From Generalized Binomial Symbol to β− and α−sequences
by Mirek Majewski
Department of Mathematics & Computer Sci.
PNG University of Technology, Papua New Guinea and
Andrzej Nowicki
Faculty of Mathematics & Computer Sci.
Nicholas Copericus University, Poland Abstract
Factorial function and binomial symbol play important role in contemporary mathematics. In this paper we explore the generalized binomial symbol and show some of its properties. We remind the definition of β−sequences and we prove that some well known sequences are β−sequences.
Finally we show how some of the enclosed here examples can be done with the help of the computer program - Scientific Notebook.
1 Introduction
Let N = {1, 2, . . .} be a set of natural numbers and gcd(n, m) denote the greatest common divisor of integers n and m. Now we introduce the two basic, for this paper, definitions.
1.1 Two basic definitions
1. Let a = {an} be a sequence of natural numbers, and a∗n denote a sequence given by the formula:
a∗n=
a1a2· · · an if n ∈ N
1 if n = 0
It is easy to notice that if a is a sequence of consecutive natural numbers, i.e. an = n for all n ∈ N, then a∗n= n!. Expression a∗n should be considered as a generalized factorial function.
2. For a given sequence a = {an} of natural numbers we define its binomial symbols as the expres- sion:
n k
a
=
a∗n
a∗k·a∗n−k if n > k 0 if n < k.
where n, k are non-negative integer numbers.
Observe that,
n n
a
=n 0
a
= 1 and n k
a
=
n n − k
a
for n > k.
Termsn k
a are refereed to as generalized binomial symbols.
Example 1 Like with ordinary binomial symbols we can build a Pascal triangle using binomial symbols of a given sequence. Let a = {2n}, thus we have
n k
a
=
Qn i=12i
Qk i=12i
∗ Qn−k
i=1 2i
= 212(n+1)n
212(k+1)k
212(n−k+1)(n−k)
= 2(12n+12)n−(12k+21)k−(12n−12k+12)(n−k)
= 2k(n−k)
Finally we may obtain the Pascal triangle
n = 1 1 1
n = 2 1 2 1
n = 3 1 4 4 1
n = 4 1 8 16 8 1
n = 5 1 16 64 64 16 1
n = 6 1 32 256 512 256 32 1
n = 7 1 64 1024 4096 4096 1024 64 1
In this example all binomial symbols of the sequence {2n} are integers. However it is not a rule.
It is easy to show an example of a sequence a for which some binomial symbols are rational. Here is one of such examples.
Example 2 Let a =n2+ 1 . Thus
n k
a
=
Qn
i=1(i2+ 1)
Qk
i=1(i2+ 1) Qn−k
i=1(i2+ 1) and
6 1
a
=37 2 ,6
2
a
= 481 5 ,5
3
a
= 221 5 ,6
5
a
= 37 2 .
1.2 About β−sequences
The above examples lead us to the following definition.
Definition We say that a is a β-sequence if all its binomial symbols n k
a are integer numbers.
Numbers of the formn k
a are well known, if a is a sequence of consecutive natural numbers, i.e.
an= n for n ∈ N. In such case a∗n= n! andn k
a= nk, where
n k
=
( n!
k!(n−k)!, if n > k, 0, if n < k.
It is also easy to show that if n > k, then nk is a natural number (the number k-element subsets of an n-element set). According to the above examples sequences {2n} and {n} are β-sequences.
Here we have a few more examples.
Example 3 Each constant sequence with natural term is a β-sequence, i.e. if c ∈ N and an = c for n ∈ N, thenn
k
a= 1 for all integers n ≥ k > 0.
Example 4 Number 2 in the example 1 can be replaced by any positive integer. This way we get the following fact: each geometric sequence with natural quotient is a β-sequence. More formally, if c ∈ N and an= cn for n ∈ N, then for any n > k, we haven
k
a = c(n−k)k.
Example 5 Product of two β-sequences is a β-sequence. If a = {an} and b = {bn} are β-sequences, then c = {anbn} is a β-sequence andn
k
a =n k
a·n k
b.
1.3 About α−sequences
Another important class of β−sequences are so called α−sequences.
Definition The sequence of natural numbers a = {an} , is an α-sequence if for any n, m ∈ N gcd(an, am) = agcd(n,m).
It is easy to prove that a constant sequence and the sequence of consecutive natural numbers are α-sequences. It is also easy to prove that if s a natural constant, then sequences {ns} and {ns} are also α-sequences.
Lemma If a = {an} is an α-sequence, then for all non-negative integers n, k, there exist integer numbers X(n, k) and Y (n, k) such that,
n + 1 k + 1
a
= X(n, k)
n k + 1
a
+ Y (n, k)n k
a
Proof Let n and k be two non-negative integers. The existence of numbers X(n, k), Y (n, k) is obvious if n 6 k. Suppose, that n > k and let d = gcd(an−k, ak+1). In such case there exist integers u, v such that
d = uan−k+ vak+1.
We used a well known property of the gcd function (see [6]). Hence
d = gcd (an−k, ak+1) = agcd(n−k,k+1)= agcd(n+1,k+1)= gcd (an+1, ak+1)
and d is a divisor of an+1. Thus an+1 = pd, where p is a natural number. Now we can easily conclude that
X(n, k) = pu, Y (n, k) = pv are numbers to be found.
With the help of the principle of mathematical induction we can prove the following theorem (see [3], pg.353 ).
Theorem Any α-sequence is a β-sequence.
Let’s show a few examples of α-sequences.
Example 6 The following sequence {Mn} = {2n− 1} we call the sequence of Mersenne numbers.
1. (a) The sequence {Mn}, is an α-sequence (see [7], pg. 373).
