1. Theoretical Background

Number sequences in sequence algebra are specified holistically. That is why this is called holistic number theory. This does not imply that traditional number theory is non-holistic as there are many famous theorems which predict number theoretic properties holistically [1]. The only reason for calling this holistic number theory is because holistic number sequences are used as algebraic variables in sequence algebraic functions and are manipulated as such [3 to 18]

Holistic number theory has its strengths and weaknesses. It can do things which traditional number theory cannot do but it fails in some applications which look easy in the latter. It is neither a cure-all nor an avant-garde breakway from traditional number theory. It is more beneficial to consider that number theory now has both reductionistic and holistic paradigms. Number theorists should treat these two as customised tools for customised applications just like surgeons who know instinctively which tool to use in which situation.

Only one axiom of Nat(z) has been added to the existing axiomatic set and yet the horizon has been greatly widened. Maybe Chaitin and Godel were right: that mathematicians should admit new axioms so long as these can help to solve problems rather than insisting on self-evident truth [2]. Many new axioms which revolutionised sciences did not come from self-evident truth. For example, how did we know that our genes are constructed as double helices before Watson and Crick's discovery? Now it looks self-evident. Was the fact that time can be compressed and stretched self-evident before Einstein? It does not look self-evident even now. If only mathematicians are more receptive to the adoption of axioms following the footsteps of empirical scientists, maybe some existing unsolved number theoretic problems which frustrated them over the past two millenia could find solutions. We are already seeing some of these happening in sequence algebra. Of course sequence algebra ushers in more unsolved problems of its own making [9]. Knowledge is recursive. The world will be very boring indeed without the challenges of unsolved problems and the discoveries of new ones. Godel's incompleteness theorem ensures that mathematicians and scientists will always be in business.


2. Order-Analysis

This is probably a unique contribution from holistic number theory.

Let us expand Nat(z) by Maple V R 3 into a finite Laurent series as shown in equation (1). Nat(z) is a primitive number sequence since it expands into a normalised sequence, i.e., all numerator coefficients have values of unities. In this expansion, the open form of Nat(z) is used because the closed form of Nat(z) = z/(z-1) would have generated a truncation term of the form O(--) which proliferates when raised to integer powers.

Nat(z):=sum(1/z^i,i=0..9);

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


If we square the above Nat(z) sequence we get an unnormalised sequence as shown in equation (2). Notice the progressive increment of the numerator coefficients or duplicity factors up to a maximum at mid-sequence. The distribuition of duplicity factors is symmetrical about the mid- term shown in bold in Nat2(z) below.

Nat2(z):=sort(expand(Nat(z)^2));

............................3.......4.......5......6......7.......8......9.....10.....9......8.....7......6......5
Nat2(z) :=1+2/z+----+----+----+----+----+----+----+----+--- +--- +--- +--- +---
..............................2.......3.......4.......5.....6.......7......8......9.....10....11....12...13...14
............................z.......z........z.......z......z.......z.......z.......z......z......z......z......z......z


........................4.......3......2........1
...................+ --- + --- + --- + ------ ..........................................................(2).
.........................15.....16.....17......18
........................z.......z.......z........z


If one is only interested in the sequence order of Nat(z), it is obvious that raising it to any integer power does not change this order. This can be easily tested by normalising Nat2(z) by Normc( ) using equation (3):

..................Normc(Nat2(z)) := Nat(z) ................................................(3).


The general order identity for Nat(z) is given by equation (4) for all positive integer values of n.


..................Normc(Nat(z)^n) := Nat(z) ..............................................(4).


In other words, Nat(z) is an order-invariant primitive sequence. Not all primitive sequence are order-invariant. We know that Even(z) is order invariant as shown by equations (5a) and (5b) but Odd(z) is order-variant as demonstrated in equations (6a) and (6b).

.....................................1........1........1.......1........1.......1.......1........1.......1.......1
............Even(z) := 1 + ---- + ---- + ---- + ---- + --- + --- + --- + --- + --- + --- ..........(5a).
.......................................2........4........6........8......10......12......14.....16....18.....20
.....................................z........z.........z........z........z........z........z........z.......z.......z


.........Normc(Even(z)^2):= Even(z) ...............................................................................(5b).


