1. The Infinity Problems
Many unsolved problems are related to enquiries on whether there is an infinity of composite
numbers, or primes, or Mersenne primes, or Fermat pimes, or Perfect numbers amongst others. This
paper describes solutions based on the use of final value theorem on holistic number sequences.
Some of the solutions are purely dialectic but others are algorithmic. In general, mixed mode
formulations are algorithmically efficient but dialectically opaque. This is exemplified by the
proof of an infinity of prime numbers by Euclid which is difficult to emulate in sequence algebra
and yet the latter succeeds in offering a mixed mode global formulation for primes. Number
theory may be one of those rare disciplines with two complementary and yet distinct theories,
one reductionistic and the other holistic. These two theories are complementary where synergistic
benefits are derived by the combined usage. This paper will demonstrate how the strengths of
both theories are harnessed to solve some infinity problems. Like all elses in numbe theory,
sequence algebra also unearths new unsolved problems. Readers who are not familiar with
sequence algebra are referred to previous publications released in the author's URL site:
http://web.singnet.com.sg/~huens/ . Reference [5] is especially written for beginners.
(a) Proof of an infinity Comp(z) [10]
The equation of divisibles is given by Comp(z) = 1/(z^i*(z^i-1))[10]. Inspection of this closed form
generating function will reveal that it contains an infinite number of individual closed forms each
of which will generate a periodic sequence with an interval of i units. The sum total of
contributions of all these sequences yields the global composite sequence. Applying final value
theorem by multiplying Comp(z) by (z-1)/z and putting z = 1 will prove that there is a no end to
Comp(z):
Final value in Comp(z) = {(z-1)/z * Comp(z)| z=1}
..................= (z-1)/z * 1/(z^i * (z-1) * (z^(i-1) + z^(i-2) + ...... + z^1 + 1))
..................= 1/(z^(i+1)*(z^(i-1) + z^(i-2) + ...... + z^1 + 1))
..................= sum(1/i, i = 2..infinity) ...................................................................(1).
Summing i from 2 to infinity, will give
Final value of Comp(z) = 1/2 + 1/3 + 1/4 + ....................+ 1/infinity ...............(2),
which is an infinite sum which can be demonstrated by the following Maple line:
evalf(sum((1/i),i=2..100000000000000000000000000000000000)); Ans= 80.16769391
In other word there is no end to Comp(z).
(b) Proof of an infinity of Prime(z)
It would be awkward to prove the infinity of primes in Prime(z) via sequence algebra due to the
presence of either mixed mode arithmetics or the presence of the Normc( ) operator [12]. In
sequence algebraic formulation, Prime(z) is written as follows:
.....................Prime(z) := Nat(z) - Normc(Comp(z)) ; ......................................(3).
The presence of the Normc( ) operator makes it difficult to continue algebraic analysis. To
overcome this, we proceed to investigate what will be the effect to the final value prediction with
and without Normc( ). We use a finite sequence for Comp(z) to carry out evaluations.
Comp(z):=series(sum(1/(z^i*(z^i-1)),i=2..12),z=infinity,12); ........................(4).
The "final value" for the finite Comp(z) sequence is 2 at the integer 10.
................................................1.......2.......2.......1........2............1
.........................Comp(z) := ---- + ---- + ---- + ---- + --- + O(---) ............................(5)
..................................................4.......6.......8.......9........10........12
................................................z........z.......z.......z..........z...........z
After applying Normc( ), the "final value" of the normalised Comp(z) sequence is 1 at the integer
10.
Normc(Comp(z)):=
................................................1.......1........1........1........1........1
......................NComp(z) := ---- + ---- + ---- + ---- + --- + O(---) .............................(6).
.................................................4.......6........8........9........10........12
................................................z.......z........z........z..........z.........z
It shows that even though the numerator coefficient is reduced to unity, it does not affect the
outcome if one is only interested to find out whether there is or there is not an infinity of
composites. Thus whether there are 10000 duplicates of the integer at infinity or only one copy
does not affect the infinity property. This finding is correct if such a sequence under Normc( ) is
investigated in isolation. When Normc( ) appears in a compound sequence formulation, the
above rule does not apply. This is because it is impossible to do arithmetics between two
sequences when one has been normalised and the other not normalised. All sequences must be
normalised before final value theorem can be applied. This is easily checked by finding the
Prime(z) sequence via two formulations in equations (8) and (9).