(b) In definition of the Mersenne numbers the number 2 can be replaced by any real number a > 1. The sequence {sn} = {an− 1} is also an α-sequence (see [7], pg. 11).
(c) Finally, if a > b are natural numbers and gcd(n, m) = 1 then the sequence {un} = {an− bn} is also an α-sequence (see [3], pg.174).
Example 7 Let f (x) be a polynomial of variable x with natural coefficients. We define a sequence {bn} as follows
bn=
f (0) if n = 0 f (bn−1) if n > 0
The sequence {bn} is an α-sequence (see [5], 1/1989 ). If f (x) = 2x + 1, then {bn} is the Mersenne sequence from example 6.
Example 8 You have probably heard of the Fibonacci sequence: 0,1,1,2,3,5,8,13,21,... in which every number after the first two is the sum of the preceding two numbers. That is
Fn=
0 if n = 0
1 if n = 1
Fn−1+ Fn−2 if n > 1
This sequence, dating from 1202 A.D. is still of such interest that there is a journal called The Fibonacci Quarterly, just for papers on it. It was proven that Fibonacci sequence {Fn} is also an α-sequence (see [8], [7], pg. 280).
Example 9 Let p and q be fixed natural numbers. We define the sequence {vn} as follows:
vn=
1 if n = 1
p if n = 2
pvn−1+ qvn−2 if n > 2.
The sequence {vn} is an α-sequence (see [4], [2]).
Example 10 Suppose that {an}, and {bn} are α-sequences, in such case {cn}, where cn= banfor all n ∈ N,is also an α-sequence. Hence sequences {usn} , {3vn− 1} , {u2n−1}are also α-sequences.
2 Final comments
1. Product of two α-sequences may not be an α-sequence. For instance, the sequence {n(2n− 1)}, is the product of two α-sequences, and is not an α-sequence. We can prove that at the same time it is a β-sequence.
2. A sequence {xn}, with natural terms, is an α-sequence if and only if gcd (xm, xn) = gcd (xm−n, xn) , for all natural numbers n, m such that m > n (see [7], pg. 282).
3. From the proven theorem it follows that if {an} is an α-sequence and s is a natural number, then product of each s consecutive terms of {an} is divisible by a∗s= a1a2· · · as.
3 Computer approach
Some of the enclosed examples contain heavy calculations, both symbolic and numerical. Most of them can be done, and some were done, using a computer tool called Scientific Notebook. The program contains two parts: interface – a kind of word processor and computing engine–Maple V. Let’s show how most of the enclosed examples can be done.
Example: to produce Mersenne sequence binomial coefficients
Step 1 To define Mersenne sequence type the formula M (n) = 2n − 1, leave cursor at the end of formula and choose in menu Maple+Define+New Definition. Since this moment SNB will remember this formula and it will use it all appropriate operations. For example, type in maths mode M (25) and click Maple+Evaluate, you will obtain M (25) = 33 554 431
Step 2 To define a Mersenne sequence binomial symbols you have to do the two following changes.
First, you cannot use the symboln k
a, as SNB doesn’t recognize it as a function. You have to use a name for this function and enclose in the round bracket variables n and k. This means you have to replacen
k
a by F (n, m). Next change you have to do inside of the binomial symbol formula.
SNB doesn’t recognize a notation M (1)M (2) · · · M (n). Instead of it we can useQn
i=1M (i). Thus the formula for binomial coefficients will obtain the form:
F (n, k) =
Qn i=1M (i)
Qk
i=1M (i) Qn−k
i=1 M (i)
Now you can simplify the formula, by clicking on Maple+Simplify, you should obtain
F (n, k) =
Qn
i=1 2i− 1
Qk
i=1(2i− 1) Qn−k
i=1 (2i− 1)
Again, leave the cursor inside of the formula and choose in menu Maple+Define+New Definition.
Step 3 Now you can use defined formulae to obtain Mersenne sequence binomial symbols. For ex- ample, type in maths mode F (5, 3) and click Maple+Evaluate, you will obtain F (5, 3) = 155.
What about other binomial symbols? Here we have: F (4, 2) = 35, F (23, 13) = 4708 070 980 050 125 241 816 734 797 968 472 558 995, etc.
Step 4 To obtain the Pascal triangle click on menu Insert+Matrix (set 4 rows and 9 columns) and fill it with appropriate values like F (1, 1), F (2, 1), F (2, 2) etc. When you put all necessary values select each of them separately and while selected press Ctrl key and click on Maple+Evaluate.
You should obtain a triangle like this:
1 1
1 3 1
1 7 7 1
1 15 35 15 1
If you wish to explore with Scientific Notebook a few more examples, obtain its evaluation copy from http://scinotebook.tcisoft.com/scinotebook/ and experiment.
References
1. G. L. Alexanderson, L. F. Klosinski, A Fibonacci analogue of Gaussian binomial coefficients, Fibonacci Quarterly, 12(1974), 129 - 132.
2. P. Doma´nski, Generalized Fibonacci Numbers (in Polish), Delta, 1(1979).
3. R. L. Graham, D. E. Knuth, O. Patashnik, Concrete Mathematics, a foundation for Computer Science, Addison Wesley Publishing, 1994.
4. V. E. Hoggatt, Fibonacci numbers and generalized binomial coefficients, Fibonacci Quarterly, 5(1967), 383 - 400.
5. Kwant (Monthly Mathematical Magazine), in Russian.
6. W. Sierpi´nski, Number Theory, in Polish,Matematical Monographs, Warsaw 1950.
7. W. Sierpi´nski, Number Theory II, in Polish, PWN, Warsaw 1959.
8. N. N. Worobjow, Fibonacci Numbers, in Russian, Popular Lessons in Mathematics - 6, Nauka, Moscow, 1978.