.......................................1........1........1.........1........1.......1.......1.......1......1
.............Odd(z):= 1/z + ---- + ---- + ---- + ---- + --- + --- + --- + --- + ----- .................(6a).
.........................................3.........5........7........9.......11.....13......15.....17......19
.......................................z.........z.........z.........z........z.......z........z........z.......z


Note that Odd(z)^2 is of even order. In fact Odd(z)^n is of even order if n is even and odd order if n is odd but the sequence itself is also shifted by n unit intervals from the original Odd(z) sequence. Therefore Odd(z) is not an order-invariant sequence. The asymmetries between Even(z) and Odd(z) is interesting. It shows that even though Even(z) and Odd(z) are concomittent, i.e., the sum of the two equals Nat(z), one can never generate the Odd(z) sequence directly from the Even(z) sequence by exponentiation. In reverse, partial Even(z) sequence can be generated by the exponentiation of the Odd(z) sequence. Equation (6b) shows that Odd(z)^2 does not start from the integer zero which is the reason why it is not order-invariant.

..................................1.......1.......1......1.......1......1......1......1.......1.......1.......1......1.....1......1
Normc(Odd(z)^2):=----+ ----+ ----+ ----+ ---+ ---+ ---+ ---+ --- + --- + --- + ---+ --- + ---
....................................2......4........6.......8.....10.....12....14....16....18.....20.....22...24....26....28
..................................z.......z........z.......z........z.......z.......z.....z.......z.......z........z.....z......z......z


. ............................1.......1......1.......1.......1
.........................+ --- + --- + --- + --- + ----- ......................................................(6b).
...............................30......32.....34.....36....38
..............................z.......z........z.......z.......z


Another interesting fact about Odd(z) is that when raised to an infinite integer power, it approaches asymptotically to Zero(z) the global null sequence.

...........................Normc(Odd(z)^infinity) -----------> Zero(z) .....................................(7).


Where.................Zero(z) = {0000000000000000............................} ........................(8).


Of course Maple V R 3 will simply report Zero(z) as a single 0. Thus we have the following observations:

(i) Nat(z) and Even(z) are order-invariant primitive sequences. It is a trivial fact that the global null sequence is also order-invariant.

(ii) Odd(z) is not order-invariant. In fact should we call Odd(z) a primitive sequence at all since its property is so different from the previous two?

(iii) One can generate a partial Even(z) sequence by raising Odd(z) to even powers but not vice versa for Even(z), i.e., one can never generate an Odd(z) sequence by raising Even(z) to any integer power. Of course Odd(z)^n = (Even(z)/z)^n but this is a different approach.

(iv) Odd(z)^n approaches Zero(z) asymptotically as n approaches infinity.


3. Lemmata, Colloraries, And Theorems

In sequence order analysis, we are interested to determine the contiguity and order-invariance of number sequences. By definition, order analysis is only applicable to number sequences generated by closed sequence algebraic generating functions. Such sequences must be normalised by Normc( ) before order analysis.

The holistic generation of a number sequence by its closed form defines the order contiguity of successive integers. For example Nat(z) = z/(z-1) will generate the contiguous integer set {1,1,1,1,1,1........} whereas Even(z) = z^2/(z^2-1) will generate the contiguous integer set {1,0,1,0,1,0,............}. Thus the contiguity of every unique number sequence is unique. The only sure test on whether the sequence order has changed is to examine its closed generating function after processing. An example will make this clear.

Example 1: Check the contiguity of Odd(z)^2.


Solution: Since Odd(z)^2 will generate an even number sequence, there is no point in choosing an Odd(z) as a reference. We should therefore choose Even(z)/z^2 as the reference sequence. If Odd(z)^2 is contiguous, it will subtract exactly from Even(z)/z^2. Equations (9) to (12) describe the procedures. Equation (12) shows that Odd(z)^2 generates Even(z)/z^2 which is an even order sequence but it does not start from integer zero. Only odd power of Odd(z) will generate odd sequences.

