__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.
__
PASCAL’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
<----------------------------------------------------->
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.
<--------------------------------------------->
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
<-------------------------------------------->
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.
<-------------------------------------------->
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)/(2x1)(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.
<--------------------------------------------->
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?
<--------------------------------------------->
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
3x3 square, and the lowest corner of the square will fall on the
20 in the 7th row. Draw a 4x4 square on the triangle and the
lowest corner of the square will fall on the 70 in the 9th row.
Draw a 2x3 rectangle on the triangle and the lowest corner will
fall on the 10 on the 6th row, either way you draw the
rectangle. A 3x5 rectangle results in 56 paths. Clever?
<---------------------------------------------->
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.
<--------------------------------------------->
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
<--------------------------------------------------------------------------------------->
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.
Continue This Article | Part I |
Part II |