MUDr. Tom Hermann Tietken
Nikolaus Graf zu Castell-Castell

Prague, Aug. 2020

The new TIETKEN-CASTELL-PRIME-ALGORITHM

Abstract

In sections 1 to 11 of the article, the implementable Castell-fact-algorithm was developed further. It factorizes large integers, yet, tends to be rejected by representatives of both business and politics due to the fact that it is ignored. Insights into data protection seem to be in demand neither from the NSA nor from Facebook and other businesses/institutions.
In the following section 12 of the article, the Tietken-Castell-Prim-Algorithm is presented, which, indirectly,
(a) Can produce prime numbers and
(b) Can identify prime numbers, and which
(c) Is also suitable for factorizing large integers (which are composed of prime numbers) after a correspondingly extensive list of numbers has been generated.

Objective

If for the factorization of large numbers large numbers are presented, which were reliably obtained by multiplying two prime numbers, the following question does not arise.
However, if it is questionable whether the present “large number” is a product of two prime factors or if it is itself a prime number (which cannot be factorized) or perhaps just a product ending with the digits 1, 3, 7 or 9, there must be a reliable way of clarifying what the present large number represents.
　
Taking a look at the many previous solutions that did not work reliably and unambiguously it is obvious that a new approach is needed here, one without uncertain multiple attempts and without error rates (Fermat, Miller-Rabin, Chinese remainder theorems, Mersenne conjecture, and many others) and without probability calculus, approximations and repetitions, etc., one which will generate more than just a “most likely prime”.

Even the RSA (Ron Rivest, Shamir, Adleman) uses only “random numbers”, which do not provide 100% prime numbers, but so-called “probable prime numbers”.

With the new idea of the TIETKEN-CASTELL-PRIME-ALGORITHM, the user is spared the above-mentioned uncertainties and inaccuracies. It unambiguously delivers “100% prime numbers” at all times and in all orders of magnitude.

Furthermore, it avoids the disadvantage, which is obvious in the course of RSA multiplication of two identical prime numbers, namely, that due to the big jumps in numbers (cf. the situation when big prime numbers are multiplied by themselves) other, smaller prime numbers, which lie below this jump, are overlooked and ignored. In short, the difficulty is that disarray arises when it comes to the question, which other prime numbers exist. If there was total clarity and transparency, no encryptor would have to multiply “probable” prime numbers by themselves, to be on the safe side.

Personal preliminary remark as regards the TIETKEN-CASTELL-PRIME- ALGORITHM

The new Tietken-Castell-prime-algorithm presented here does not only allow identifying prime numbers in an indirect way, it can also produce them in an indirect way.
The solution that proposes the new procedure mentioned above is simple and rather obvious.
　
To produce a prime number, the length of which would fill a dozen file folders, may be an impressive demonstration of the quantitative possibilities of computing machines. However, apart from some initial thoughts about the algorithm for the still relatively small prime numbers at the beginning, it has nothing to do with qualitatively increasing human thinking (see Gimps and 400-year-old Mersenne numbers and others). It is a demonstration, which involves high electricity costs and minimal human effort (one, which, currently, is still useless).
　

　
Verbal explanation of the TIETKEN-CASTELL-PRIME-ALGORITHM

The basic idea as regards the present Tietken-Castell-prime-algorithm is to indirectly achieve the still existing impossibility of quickly recognizing and producing prime numbers.
　
(a)
If prime numbers cannot be produced in a simple and correct way, they can be produced by using counterparts which are created by the differential method and which – in the ideal scenario (as it is the case here) – are exactly calculable. Here, these are the well-ordered counting factors which, on the one hand, remain constant and, on the other hand, the counted factors of the non-prime numbers, which are cardinally increasing in two steps and are also increasing in a well-ordered and comprehensible way, and/or of the products close to the prime numbers searched for.
　
Then, the prime numbers are the “remainders”. They are the numbers for which there are no factors!

(b)
If prime numbers can be “produced” in this indirect way, then it is in this way that these prime numbers can be identified by their position.
Because of the simplicity and due to the fact that all steps and procedures remain the same, there is a sequence of prime numbers that is built up in the numerical values, which shows neither errors, nor gaps, i.e. one that captures and makes recognizable every prime number.
　
The Tietken-Castell-prime-algorithm, once set in motion, can “automatically”, i.e. without further human intervention and without any further mental effort, build up a register that (without new input) grows steadily larger and, since it is ordered by numerical values, can be accessed at any time!

