Unsolved Problems In Sequence Algebra

by

Huen Y.K.

CAHRC, P.O.Box 1003, Singapore 911101
http://web.singnet.com.sg/~huens/
email: huens@mbox3.singnet.com.sg

(A short communication - 1st released: 6/6/97. Revised:5/7.)


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Abstract

In number theory, frequently solving one problem leads to discoveries of more unsolved problems. This is the same with sequence algebra. This paper describes seven unsolved problems in sequence algebra. It is probable that some solutions could only come from conventional number theory such as the topic of factorisation. Other problems are peculiar to sequence algebra and can only be solved within this domain. The best strategy is to keep both traditional method and sequence algebraic method open whilst in search of solutions. Although the stress is on algebraic solutions, if these cannot be solved, even mixed mode solutions should be considered acceptable.


1. Introduction


This paper describes seven unsolved problems in sequence algebra encountered by the author mostly from previous publications [3 to 11].


(a) Unsolved Problem 1: Direct formulation for Prime(z)


The author succeeded in deriving a direct global formulation for the composite number sequence given by equation (1) [ 9]:


...............................................................1
.......................................Comp(z) = ------------- .............................................(1).
...........................................................i...i
.........................................................z..(z..-.1)

The development was based on the historical Sieve of Erastothenes with a twist [1]. Instead of attacking the prime sequence directly, the author chose to attack the composite sequence. This turned out to be much easier as evidenced by the simple formulation.

From the sequence identity given in equation (2), one hopes to find the global prime sequence given by equation (3). Note that equation (3) does not involve any primality test and the predictions are determinstic.

...............Nat(z) = Normc(Comp(z)) + Prime(z) + 1 + 1/z ................(2)

This leads to


................Prime(z) = Nat(z) - Normc(Comp(z)) - 1 - 1/z ...................(3).

Comp(z) in equation (1) is called an analytical sequence whereas Prime(z) in equation (3) is called an algorithmic sequence. If sequences are in analytical format, one could continue to manipulate and build higher analytical sequences. On the other hand if a sequence is algorithmic, one could still develop efficient computer programs using it but continued analysis cannot proceed without the use of the operator Normc( ). Normc( ) itself is algorithmic rather than analytical. In order to solve this problem, one should find a direct sequence formulation for Prime(z). Of course one could write down Prime(z) if a primality test is admitted as shown in equation (4).

Prime(z)=sum(1/z^i*(isprime(i)-false)/(true-false),i=2..100);

........................1......1.........1........1.......1.......1.......1.......1.......1.......1.......1.......1.......1.......1
...Prime(z) := ---- + ---- + ---- + ---- + --- + --- + --- + --- + --- + --- + --- + --- + --- + ------
.........................2......3.........5........7......11.....13.....17......19.......23.....29.....31.....37.....41.....43
........................z......z.........z........z.........z.......z.......z.........z.........z.........z.......z.......z.......z.......z

............1......1.......1........1.......1.......1.......1.......1.......1.......1.......1
.......+ --- + --- + --- + --- + --- + --- + --- + --- + --- + --- + ------ .......................(4).
.............47.....53....59....61.....67.....71.....73.....79.....83.....89.....97
............z......z.......z........z........z.......z.........z.......z.......z.......z.......z

(Abstracted from Maple V R 3's help page for Isprime: The function isprime is a probabilistic primality testing routine. It returns false if n is shown to be composite within one strong pseudo- primality test and one Lucas test and returns true otherwise. If isprime returns true, n is ``very probably'' prime - see Knuth ``The art of computer programming'', Vol 2, 2nd edition, Section 4.5.4, Algorithm P for a reference and H. Reisel, ``Prime numbers and computer methods for factorization''. No counter example is known and it has been conjectured that such a counter example must be hundreds of digits long.)

Primality tests are probabilistic which are efficient but are algorithmic formulations which cannot be embedded in analytical formulations for further manipulations. This is because such tests return logical switches between true and false which function like mixed mode arithmetics in sequence algebra.



Problem Statement 1

Find a direct formulation for Prime(z) without using any primality test and without relying on Comp(z).
[On a scale where 0 = very easy and 100 = almost impossible, the degree of difficulty is about 95.]