Odd(z):=sum(1/z^(2*i-1),i=1..9);


...........................................1.........1.........1........1........1......1.......1.......1.......1
..................Odd(z) := 1/z + ---- + ---- + ---- + ---- + --- + --- + --- + --- +---- ..........(9).
..............................................3.........5........7........9.......11....13.....15.....17.....19
...........................................z..........z.........z........z.........z......z........z.......z.......z


Odd2(z):=sort(expand(Odd(z)^2));


........................1.......2.........3.......4........5......6.......7.......8.......9......10......9.......8.......7......6
...Odd2(z) := ---- + ---- + ---- + ---- + --- + --- + --- + --- + --- + --- + --- + --- + --- + ---
..........................2.......4........6........8........10.....12....14....16......18.....20....22.....24.....26....28
........................z........z........z........z........z.........z......z.......z.........z.......z.......z.......z.......z.......z


......................5......4......3......2......1
................+ --- + --- + --- + --- + --- ..............................................(10).
......................30......32....34....36....38
.....................z........z.......z......z......z


Even(z)/z^2:=series(1/(z^2-1),z=infinity,39);

...........1........1........1........1......1.......1......1.......1.......1.......1.......1.......1.......1.......1
.....= ---- + ---- + ---- + --- + --- + --- + --- + --- + --- + --- + --- + --- + --- + ----
.............2........4........6........8......10.....12.....14.....16....18.....20.....22.....24....26....28
...........z........z.........z........z.......z........z.......z.......z........z.......z.......z........z.......z.......z


...............1......1........1.......1......1
...........+ --- + --- + --- + --- + --- ....................................................(11).
................30.....32......34.....36.....38
...............z.......z........z.......z.......z


Error(z):= Even(z)/z^2 - Normc(Odd2(z)) = 0. .............................(12).

Followed immediately are some lemmata, colloraries, and theorems relevant to order analysis.


Lemma 1: Nat(z) is order invariant.


.............Proof: Normc(Nat(z)^n) = Nat(z). ...............................................(13).


Lemma 2: Even(z) is order invariant.


.............Proof: Normc(Even(z)^n) = Even(z). .............................................(14).


Lemma 3: Odd(z) is not order invariant.


..............Proof: Normc(Odd(z)^n) /= Odd(z). .........................................(15).


This resultant sequence can be either odd or even depending on whether n is odd or even but it is shifted from 1/z by n unit intervals. Therefore Odd(z) is not order-invaraint.


Corollary 1: Normc(Odd(z)^infinity) -----> Zero(z). .................................(16).


.............Proof: Odd(z) would have shifted by an infinity number of unit intervals to the right so that what is left approaches asymptotically to Zero(z).


Corollary 2: Zero(z) is order invariant.


............Proof: This is because Zero(z) starts from integer 0 and has uniform intervals.


Lemma 4: Every integer number can be represented by a CGF(z), i.e. a closed canonical generating function.


............Proof: Take any arbitrary integer number represented by N(k) = 1/z^k. This can be represented using the CGF(z) as follows:


N(k) = 1/(z^k-1) - 1/(z^k*(z^k-1)) ...............................................(17).


Test: Find the CGF(z) of N(22307).


N(22307):=series(1/(z^22307-1)-1/(z^22307*(z^22307-1)),z=infinity,1223800);


Note that even though 1223800 terms of expansion are asked for, only one integer is generated as given below:


............................................................1......................1
...................................N(22307) := ---------- + O(-----------) ...........................(18).
.............................................................22307...............44614
...........................................................z........................z


Lemma 5: The difference between two order-invariant sequences cannot be order- invariant.


Proof: By definition, sequences are always assumed normalized in sequence analysis. Since both sequences have a zero integer each in the first term subtraction will eliminate it. Then the resultant difference sequence cannot be order invariant.


Lemma 6: All sums formed by sequences with Zero(z) remain unchanged. All products of sequences with Zero(z) become Zero(z).