Nat(z):=series(1/(z-1),z=infinity,12);
..........................1.......1.........1.......1.........1.........1.......1.........1........1........1.........1
Nat(z) := 1/z + ---- + ---- + ---- + ---- + ---- + ---- + ---- + ---- + --- + --- + O(---) ..............(7).
............................2......3.........4........5........6..........7.......8.........9.......10.......11.......12
..........................z.......z..........z........z.........z.........z.......z..........z........z..........z..........z
Prime(z):=Nat(z)-Comp(z);
........................................1.......1..........1.......1.......1.......1.......1.......1
...........Prime(z) := 1/z + ---- + ---- + ---- - ---- + ---- - ---- - --- + --- ........................(8).
..........................................2.......3........5.......6..........7.......8......10......11
........................................z........z........z.......z..........z.......z.........z.......z
Prime(z)= Nat(z)-Normc(Comp(z)).
.....................................1........1........1.......1.........1..........1
...........Prime(z)=1/z + ---- + ---- + ---- + ---- + --- + O(---) .....................................(9).
........................................2........3.........5.......7........11.........12
......................................z........z.........z.......z.........z...........z
Fortunately classical number theory comes to the rescue. Euclid has already proven that there is
an infinity of primes[1]. This is a synergetic partnership between reductionistic and holistic
number theory. If we accept both theories as being covered by the same axiomatic set, then
there is no need to find a separate sequence algebraic proof of the infinity of primes. Note that
the admission of sequence algebra into number theory only requires the acceptance of Nat(z) as
an axiom which is not defined in traditional number theory. This is justified by the number of
new results found including the solutions of some previously unsolved number theory problems
[10 to 12].
Here I quote Weyl (1949): "A truly realistic mathemtics should be conceived, in line with physics,
as a branch of the theoretical construction of the one real world, and should adopt the same
sober and cautious attitude toward hypothetic extensions of its foundations as is exhibited by
physics." [4].
(c) Proof of an infinity of Even(z) and Odd(z)
It appears that one should not be called upon to prove an infinity of Nat(z) since this is an axiom although it
appears that this is provable using final value theorem. We will comment on this after proving
that there is an infinity of Even(z) and also of Odd(z) below:
Even(z) = z^2/(z^2-1):
Applying final-value theorem, we get
Final value of Even(z) = {(z-1)/z*z^2/(z^2-1)|z=1}
..................= (z-1)/z * z^2/((z-1)*(z+1)) = z/(z+1) =1/2. ......................(10).
Remember that numerators of Even(z) will alternate between 0 and 1 indefinitely. Final value theorem
predicts that the infinity value is 1/2 which looks strange but it is the truth. This is because (z-1)
is a diffferentiator in sequence algebra. Within two unit intervals, there is a zero numerator and a
unity numerator. Therefore the average magnitude is 1/2. Final value simply predicts the rate
of change per unit interval which is 1/2. It proves that at infinity the final value is still finite.
Therefore there is infinity in Even(z).
Odd(z) = z/(z^2-1):
Applying final-value theorem, we get
Final value of Odd(z) = {(z-1)/z*z/(z^2-1)|z=1}
..............= (z-1)/z * z/((z-1)*(z+1)) = 1/(z+1) =1/2. .................................(11).
Remember that like Even(z), the numerators of Odd(z) will also alternate between 0 and 1
indefinitely. Therefore the average magnitude is 1/2. Final value simply predicts the rate of
change per unit interval which is 1/2. Therefore there is infinity in Odd(z).
Having proved infinities for Even(z) and Odd(z), we find that the final value of Nat(z) = Even(z) +
Odd(z) = 1 which is correct. This can also be proved directly using final value theorem. The
reason is that this property is not specifically defined in the axiom. The axiom for Nat(z) simply
defines the orderliness of integer numbers. The final value is not part of the assertion.