(b) Unsolved Problem 2: Proof of an infinity of Prime(z) via sequence algebra

There are two types of sequence formulations, viz., analytical sequence formulation and algorithmic sequence formulation. Whenever the Normc( ) operator or mixed mode arithmetics appears, then the sequence formulation is algorithmic rather than analytical. The problem is directly related to unsolved problem 1. If problem 1 is solved, the solution of the present problem automatically follows using final value theorem [8]. At present, the proof of an infinity of Prime(z) is dependent on Euclid's prime number theorem [1].



Problem Statement 2

Prove that there is an infinity of Prime(z) via sequence algebra.
[On a scale where 0 = very easy and 100 = almost impossible, the degree of difficulty is about 95.]




(c) Unsolved Problem 3: Automatic normalisation of sequences in additions, subtractions and multiplications

Sequences which contain only unity values for numerator coefficients are labelled as normalised sequences. Primitive sequences such as Nat(z), Even(z) and Odd(z) are normalised sequences. Examples of Nat(z) and Even(z) are shown in expanded form in equations (5) and (6).

.........................................1.......1.......1.........1........1........1........1........1............1
........Nat(z) := 1 + 1/z + ---- + ---- + ---- + ---- + ---- + ---- + ---- + ---- + O(---) .............(5).
..........................................2.......3.......4..........5...........6.......7.......8.......9...........10
.........................................z.......z.......z..........z..........z..........z.......z.......z............z

..............................................1.......1......1..........1...........1
...................Even(z) := 1 + ---- + ---- + ---- + ---- + O(---) .................................(6).
...............................................2.......4.......6..........8..........10
..............................................z........z......z..........z...........z

In sequence addition such as in equation (7), the numerators are no more normalised. How does one write a direct formulation for addition of these two sequences such that the resultant output is automatically normalised?

SumNatEven(z) := Nat(z) + Even(z)

.................................2......1......2.......1.......2.......1.......2.......1............1
...........= 2 + 1/z + ---- + ---- + ---- + ---- + ---- + ---- + ---- + ---- + 2 O(---) ........(7).
..................................2......3......4.......5........6.......7.......8.......9............10
..................................z......z......z.......z........z......z........z.......z............z

If the normalised form for Nat(z)+Even(z) can be found, one could immediately derive its closed form as shown in equation (8):

........................Nat(z) + Even(z) = Nat(z) ............................................. (8).

This means that Even(z) is a subset of Nat(z) and the order of the summation is already determined by Nat(z).

Equation (9) shows sequence cross multiplication of Even(z)^2.

...................................................2......3......4.........5......4........3.........2......1
...........Evensquared(z) := 1 + ---- + ---- + ---- + ---- + --- + --- + --- + --- ................(9).
....................................................2.......4......6.........8.....10.......12......14....16
...................................................z......z......z.........z........z........z.........z.......z

This means that the order of Even(z)^2 is exactly the same as Even(z).

If Evensquared(z) is normalised, one could write down the closed form as:

.............Normc(Evensquared(z)) = Even(z) ...................................(10).

In fact there is a general identity for Even(z)^n given by:

.....................(Normc(Even(z)^n) = Even(z) .....................................(11).

The above analyses look strange but are useful in sequence order analysis where we are only interested in the order properties of sequences and not the numerator values. For example, in equation (11), it is shown that Even(z)^n has the same order as Even(z) even though the numerator values can be quite different.



Problem Statement 3

Find an algebraic method of automatic normalisation without using the operator Normc( ).
[On a scale where 0 = very easy and 100 = almost impossible, the degree of difficulty is about 90.]




(d) Unsolved Problem 4: Find a direct formulation for Oddcomp(z).

We know that

.........Normc(Comp(z)) = Evencomp(z) + EvenOddcomp(z) ...................................(12).

But EvenOddcomp(z) contains both even and odd composite terms. We need to abstract the odd terms from this sequence to get the pure odd sequences called Oddcomp(z).

Therefore ........EvenOddcomp(z) = Evencompx(z) + Oddcomp(z) ....................................(13).

Therefore

........Oddcomp(z) = Normc(Comp(z)) - Even(z) - Evencompx(z).........................(14).

This is an algorithmic formulation.



Problem Statement 4

Write down the direct formulation for Oddcomp(z) in sequence algebra.
[On a scale where 0 = very easy and 100 = almost impossible, the degree of difficulty is about 80.]