Numerical examples of the TIETKEN-CASTELL-PRIME-ALGORITHM

The Tietken-Castell Prim Algorithm builds up per series of ten only with the final digits 1, 3, (5), 7, 9.
The digit 5 is written in brackets here, as it is an exception. It can never become a prime number (except as a single digit), even with added decimal digits, since numbers with the final digit 5 are always divisible by 5. However, in this Tietken-Castell system, it is counted in, so as to maintain a regular increment and to avoid having to take a double step every time the final number “5” is reached, where the “5” is skipped.
　
The following diagram shows how the “Tietken-Castellre-gister” (that is the stock of numbers that gradually builds up) increases in numerical value, i.e., here, gradually and in a constant order. According to our premise (of a decimal system that remains the same), it can continue indefinitely!

(a)
In each of the unlimited number of lines, only numbers with the final digits 1, 3, 5, 7 or 9 are entered. This is because each prime number must have 1, 3, 7 or 9 (the exception 5 is omitted) as a final digit.
　
However, not every number with this final digit is a prime number. In order to cover all prime numbers without gaps, all numbers with one of the four final prime digits must be included in the register. Among them are also the prime numbers. They can be identified by the fact that there are no factors for them.
　
(b)
　
The lines of the Tietken-Castell register always look the same. All lines have the final digits 1, 3, 5, 7 and 9. And each additional line that is added gets an additional ten-decimal digit.
Every newly added row, receives an additional decade with 1, 3, (5), 7, 9 on the end.
1 3 5 7 9
11 13 15 17 19
21 23 25 27 29
31 33 35 37 39
41 43 45 47 49
51 53 55 57 59
61 63 65 67 69
71 73 75 77 79
etc.
or:
1001 1003 1005 1007 1009
1011 1013 1015 1017 1019
1021 1023 1025 1027 1029
etc.
or:
2381 2383 2385 2387 2389
2391 2393 2395 2397 2399
2401 2403 2405 2407 2409
2411 2413 2415 2417 2419
etc.
 
(c) 

Thus, it is claimed here that there is no upper limit to the numbers!
However, we are not looking for a prime number with about 2 million digits, but for the time being “only” 1,000 to 2,000 digit “big numbers”, which were created by multiplying two prime numbers.
　
According to our estimation, this task can be accomplished easier and faster (even under the premise that prime numbers become rarer and rarer with increasing numbers) compared to a confusing pool of possibilities of finding prime numbers at random and then having the difficulty of identifying the given number reliably as a prime number.
　
The Tietken-Castell-Prime-Algorithm necessary for that purpose has to count up without gaps, add up continuously, keep an uninterrupted connection to the previous factors and also multiply between two prime numbers and remember their results as “big numbers” to be able to recall them later with their supplied factors!

(d)

In order to obtain only numbers with the final digits 1, 3, 7 or 9 in the register, all numbers ending in 0, 2, 4, 6 or 8 are excluded from the register. The 5 is counted in the register, however, it plays no additional role.

(e)

Each number of the above-mentioned register makes itself available as a constant factor for a specific number series of further numbers (in intervals of 2 times the respective number).
　
Whenever a number turns out to be a prime number because of missing factors which could have produced it, it too is multiplied by itself and starts a new and unlimited series of numbers in the register from the place of the resulting product (i.e. a “large number” of two prime numbers).
　
Prime numbers are numbers that occur in the Tietken-Castell-register without factors. However, if factors (two or more) are involved in the creation of the number in question, it is not a prime number.
　
Since the corresponding factors lie “left” in the past, the product is already calculated there. If, then, a number appears that has been calculated in the past, it is already known as a product, together with its factors. In this way, it can also be determined whether the number in question is a “large number”, i.e. a product of two prime numbers.

(f)

In the following, the effect of the numbers in the Tietken-Castell-Prime-Algorithm is illustrated. Inevitably, there are small numbers at the outset. However, the principle of procedure and use continues indefinitely upwards, just as it did at the beginning, because the laws of the decimal system remain the same for both small and large numbers, and the Tietken-Castell-prime-algorithm's procedure is always the same and does not change either.