Therefore there is no inconsistency here.
(d) Proof of infinity of Evencomp(z) and Oddcomp(z)
From the sequence identity in equation (12), we know that Comp(z) and Evencomp(z) have
explicit formulations but the identity will only be correct if normalised sequences are used.
Normc(Comp(z)):= Evencomp(z) + Normc(Oddcomp(z)); ...........................(12).
Evencomp(z):=series(z^(-2)/(z^2-1),z=infinity,12);
....................................................1.......1.......1.........1...........1
........................Evencomp(z) := ---- + ---- + ---- + --- + O(---) ...................................(13).
......................................................4.......6.......8........10..........12
....................................................z.......z.......z..........z...........z
Final value of Evencomp(z): (z-1)/z * (1/(z^2*(z^2-1)))
..........................................=(z-1)/z * (1/(z^2*(z-1)*(z+1)))
..........................................= 1/2.....................................................................(14).
Therefore there is an infinity of Evencomp(z).
Now Normc(Oddcomp(z)) = Normc(Comp(z)) - Evencomp(z)
.....................= Normc(1/(z^i*(z^i-1)) - 1/(z^2*(z^2-1)) .................................(15).
There is no way one could predict infinity of Oddcomp(z) via this route in view of the presence of
Normc( ) operator. We also can express alternatively
Oddcomp(z) = Odd(z) - Prime(z) but again this will not facilitate solutions since
Prime(z) = Nat(z) - Normc(Comp(z)) which again contain the Normc( ) operator.
Thus the infinity of Oddcomp(z) is a new unresolved problem.
(e) Proof of an infinity of Mersenne Numbers: Mersennenumb(z)
Definition: When 2^n -1 is prime, it is said to be a Mersenne prime. In general, it is a
Mersenne number.
To predict infinity of Mersenne numbers we proceed as follows:
Let f(i) = 2^n - 1. Then feeding this function into the canonical generating function [6] or
CGF(z):= 1/(z^f(i)-1) - 1/(z^f(i)*(z^f(i)-1)); .............................................(16).
Applying final value theorem to the above equation, where we let
Where (.......) = (z^(f(i)-1) + z^(f(i)-2) + ..................+ z + 1), we get
Final value of {Mersennenumb(z)|z=1}
............................= {(z-1)/z * ( 1/((z-1)*(.......)) - 1/(z^f(i)*(z-1)(.......))}
............................= 1/(z*(.......)) - 1/(z^(f(i)+1)*(.......)}
............................= 1/(.......){1/z - 1/ (z^(f(i)+1)}
............................= 0................................................................................(17).
This is totally unexpected but there is a pitfall in final value theorem since it predicts rates and
not absolute magnitudes. Zero can mean 1/infinity but as a rate it is finite. It measures the
average magnitude in the last interval which of course is infinity itself. This can be seen in the
second last line in equation (17) where within the { }, the first term is 1/z and the second term is
1/(z^(infinity+1)). Therefore the interpretation here should be that there is infinity in Mersenne
numbers.
The global formulation for Mersenneprime(z) can only be written down in mixed mode as shown
in equation (18):
Mersenneprime(z) : = { AND Normc(Mersennenumb(z)) Normc(Prime(z)) } ..........(18).
Again due to the presence of Normc( ) operator, the prediction cannot be made. However, a
close form mixed mode generating formulation is obtained via sequence algebra which is
algorithmic. An infinity in Mersenneprime(z) remains an open problem.
(f) Proof of an infinity of Fermat numbers: Fermatnumb(z) [12]
The proof of an infinity for Fermat numbers can proceed along the line for Mersennenumb(z)
but is more elaborate. For completeness, we reproduce the proof here.
Let f(i) = 2^(2^(2^n) - 1. Then feeding this function into the canonical generating function or
CGF(z):= 1/(z^f(i)-1)) - 1/(z^f(i)*(z^f(i)-1))); ................................................(19).
2^(2^n) grows big very fast but it is still only a numeric integer when expanded in the number
sequence and the steps are no different from that for Mersenne number.