(e) Unsolved Probelm 5: Factorisation of finite primitve sequences

Nat(z), Even(z) and Odd(z) are defined as global primitive sequences since these are always normalised, i.e., all numerator coefficients have unity values. The unsolved problem relates to factorisability of these sequences. It is found that finite Nat(z) sequences with even number of terms can be factorised but not those with odd number of terms as shown from equation (15a) to (15f): However finite sequences with intervals greater than unity can all be factorised whether there are odd or even number of terms as shown from equation (16 to 19).

....................................................2
..................................................z + z + 1
..........factor(1+1/z+1/z^2) := -------------------- ........................................(15a).
............................................................2
...........................................................z

..................................................................... 2
..............................................................(z + 1) (z + 1)
..........factor(1+1/z+1/z^2+1/z^3) := --------------------- ............................(15b).
...........................................................................3 ..........................................................................z

....................................................................4.....3.....2
..................................................................z + z + z + z + 1
.......factor(1+1/z+1/z^2+1/z^3+1/z^4); -----------------------...........................(15c).
..............................................................................4 ............................................................................z

A:=factor(1+1/z+1/z^2+1/z^3+1/z^4+1/z^5);

............................................................2...............2
..............................................(z + 1) (z - z + 1) (z + z + 1)
......................................= -------------------------------------..........................(15d).
.....................................................................5
...................................................................z

A:=factor(1+1/z+1/z^2+1/z^3+1/z^4+1/z^5+1/z^6);

...............................................6.....5.....4.....3.....2
..............................................z..+.z..+.z..+.z..+.z..+.z.+.1
...................................A := -------------------------------- ...............................(15e).
....................................................................6
..................................................................z

A:=factor(1+1/z+1/z^2+1/z^3+1/z^4+1/z^5+1/z^6+1/z^7);

..............................................................2..........4
................................................(z + 1) (z + 1) (z + 1)
...................................A := --------------------------- ....................................(15f).
..................................................................7
................................................................z

On the other hand all Even(z) and Odd(z) with number of terms greater than two can be factorised. The conjecture states that all normalised sequences above Nat(z) with uniform intervals can be factorised.

Here are some examples:

factor(1/z^1+1/z^3+1/z^5+1/z^7+1/z^9+1/z^11);

................................2...............2..........2...............4....2
.............................(z..-.z.+.1) (z..+.1) (z..+.z.+.1) (z..-.z..+.1)
......................... ------------------------------------------ ..............................(16).
...........................................................11
..........................................................z

factor(1+1/z^4+1/z^8+1/z^12+1/z^16+1/z^20+1/z^24);

...6....5....4.....3.....2...............6....5.....4.....3.....2................12...10....8....6.....4....2
(z..-.z..+.z..-.z..+.z..-.z.+.1) (z..+.z..+.z..+.z..+.z..+.z.+.1) (z...-.z...+.z..-.z..+.z..-.z..+.1)
-----------------------------------------------------------------------------------...............(17).
...................................................................24
.................................................................z

factor(1/z^1+1/z^6+1/z^11+1/z^16+1/z^21+1/z^26);

....................2...............2..............4.....3.....2..............8......7....5....4....3
......(z.+.1) (z..+.z.+..1)(z..-.z.+.1)(z...-.z..+.z..-.z.+.1)(z...+.z..-.z..-.z..-.z..+.z.+.1)
............8....7.....5....4.....3........../..26
..........(z - z + z - z + z - z + 1) /..z ...........................................................(18).
............................................../



factor(1+1/z^5+1/z^10+1/z^15+1/z^20+1/z^25+1/z^30);

.....6.....5.....4....3.....2
..(z..+.z..+.z..+.z..+.z..+.z.+.1)

........24...23....19....18....17...16...14...13....12...11....10....8....7....6.....5
......(z...-.z...+.z...-.z...+.z...-.z...+.z...-.z...+.z...-.z...+.z...-.z..+.z..-.z..+.z..-.z.+.1)

........./..30
......../..z .......................................................................................................(19).
......./



Problem Statement 5

Prove that all normalised sequences above Nat(z) with uniform intervals can be factorised but only finite Nat(z) sequences with even number of terms can be factorised.
[On a scale where 0 = very easy and 100 = almost impossible, the degree of difficulty is about 80.]