(g)

1st line, 1st digit: 3

The 3 is multiplied by itself, like all numbers of the register. It results in 9. From here on, the first series of numbers begins, which runs through all the numbers of the Tietken-Castell-register. At the same time, the numbers of the register are becoming larger and larger.
　
This 3 remains as a counting constant and forms new numbers first with itself, i.e. with the 3, and then with the odd numbers 5, 7, 9, 11, 13, (15), 17, 19, 21, 23, (25), 27, 29 31, 33,(35), 37, 39 etc., which are sometimes normal products, but sometimes also products of two prime numbers.
3 is a prime number because in the register it has no factors from the left.
If it is multiplied with e.g. 11 or 13, to which the same applies, the resulting product 33 or 39 represents such “large numbers” suitable for encryption.
　
As the number series become more and more frequent and run indefinitely through the entire register, there may be multiple assignments for numbers where number series overlap. This means that the number in question has several factors. In this context, the most important thing for the Tietken-Castell-Prime-Algorithm is to determine whether it has factors at all. If it does have factors, it can be determined whether these factors were prim and whether the product is suitable as a “large number” for cryptography. If it has no factors, it is (as already emphasized) itself a prime number. In this case, there can be no multiple assignments.

(h)

1st line, 2nd digit: 5

The 5 is a prime number because, like the 3, no factors lead to it.
Its series of numbers starts from 5 * 3 = 15, but with its added decimal digits it will never be a prime number again, since it will always be divisible by 5.
　
(i)

1st line, 3rd digit: 7

The 7 is a prime number and with 7 * 3 (= 21) as a series of 7s starts marching through the Tietken-Castell-register in an unlimited way.

(i)

1st line, 4th digit: 9

The 9 is an intermediate stage of the 3 series and therefore not a prime number. However, together with decimal places it forms an important prime final digit and often results in prime numbers itself, e.g. 19, 29, 59, 79, 89, 109, 139, 149, 179, 199, 229 239, 269, 349, 359, 379, 389 and an unlimited number of others!
　
The series of 9, which will run through the Tietken-Castell-register, begins with 9 * 3 (= 27) and will form the first numbers in the series of 9, with 9 as a constant factor and the other odd factors, which are the same for all series of numbers (in addition to 3 it is 5, 7, 9, 11, 13, 15, 17, 19 and so on, unlimited).
　
Since the digit 9 is not a prime number, no “large numbers” needed for cryptography can be formed, although the second factors 11, 13, 17, 19, 23, 29 and 31 are prime numbers and would be suitable for this purpose. This initial situation changes in connection with decimal places. Already from the above-mentioned 19, 29, 59, 79, 89, 109 etc. the prime end digit 9 is ‘back in the race’ as part of prime numbers.

(k)

Summary concerning the first line:

Already with just the 4 examples mentioned above, it can be seen that the factors (as far as they exist) follow a regular order. On the one hand, there are numbers with always the same final digits in the same order (1, 3, (5), 7, 9) and on the other hand, there are always the same odd numbers cardinally incremented in steps of 2.

(l)

2nd line, 1st to 5th digit: 11 to 19

The 2nd line is calculated just like all other unlimited lines, like the 1st line.
　
(1)

11 is multiplied by 3, 5, 7, 9,11 etc. Multiplication by itself yields 11 * 11 = 121 (i.e. a “large number”), following which 11 continues its series indefinitely with 11 * 13, 11 * 15, 11 * 17, 11 * 19, 11 * 21, etc. For an encryption, it would not be necessary here to multiply a prime number like 11 by itself, since the other prime numbers are known so as to form a “large number” together with them.
　
(2)

13 is multiplied by 3, 5, 7, 9, 11,13 etc. Multiplication by itself yields 13 * 13 = 169 (i.e. a “large number”), following which 13 continues its series indefinitely with 13 * 15, 13 * 17, 13 * 19, 13 * 21, 13 * 23, etc. For an encryption, it would not be necessary here to multiply a prime number like 13 by itself, since the other prime numbers are known so as to form a “large number” together with them.
　
(3)

15 is multiplied by 3, 5, 7, 9, 11, 13, 15, 17, 19, etc., but can never be used to create a “large number” for RSA cryptography, since it is only prime as a single digit (5).
　
(4)