.........Proof: Although Maple will not output the null sequence, strictly a global null sequence should be represented as shown in equation (19):


......................................0........0........0........0........0........0.........0........0.......0
...................Zero(z) := ---- + ---- + ---- + ---- + ---- + ---- + ---- + ---- + --- ..................(19).
........................................0.........1........2........3.......4.......5........6........7........8
......................................z.........z........z.........z........z........z........z........z.........z


Whatever sequence is summed with the above sequence, nothing will change. However any sequence multiplied by Zero(z) will become Zero(z) since the numeator products are all zeroes.


Lemma 7: All sequences in the general form z^i/(z^i-1) are order invariant.


.........Proof: An order-invariant sequence must have a zero integer as the first term and must have uniform intervals. This is satisfied by the above general form. Therefore all sequences represented by this form will be order invariant.


Tests: It can be seen that i determines the intervals between successive integers but all expanded sequences start from 1/z^0 or simply the zero integer. Note that 1/z^0 is not the same as 0/z^0 where the latter actually represents the absence of a zero so that such a sequence will not be order-invariant.


N(z):=series(z^2/(z^2-1),z=infinity,20);


..........................................1........1........1........1........1.......1.......1......1.......1............1
......................N(z) := 1 + ---- + ---- + ---- + ---- + --- + --- + --- + --- + --- + O(---) .....(20).
............................................2........4........6........8.......10......12....14.....16.....18.........20
..........................................z.........z.........z........z........z........z.......z........z........z...........z


N(z):=series(z^3/(z^3-1),z=infinity,20);


Note that when i is odd, the output sequence contains alternative odd and even terms. Such a sequence is neither odd nor even.


...................................................1........1.........1.......1.......1.......1...........1
...............................N(z) := 1 + ---- + ---- + ---- + --- + --- + --- + O(---) .................(21).
....................................................3..........6........9.......12......15.....18.......21
...................................................z.........z.........z........z........z........z..........z


N(z):=series(z^4/(z^4-1),z=infinity,30);


........................................1........1.......1.......1.......1.......1.......1..........1
....................N(z) := 1 + ---- + ---- + --- + --- + --- + --- + --- + O(---) ....................(22).
..........................................4........8......12......16......20......24.....28.......32
........................................z.........z.......z........z........z........z.......z...........z


N(z):=series(z^5000/(z^5000-1),z=infinity,20000);


.................................................1.........1.............1....................1
...........................N(z) := 1 + ----- + ------ + ------ + O(--------------) .....................(23).
..................................................5000.....10000.....15000.........20000
................................................z...........z.............z.................z


Theorem 1: Normc(Even(z)*Odd(z)) = Odd(z).


Before the proof, let us test with Maple V R 3 a short example:


Test:


Even(z):=sum(1/z^(2*i),i=0..10);


..............................1........1........1........1........1.......1......1.......1.......1.......1
.....Even(z) := 1 + ---- + ---- + ---- + ---- + --- + --- + --- + --- + --- + ---............(24).
................................2........4........6.........8......10......12.....14....16.....18......20
..............................z.........z........z.........z........z........z.......z.......z.......z........z


Odd(z):=sum(1/(z^(2*i-1)),i=1..10);


...................................1.........1.......1........1........1.......1.......1......1......1
........Odd(z) := 1/z + ---- + ---- + ---- + ---- + --- + --- + --- + --- + --- ................(25).
.....................................3........5........7.........9.......11.....13.....15.....17....19
...................................z.........z........z.........z........z........z.......z.......z.......z


sort(expand(Even(z)/z));


This shows that this will generate a normalised Odd(z) sequence.


.........................1.........1.......1.........1.......1........1......1.......1......1.......1
...............1/z + ---- + ---- + ---- + ---- + --- + --- + --- + --- + --- + --- ..................(26).
...........................3.........5.......7.........9......11......13....15.....17....19.....21
.........................z.........z........z..........z.......z........z.......z.......z.......z........z


. sort(expand(Even(z)*Odd(z)));


The output sequence is unnormalised which is still a odd sequence. If this is normalised by Normc( ), it is still an Odd(z) sequence.