Comments: An open sequence with uniform intervals can be expressed as:

F(z) = sum(1/z^(m + n*i), i = 0..ub);

where m is the initial term and n is the interval and i is the range with ub its upperbound.

For example :........Nat(z) = sum(1/z(0+i),i=0..ub);
..............................Even(z) = sum(1/z^(0+2*i),i=0..ub);
..............................Odd(z) - sum(1/(z^(1+i),i=0..ub); ...........................(20).





(f)Unsolved Problem 6: Find the direct formulation for Perfectnumb(z). [2]

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.

Perfectnumb(z) can be detected from the following sequence formulation but it is not specific to Perfect numbers.

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).

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 Perfectnumb(z).



Problem Statement 6

Find an explicit sequence generating function which will generate all the Perfect numbers as shown in equation (24).
Perfect(z) := 1/z^6 + 1/z^28 + 1/z^496 + ......... ...................................(24).
[On a scale where 0 = very easy and 100 = almost impossible, the degree of difficulty is about 95.]




(g)Unsolved Problem 7: Amicable Pairs [2]

Definition: An amicable pair is a pair of numbers each of which equals the sum of the other's proper divisors; the members of amicable pairs are also called amicables.

The smallest amicable pair is a (220, 284). It is interesting to note that expansions 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
........+ -------- + -------- + -------- ............................................................................(25).
...................497........498........499
..................z............z.............z



Problem Statement 7

Find the explicit generating function which will generate all the amicable pairs.
[On a scale where 0 = very easy and 100 = almost impossible, the degree of difficulty is about 98.]




(g)Unsolved Problem 8: Every finite Nat(z) sequence in which the number of terms equal to a prime integer value is unfactorisable.

Definition: An finite Nat(z) sequence is a normalised sequence. Such sequences which contain numbers of terms which are primes are found to be unfactorisable. There is no rigorous mathematical proof for this. Some examples are shown below.

factor(1/z + 1):=(z + 1)/z

factor(1/z^2 + 1/z + 1):=(z^2+z+1)/z^2

factor(1/z^4 + 1/z^3 + 1/z^2 + 1/z + 1 ) := (z^4 + z^3 + z^2 + z + 1)/z^4

factor(1/z^6+1/z^5+1/z^4+1/z^3+1/z^2+1/z+1):=(1/z^6+1/z^5+1/z^4+1/z^3+1/z^2+1/z+z+1)/z^6


Problem Statement 8

Every finite Nat(z) sequence with prime number of terms is unfactorisable.
[Scale of difficulty: This is probably not too difficult. There might be a proof somewhere which the author is not awared of. Readers are welcome to report if they come across it.]




2. Conclusions

The above are labelled unsolved since no explicit generating functions have yet been found at present. Sequence algebra is holistic. The properties of interest are holistic properties which are not normally treated in traditional number theory. It shows that with the introduction of a new axiom, we can solve new problems but we also bring in new unsolved problems.

3. References

1. Burton M. Burton: Elementary Number Theory, Third Edition, WCB publishers, 1994, 17 to 48 to 50.

2. Meows David : Perfect, amicable and socialbe numbers, http://xraysgi.ims.uconn.edu:8080/amicable.html

3. Huen Y.K.: A Simple Introduction To Sequence Algebra, URL site: http://web.singnet.com.sg/~huens/

4. Huen Y.K.: The Canonical Generating Function or CGF(z) - a Swiss-knife function. URL site: http://web.singnet.com.sg/~huens/ .

5. Huen Y.K.: Information Contents Of Number Theoretic Functions. URL site: http://web.singnet.com.sg/~huens/ .

6. Huen Y.K.: In Search Of Exotic Arithmetic Operators, URL site: http://web.singnet.com.sg /~huens/ .

7. Huen Y.K.: Visual Solutions Of Number Theoretic Functions in Multidimensional Sequence Space, URL site: http://web.singnet.com.sg /~huens/ .

8. Huen Y.K.: Final Value Theorems Applied To Number Sequences -- its strengths and weaknesses, URL site: http://web.singnet.com.sg /~huens/ .

9. 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.

10. Huen Y.K.: Some Interesing Properties Of The Natural Number System, Int. J. Math. Educ. Sci. Technol., 1996, VOL.27, NO. 5, 685-691.

11. 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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