Applying final value theorem to the above equation, where we let
Where (.......) = (z^(f(i)-1) + z^(f(i)-2) + ..................+ z + 1), we get
Final value of CGF(z) = {(z-1)/z * ( 1/((z-1)*(.......)) - 1/(z^f(i)*(z-1)(.......))}
..............................= 1/(z*(.......)) - 1/(z^(f(i)+1)*(.......)}
..............................= 1/(.......){1/z - 1/ (z^(f(i)+1)}
..............................= 0............................................................................(20).
Again using the same argument in Mersennenumb(z), there is infinity in Fermatnumb(z).
Again because of the presence of Normc( ) in the Fermat prime expression, it is impossible to
solve the infinity problem. Therefore infinity in Fermatprime(z) is still an open problem.
(g) Proof of an infinity of Perfect numbers [3]
Surprisingly the infinity problem for Perfect numbers needs not follow those of Mersenne and
Fermat numbers since there is a closed generating function for it.
Definition: A positive integer n is called a perfect number if it is equal to the sum of all of its
positive divisors, excluding n itself.
For example, 6 is the first perfect number because 6=1+2+3. The next is 28 = 1+2+4+7+14.
The next two are 496 and 8128. Note that Perfectnumb(z) includes both Perfect numbers and
nonPerfect numbers. The Perfect numbers are shown in bold in the output list in equation (23).
it takes the form m/z^n where m = n indicates a Perfect number. NonPerfect numbers are those
where m /= n. The symbol for real Perfect numbers is given by Perfect(z).
Perfectnumb(z) := series(sum( i/(z^i*(z^i-1)),i=2..ub),z=infinity, ub); .......................(21).
where ub is an integer upperbound. Notice that Perfectnumb(z) has the same form as Comp(z)
with the exception that there is a variable i in the numerator. We test a finite expansion with the
above expression:
Perfectnumb(z):=sort(series(sum(i/(z^i*(z^i-1)),i=1..500),z=infinity,500)); ..................(22).
Perfect numbers for the first three Perfect numbers are shown in bold.
..................................1..........1.........1.......3........1.......6.........1.........7.......4........8......1.......16
Perfectnumb(z) := O(----) + ---- + ---- + ---- + ---- + ---- + ---- + ---- + ---- + --- + --- + ---
....................................500.......2........3.......4.........5.......6.........7........8.......9......10.....11.....12
...................................z..........z.........z.......z..........z.......z.........z.........z........z.......z........z........z
...........1.....10......9.....15......1.....21......1......22.....11.....14.....1.....36.....6.......16.....13.....28
......+ --- + --- + --- + --- + --- + --- + --- + --- + --- + --- + --- + --- + --- + --- + --- + ---
............13...14.....15....16......17....18......19....20......21.....22....23......24.....25.....26....27....28
...........z......z.......z......z.........z......z.........z.....z.......z........z.......z.......z.......z.......z.......z......z
...............................................................................................................496
......+ ............................................................................................... + ------
................................................................................................................496
..............................................................................................................z
.....................................1
......+ ...................O(-------)
.......................................500 ............................................................................(23).
.....................................z
There is no explicit formulation for Perfect(z), only in expanded form as shown in equation (24):
Perfect(z) := 6/z^6 + 28/z^28 + 496/z^496 + ......... .......................................(24).
The infinity problem for Perfect number is solved directly without using CGF(z) because it is
already in closed form as follows:
Final value of Perfectnumb(z):= (z-1)/z * (i/(z^i*(z^i-1)))
..............................= (z-1)/z * i/( z^i * (z-1) * (z^(i-1) + z^(i-2) + .....+ z + 1)
..............................= 1/z * sum( i/(z^i * (z^(i-1) + z^(i-2) + .....+ z + 1)),i=2..infinity)
..............................= 1/z * sum( (..........),i=2..infinity) ................................(25).
when
..............i = 2: (.....) = z + 1
..............i = 3: (.....) = z^2+z +1
..............i = 4: (.....) = z^3+z^2+z+1.................................................................(26).
Therefore F.V. of Perfectnumb(z) = 2/2 + 3/3 + 4/4 + ........