17 is multiplied by 3, 5, 7, 9, 11,13, 15, 17 etc. Multiplication by itself yields 17 * 17 = 289 (i.e. a “large number”), following which 17 continues its series indefinitely with 17 * 18, 17 * 19, 17 * 21, 17 * 21, 17 * 23, etc. For an encryption, it would not be necessary here to multiply a prime number like 17 by itself, since the other prime numbers are known so as to form a “large number” together with them.
　
(5)

19 is multiplied by 3, 5, 7, 9, 11,13, 15, 17, 19 etc. Multiplication by itself yields 19 * 19 = 361 (i.e. a “large number”), following which 19 continues its series indefinitely with 19 * 21, 19 * 23, 19 * 25, 19 * 27, 19 * 29, etc. For an encryption, it would not be necessary here to multiply a prime number like 19 by itself, since the other prime numbers are known so as to form a “large number” together with them.
　
(6)

21 is multiplied by 3, 5, 7, 9, 11,13, 15, 17, 19, 21 etc. Multiplication by itself yields 21 * 21 = 441 (i.e. a “large number”), following which 21 continues its series indefinitely with 21 * 23, 21 * 25, 21 * 27, 21 * 29, 21 * 31, etc.

(m)

Summary concerning the 1st to indefinitely “last” line:

The factors shown above prove the order, calculability and correctness of this Tietken-Castell-prime-algorithm.
None of the complicated and unreliable procedures mentioned above are used in this Tietken-Castell prim algorithm to determine whether a more or less randomly found or unreliably calculated number is a prime number.
　
If the numbers 21 and 27 can be seen in a line, here e.g. the third line, then it is known from the first line that the 21 belongs to the series of 3, stemming from 3 * 7 (or written in the addition form, i.e. to 3 + 6 + 6+ 6) and the 27 belongs to the series of 9, stemming from 3 * 9 (or written in the addition form: 9 + 18).
However, no factors lead to the numbers 23 and 29 in the same line (it si the third line here).

(n)

The distances: 

For the algorithm, it can be easier to have to perform additions only. Because it is striking that the distances between all numbers in a series remain the same because of the first multiplication with itself (i.e. a multiplication by 2).
It is always exactly twice that it forms the first counting factor. Therefore, after the first multiplication, the algorithm could also form the following numbers in a number series by adding the distances that are always the same.
　
(1) 3 forms the distance 6 to 9, so after this 9 the numbers 15 (9=6), 21 (=15+6), 27 (21+6), 33, 39, 45, 51, 57, 63, 69, 75 and so on.

(2) 5 increases in steps of 10 (consisting of 2 * 5).

(3) The same applies to 7, where the distances of the numbers in this series will be 2 * 7 = 14 (49, 63, 77, 91, 105, 119, etc.).

Summary

To the right of all the numbers in this graph are the factors that provide retrievable information about their subsequent products. The algorithm stores them.
To the left of these subsequent products, their factors are noted down once more in order to demonstrate the relationship to them.
　
Prime numbers, as we now know, have no indication of factors on their left side. However, in this partial list, which only shows the path of the 3, i.e. is still incomplete, the absence of factors to the left of the number does not tell us yet what kind of number it is (i.e. prime number, “large number” or a simple product).
　
There are already many double assignments of single numbers at the beginning of the register, i.e. overlapping series of numbers – e.g. the series of 3 (with 3*21) meets with the series of 7 (with 7*9) in the number 63. In the second case (7 * 9), 63 would be suitable as a “large number”.
(It would have to be checked whether such overlaps of several pairs of factors in one number increase cryptographic security of a “large number”, since it is less clearly factorizable.)

(A)

The first 120 numbers, taking into account the series of 3 only:

