n^4 – 6n^3 + 23n^2 – 18n + 24)/24.

Checking our projected 10th term:

N(10) = 10^4 – 6(10^3) + 23(10^2) – 18(10) + 24)/24 = 10,000 – 6000 + 2300 – 180 + 24)/24 = 256.

If we draw a line cutting off the numbers in Pascal’s triangle shown below, we find some amazing results.

Row

0……………………………………………………………………….1

1……………………………………………………………………1……1

2…………………………………………………………………1…..2…..1

3………………………………………………………………1….3……3…..1

4……………………………………………………………1….4…..6……4…..1

5…………………………………………………………1….5…10….10…..5…………………………….1

6………………………………………………………1….6…15….20….15……………………………6…..1

7……………………………………………………1….7…21…35…35……………………………21…..7…..1

8…………………………………………………1….8…28…56…70…………………………….56…28….8…..1

9………………………………………………1….9…36..84..126……………………………126..84….36…..9…..1

Summing the remaining numbers in each row, we obtain 1, 2, 4, 8, 16, 31, 57, 99, 163, etc.

__PELL NUMBERS__

The Pell numbers or the Pell sequence – Pn: 1, 2, 5, 12, 29, 70, 169,…….n, (P1 = 1 and P2 = 2) derives from the recursion formula (Pn = 2P(n-1) + P(n-2) and the Binet expression Pn = [(1 + sqrt2)^n – (1 – sqrt2)^n]/2sqrt2 while Qn: 1, 3, 7, 17, 41, 99, 239,…….n, (Q1 = 1 and Q2 = 3) derives from the recursion formula Qn = 2Q(n-1) + Q(n-2) and the Binet expression Qn = [(1 + sqrt2)^n + (1 – sqrt2)^n]/2.

__PENTATOPE NUMBERS__

The pentatope numbers are the sums of the tetrahedral numbers which are the sums of the triangular numbers. The triangular numbers are 1, 3, 6, 10, 15, 21, 28, 36, 45, 55, etc. The tetrahedral numbers are 1, 4, 10, 20, 35, 56, 84, 120, 165, 220, etc. Therefore, the pentatope numbers are 1, 5, 15, 35, 70, 126, 210, 330, 495, 715, etc.