................................= infinity .......................................................................(27).
The global generating function for Perfect(z) can be written down but it involves mixed mode
arithmetics given as follows:
Perfect(z) := NOT ( i/(z^i*(z^i-1)) -- i/z^i ); ..............................................(28).
What the above formulation does is that whenever the first term in the bracket equals the second
term, that term will be zero. Otherwise they are finite. Therefore when the whole expanded
sequence is complemented by NOT, only Perfect numbers will remain. Thus the infinity problem
of Perfect(z) itself is an open question until an arithmetic closed form formulation for Perfect(z)
can be derived.
(h) Proof of an infinity of Amicable Pairs [3]
Definition: An amicable pair is a pair of numbers each of which equals the sum of the
other's proper divisors; the members of amicalbe pairs are also called amicables.
The smallest amicable is a (220, 284). It is interesting to note that the expansion of
Perfectnumb(z) also include amicables.
...................................1.........1........1........3........1........6.........1.......7..........4.......8.......1.......16
Perfectnumb(z) := O(----) + ---- + ---- + ---- + ---- + ---- + ---- + ---- + ---- + --- + --- + ---
...................................500......2........3........4.........5........6..........7........8..........9......10.....11.....12
..................................z..........z..........z........z.........z.......z...........z..........z...........z.......z.......z......z
...........1.......166.....75......110......49.....384.....39....112.....77......284.....31.....234......1
......+ ---- + ---- + ---- + ---- + ---- + ---- + ---- + ---- + ---- + ---- + ---- + ---- + ----
............211....212....213.....214.....215.....216....217.....218.....219....220....221.....222....223
...........z........z........z.........z...........z........z.........z........z............z........z.........z........z........z
......+ ................................................................................................................+.......
.........396........1.....142....137.....440......1......294.....1......220....195....218......49.......531
......+ ---- + ---- + ---- + ---- + ---- + ---- + ---- + ---- + ---- + ---- + ---- + ---- + ----
............276....277....278....279....280....281....282....283....284.....285.....286.....287.....288
...........z.........z........z.........z........z........z........z..........z.........z.........z...........z........z........z
.................79.........510...........1
........+ -------- + -------- + -------- ............................................................................(29).
...................497........498........499
..................z............z.............z
Other than the above curious fact, there is no specific generating function special to amicable
numbers. Neither is there any suitable method by which amicables can be filtered from the
Perfectnumb(z) sequence. That there is an infinity of amicables is still an open question.
2. Conclusions
Sequence algebra is able to advance the infinity problems for some famous sequences but at
the same time it uncovers problems with mixed mode arithmetics. In general, if there is no
closed form generating function of the pure arithmetic type, the infinity problem cannot be
solved. There is one exception however since a infinity of primes has been solved by Euclid
more than 2000 years ago using a traditional method [1]. Other than that the infinity problems
for Odd composites, Mersenne primes, Fermat primes, Perfect number and Amicable numbers
still remain open problems. Sequence algebra has pinpointed where the problem lies but at
present there is no solution to the mixed mode problem. The author advocates that this is an item of priority in
mathematical research.
3. References
1. Burton M. Burton: Elementary Number Theory, Third Edition, WCB publishers, 1994, 17 to
48 to 50.
2. Caldwell C.K.
3. Meows David
4. Weyl, H. (1949): Philosoophy of Mathematics and Natural Science, Princeton University
Press, Princeton, New Jersy.
5. Huen Y.K.
6. Huen Y.K.
7. Huen Y.K.
8. Huen Y.K.
9. Huen Y.K.
10. Huen Y.K.: A Matrix Map for Prime and Non-prime Numbers, INT. J. Math. Educ. Sci.
Technol., 1994, VOL. 25, NO.6, pp 913-920.
11. Huen Y.K.: Some Interesing Properties Of The Natural Number System, Int. J. Math. Educ.
Sci. Technol., 1996, VOL.27, NO. 5, 685-691.
12. Huen Y.K.: Visual algebra and its applications, INT. J. Math. Educ. Sci. Technol.,1996,
VOL.??, NO.?, ???-??? (In the press as proof paper mes 100421).
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