................2.......3.......4.......5.......6......7.....8.......9......10.....10......9......8......7......6......5
.....1/z + ----+ ----+ ----+ ----+ ---+ ---+ --- + --- + --- + --- + --- + --- + --- + --- + ---
..................3........5.......7.......9......11....13.....15....17.....19.....21.....23.....25....27.....29....31
................z........z........z.......z.......z.......z.......z.......z.......z.......z........z.......z.......z.......z......z


.. ................4......3......2......1
............+ --- + --- + --- + ---
.................33.....35.....37.....39
................z......z.......z.......z .......................................................................(27).


Proof: In Even(z)* Odd(z), all that is required is to multiply Even(z) by 1/z, i.e., the first term from Odd(z). The remaining terms from Odd(z) will increase duplicity but will not alter the sequence order which is already set by multiplying Even(z) by the first term of 1/z.


Theorem 2: Normc(Nat(z)*Even(z)) = Nat(z)


Proof: Every even term multiplied into Nat(z) will convert the even term to odd and the odd to even. The only items affected are the duplicity factors. However, since the product sequence is normalised, then Nat(z) will still remain as Nat(z) which proves the above theorem.


Test:


...............................1........1........1........1........1.......1......1......1.......1.......1
......Even(z) := 1 + ---- + ---- + ---- + ---- + --- + --- + --- + --- + --- + --- ..........(28).
.................................2.........4.......6.........8......10......12....14....16.....18.....20
...............................z.........z........z.........z........z........z.......z.......z.......z.......z


..........................1........1........1........1.......1.........1.........1.......1.......1.......1.......1.......1.......1
Nat(z) := 1/z + ---- + ---- + ---- + ---- + ---- + ---- + ---- + ---- + --- + --- + --- + --- + ---
............................2........3........4.........5.......6.........7.........8.......9......10.....11.....12.....13.....14
..........................z.........z........z.........z.......z.........z..........z........z........z.......z.......z......z......z


..........1......1.......1.......1.......1.......1
......+ --- + --- + --- + --- + --- + --- ................................................................(29).
...........15.....16.....17.....18.....19.....20
..........z.......z........z.......z........z.......z


Nat(z)*Even(z):=


..........1.........2........2........3........3.........4........4.........5.......5.......6.......6.......7.......7......8......8
1/z + ---- + ---- + ---- + ---- + ---- + ---- + ---- + ---- + --- + --- + --- + --- + --- + --- + ---
.............2.........3........4.........5.......6.......7.........8..........9......10....11.....12.....13.....14.....15....16
...........z..........z........z........z........z........z..........z.........z........z........z.......z.......z.......z.......z.......z


........+....................................................................+................


...........4......4.......3......3.......2......2.......1.......1
.....+ --- + --- + --- + --- + --- + --- + --- + --- .................................................(30).
..........33.....34.....35.....36.....37.....38.....39......40
.........z.......z........z.......z........z........z.......z........z


It can be seen that Normc(Nat(z)*Even(z)) = Nat(z) which validates the theorem.


Theorem 3: Normc(Nat(z)*Odd(z)) = Nat(z)/z


Proof: Although Nat(z) is order invariant, Odd(z) is not. Therefore when the first term of Odd(z), i.e. 1/z is multiplied into Nat(z), it will shift Nat(z) by one unit interval to the right. The remaining terms from Odd(z) could contribute either even or odd terms but these do not alter the order of Nat(z), only the duplicities.


Theorem 4: Normc(Nat(z)*Comp(z)) = Nat(z)/z^4


Proof: The first term in Comp(z) is 1/z^4. When multiplied into Nat(z), it shifts the Nat(z) sequence by 4 unit intervals to the right. The remaining terms from Comp(z) only contribute to the duplicity factors and not the order. Thus the output sequence is Nat(z)/z^4.


Theorem 5: Normc(Nat(z)*Prime(z)) = Nat(z)/z^3


Proof: The proof is similar to Theorem 6 with the exception that the first term from Prime(z) is 1/z^3. Therefore the output sequence is Nat(z)/z^3.