1 3(3*3=9) 5(3*5=15) 7(3*7=21) (3*3=9)9(3*927)
11(3*11=33)13(3*13=39) (3*5=15)15(3*15=45) 17(3*17=51) 19(3*19=57)
(3*7=21)21(3*21=63) 23(3*23=63) 25(3*25=75)(3*9=27)27(3*27=81) 29(3*2=87)
31(3*31=93)(3*11=33)33(3*33=99)35(3*35=105)37(3*37=111)(3*13=39)39(3*117)
41(3*41=123) 43(3*43=129) (3*15=45)45(3*45=135) 47(3*47=141) 49
(3*17=51)51(3*51=153) 53(3*53=159) 55(3*55=165) (3*19=57)57(3*57=171) 59(3*59=177)
61(3*61=183) (3*21=63)63(3*63=189) 65(3*65=195) 67(3*67=201) (3*23=69) 69(3*69=207)
71(3*71=213) 73(3*73=219) (3*25=75)75(3*75=225) 77(3*77=231) 79(3*79=237)
(3*27=81)81(3*81=243) 83(3*83=249) 85(3*85=255) (3*29=87)87(3*87=261) 89(3*89=267)
91(3*91=273) (3*31=93)93(3*93=279) 95(3*95=285) 97(3*97=291) (3*33=99)99(3*99=297)
101(3*101=303) 103(3*103=309) (3*35=105)105(3*105=315) 107(3*107=321) 109(3*109=327)
(3*37=111)111(3*111=333) 113(3*113=339) 115(3*115=345)
(3*117=351)117(3*117=351)119(3*119=357)
121(3*121=363)(3*41=123) 123(3*123=369) 125(3*125=375) 127(3*127=381)
(3*43=129)129(3*129=387)
131(3*131=393) 133(3*133=399) (3*45=135)135(3*135=405) 137(3*137=411) 139(3*139=417)
(3*47=141)141(3*141=423) 143(3*143=429)145(3*145=435) (3*49=147)147(3*147=441)
149(3*149=447)

B)

The first 120 numbers, taking into account the series of 5 only:
(no prime end digit, but odd. Distances in steps of 10):

1 3(5*3=15) 5(5*5=25) 7(5*7=35) 9(5*9=45)
11(5*11=55) 13(5*13=65) (5*3=15) 15(5*15=75) 17(5*17=85) 19(5*19=95)
21(5*21=105) 23(5*23=115) (5*5=25)25(5*25=125) 27(5*27=135) 29(5*29=145)
31(5*31=155) 33(5*33=165) (5*7=35)35(5*35=175) 37(5*37=185) 39(5*39=195)
41(5*41=205) 43(5*43=215) (5*9=45)45(5*45=225) 47(5*47=235) 49(5*49=245)
51(5*51=255) 53(5*53=265) (5*11=55)55(5*55=275) 57(5*57=285) 59(5*59=295)
61(5*61=305) (7*9=63)63(5*63=315) (5*13=65)65(5*65=325) 67(5*67=335) 69(5*69=345)
71(5*71=355) 73(5*73=365) (5*15=75)75(5*75=375) (7*11=77)77(5*77=385) 79(5*79=395)
81(5*81=405) 83(5*83=415) (5*17=85)85(5*85=425) (5*29=87)87(5*87=435) 89(5*89=445)
(7*13=91)91(5*91=455) 93(5*93=465) (5*19=95)95(5*95=475) 97(5*97=485) 99(5*99=495)
101(5*101=505) 103(5*103=515) (5*21=105)105(5*105=525) 107(5*107=535) 109(5*109=545)
111(5*111=555) 113(5*113=565) (5*23=115)115(5*115=575) 117(5*117=585) 119(5*119=595)
121(5*121=605) 123(5*123=615) (5*25=125)125(5*125=625) 127(5*127=635) 129(5*129=645)
131(5*131=655) 133(5*133=665) (5*27=135)135(5*135=675) 137(5*137=685) 139(5*139=695)
141(5*141=705) 143(5*143=715) (5*29=145)145(5*145=725) 147(5*147=735) 149(5*149=745)

(C)

The first 120 numbers, taking into account the series of 7 only:

1 3(7*3=21) 5(7*5=35) 7(7*7=49) 9(7*9=63)
11(7*11=77) 13(7*13=91) 15(7*15=105) 17(7*17=119) 19(7*19=133)
(7*3=21)21(7*21=147) 23(7*23=161) 25(7*25=175) 27(7*27=189) 29(7*29=203)
31(7*31=217) 33(7*33=231) (7*5=35)35(7*35=245) 37(7*37=259) 39(7*39=273)
41(7*41=287) 43(7*43=301) 45(7*45=315) 47(7*47=329) (7*7=49)49(7*49=343)
51(7*51=357) 53(7*53=371) 55(7*55=385) 57(7*57=399) 59(7*59=413)
61(7*61=427) (7*9=63)63(7*63=441) 65(7*65=455) 67(7*67=469) 69(7*69=483)
71(7*71=497) 73(7*73=511) 75(7*75=525) (7*11=77)77(7*77=539) 79(7*79=553)
81(7*81=567) 83(7*83=581) 85(7*85=595) (7*29=87)87(7*87=609) 89(7*89=623)
(7*13=91)91(7*91=637) 93(7*93=651) 95(7*95=665) 97(7*97=679) 99(7*99=693)
101(7*101=707) 103(7*103=721) (7*15=105)105(7*105=735) 107(7*107=749) 109(7*109=763)
111(7*111=777) 113(7*113=791) 115(7*115=805) 117(7*117=819) (7*17=119)119(7*119=833)
121(7*121=847) 123(7*123=861) 125(7*125=875) 127(7*127=889) 129(7*129=903)
131(7*131=917) (7*19=133)133(7*133=931) 135(7*135=945) 137(7*137=959) 139(7*139=973)
141(7*141=987) 143(7*143=1001) 145(7*145=1015) (7*21=147)147(7*147=1029) 149(7*149=1043)
 