Theorem 6: Normc(Comp(z)^2) = Even(z)(1/z^8+1/z^13);


Proof: We know that Comp(z) contains the sequence Even(z)/z^4 but we do not know the function for Oddcomp(z) but we know the first term starts from 1/z^9. When squaring Comp(z), the even sequence will start from Even(z)/z^8 and the odd sequence will start from Even(z)/z^13. Hence the above identity is proved.


Theorem 7: Oddcomp(z) := Normc(Comp(z)) - Even(z)/z^4


Proof: All even integers above 2 are composites. What are left are the odd composites.


Theorem 8: Odd(z) := Prime(z) + Oddcomp(z) + 1/z


Proof: Prime(z) starts from 1/z^3 and is always odd but never composite. Oddcomp(z) is always composite but never prime. Therefore the sum of the two sequences must add up to Odd(z) with the exception of the term 1/z which represents integer 1.


5. Some Interesting Identities


Here are some interesting sequence order identities arising from theorems proved in the last section. Most of the proofs are very short and can be completed mentally if you know how. Remember that you are handling normalised sequences in order analysis. The short form N( ) will represent Normc( ).

(i) Prove that N((Nat(z)-Even(z))*(Nat(z)-Odd(z))) := Odd(z).
Proof: Nat(z)-Even(z) = Odd(z) and Nat(z)-Odd(z) = Even(z). Then by Theorem 1, N(Odd(z)*Even(z)) = N(Odd(z)) = Odd(z).

(ii) Prove that (N(Even(z)^3)-N(Odd(z)^4)) := Even(z)*(1 - 1/z^4)
Proof: N(Even(z)^3) = Even(z) and N(Odd(z)^4)=Even(z)/z^4. Therefore (N(Even(z)^3)-N(Odd(z)^4))=Even(z)*(1-1/z^4).

(ii) Prove that N((Nat(z)-1)*N(Nat(z)+1))):= Nat(z)/z
Proof: Nat(z)-1 = Nat(z)/z and N(Nat(z)+1) = Nat(z). Therefore N((Nat(z)-1)*N(Nat(z)+1))):= N(Nat(z)^2/z) = Nat(z)/z.

(iv) Prove that N((Even(z) - 1)*(Odd(z) - 1/z)):= Odd(z)/z^4
Proof: Even(z) - 1 = Even(z)/z^2 and Odd(z)-1/z = Odd(z)/z^2. Therefore N((Even(z) - 1)*(Odd(z) - 1/z)):= (Even(z)*Odd(z))/z^4 which by Theorem 1 is Odd(z)/z^4.


6. Conclusions


This collection of lemata, corollaries, theorems, and identities are not exhaustive but these reveal some interesting order properties on primitive sequences. Order analysis cannot be carried out properly without the use of the normalising operator Normc( ). Order analysis is analogous to dimensional analysis in mechanics.


7. References


1. Burton M. D.: Elementary Number Theory, Third Edition, WCB publishers, 1994, most parts of this book.

2. Chaitin G.J.: Godel's Theorem and Information, International Journal of Theoretical Physics 22 (1982), pp 941-954.

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.: Unsolved Problems In Sequence Algebra, URL site: http://web.singnet.com.sg /~huens/ .

10. Huen Y.K.: Explicit Formulation For Modular Arithmetic In Sequence Algebra, URL site: http://web.singnet.com.sg /~huens/ .

11. Huen Y.K.: Cyclic Generating Functions In Sequence Algebra, URL site: http://web.singnet.com.sg /~huens/ .

12. Huen Y.K. : Methods Of Developing Sequence Algebraic Formulations For Comp(z) and Prime(z). URL site: http://web.singnet.com.sg /~huens/.

13. Huen Y.K. : Information Contents Of Hypothetical DNA Sequences. URL site: http://web.singnet.com.sg/~huens/ .

14. Huen Y.K. : Composite Number Sequence Challenge 1/97. URL site: http://web.singnet.com.sg/~huens/ .

15. 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.

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

17. 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).

18. Huen Y.K.: Twin primes revisited: INT. J. Math. Educ. Sci. Technol., 1997, VOL.??,NO.?, ???-???. (In the press as proof paper mes 100488).

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