(D)

The first 120 numbers, taking into account the series of 9 only:
Re (A) to (D)

1 3(9*3=27) 5(9*5=45) 7(9*7=63) 9(9*9=81)
11(9*11=99) 13(9*13=117) 15(9*15=135) 17(9*17=153) 19(9*19=171)
21(9*21=189) 23(9*23=207) 25(9*25=225) (9*3=27)27(9*27=243) 29(9*29=261)
31(9*31=279) 33(9*33=297) 35(9*35=315) 37(9*37=333) 39(9*39=351)
41(9*41=369) 43(9*43=387) (9*5=45) 45(9*45=405) 47(9*47=423) 49(9*49=441)
51(9*51=459) 53(9*53=477) 55(9*55=495) 57(9*57=513) 59(9*59=531)
61(9*61=549) (9*7=63)63(9*63=567) 65(9*65=585) 67(9*67=603) 69(9*69=621)
71(9*71=639) 73(9*73=657) 75(9*75=675) 77(9*77=693) 79(9*79=711)
(9*9=81)81(9*81=729) 83(9*83=747) 85(9*85=765) 87(9*87=783) 89(9*89=801)
91(9*91=819) 93(9*93=837) 95(9*95=855) 97(9*97=873) (9*11=99)99(9*99=891)
101(9*101=909) 103(9*103=927) 105(9*105=945) 107(9*107=963) 109(9*109=981)
111(9*111=999) 113(9*113=1017) 115(9*115=1035) (9*13=117)117(9*117=1053) 119(9*119=1071)
121(9*121=1089) 123(9*123=1107) 125(9*125=1125) 127(9*127=1143) 129(9*129=1161)
131(9*131=1179) 133(9*133=1197) (9*15=135)135(9*135=1215) 137(9*137=1233) 139(9*139=1251)
141(9*141=1269) 143(9*143=1287) 145(9*145=1305) 147(9*147=1323) 149(9*149=1341)

Comments on the four tables above:
　
The prime end digit 1 is omitted as a factor. If it were used as a factor, it would affect all numbers in the register without changing their numerical value, but the prime numbers – due to the operations "1 * 3", "1 * 7", "1 * 9", "1 * 11", "1 * 13", etc. – would become products and would lose their status as prime numbers.
The series of numbers thus begin with the 3, followed by the 7, 9, 11, 13, 15, 17, 19, 21 and so on indefinitely.
The three tables, each reaching 119 (which is actually a single table, the presentation in three parts was chosen for clarity), show how already the 3, 7 and 9 number series fill the register with information for the numbers 3 to 119. There is no number in it about which there was insufficient information as regards the questions to be answered here.

Concluding remark: 

As a side effect, the factorization of “large numbers” (i.e. of products of prime factors) could be omitted if the register is large enough. (The algorithm for this factorization, the CASTELL-FACT ALGORITHM, was already found by us in October 2019).

The lower case multiplication to the right of the numbers concerns information for subsequent numbers. The lower case multiplication on the left side of each
non-prime number concerns the factors that “come from the left” and have created the present product by multiplication.
These factors, if they are prime, have created a “large number” for the given number, which can be used for RSA cryptography, but they are also the prime numbers that are supplied here when factorizing.

Tom Hermann Tietken,
MUDr. (Charles-University Prag)
tom.tietken@gmail.com

Nikolaus Graf zu Castell-Castell,
Dipl. Vw. (University Hamburg)
nikolaus.castell@mail.com

Prague Research Institute
Zug (CH) und Prague (CR)
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