20191125, 18:28  #1 
Nov 2016
2^{2}×3×5×47 Posts 
Minimal set of the strings for primes with at least two digits
https://primes.utm.edu/glossary/page...t=MinimalPrime
In 1996, Jeffrey Shallit [Shallit96] suggested that we view prime numbers as strings of digits. He then used concepts from formal language theory to define an interesting set of primes called the minimal primes: A string a is a subsequence of another string b, if a can be obtained from b by deleting zero or more of the characters in b. For example, 514 is a substring of 251664. The empty string is a subsequence of every string. Two strings a and b are comparable if either a is a substring of b, or b is a substring of a. A surprising result from formal language theory is that every set of pairwise incomparable strings is finite [Lothaire83]. This means that from any set of strings we can find its minimal elements. A string a in a set of strings S is minimal if whenever b (an element of S) is a substring of a, we have b = a. This set must be finite! For example, if our set is the set of prime numbers (written in radix 10), then we get the set {2, 3, 5, 7, 11, 19, 41, 61, 89, 409, 449, 499, 881, 991, 6469, 6949, 9001, 9049, 9649, 9949, 60649, 666649, 946669, 60000049, 66000049, 66600049}, and if our set is the set of composite numbers (written in radix 10), then we get the set {4, 6, 8, 9, 10, 12, 15, 20, 21, 22, 25, 27, 30, 32, 33, 35, 50, 51, 52, 55, 57, 70, 72, 75, 77, 111, 117, 171, 371, 711, 713, 731} Besides, if our set is the set of prime numbers written in radix b, then we get these sets: Code:
b, we get the set 2: {10, 11} 3: {2, 10, 111} 4: {2, 3, 11} 5: {2, 3, 10, 111, 401, 414, 14444, 44441} 6: {2, 3, 5, 11, 4401, 4441, 40041} Now, let's consider: if our set is the set of prime numbers >= b written in radix b (i.e. the prime numbers with at least two digits in radix b), then we get the sets: Code:
b, we get the set 2: {10, 11} 3: {10, 12, 21, 111} 4: {11, 13, 23, 31, 221} 5: {10, 12, 21, 23, 32, 34, 43, 111, 131, 133, 313, 401, 414, 14444, 30301, 33001, 33331, 44441, 300031} 6: {11, 15, 21, 25, 31, 35, 45, 51, 4401, 4441, 40041} 7: {10, 14, 16, 23, 25, 32, 41, 43, 52, 56, 61, 65, 113, 115, 131, 133, 155, 212, 221, 304, 313, 335, 344, 346, 364, 445, 515, 533, 535, 544, 551, 553, 1112, 1211, 1222, 2111, 3031, 3055, 3334, 3503, 3505, 3545, 4504, 4555, 5011, 5455, 5545, 5554, 6034, 6634, 11111, 30011, 31111, 33001, 33311, 35555, 40054, 300053, 33333301} 8: {13, 15, 21, 23, 27, 35, 37, 45, 51, 53, 57, 65, 73, 75, 107, 111, 117, 141, 147, 161, 177, 225, 255, 301, 343, 361, 401, 407, 417, 431, 433, 463, 467, 471, 631, 643, 661, 667, 701, 711, 717, 747, 767, 3331, 3411, 4043, 4443, 4611, 5205, 6007, 6101, 6441, 6477, 6707, 6777, 7461, 7641, 47777, 60171, 60411, 60741, 444641, 500025, 505525, 3344441, 4444477, 5500525, 5550525, 55555025, 444444441, 744444441} Can someone complete my base 7 and 8 set? Also find the sets of bases 9 to 36. 
20191126, 05:46  #2 
Romulan Interpreter
Jun 2011
Thailand
3^{3}×347 Posts 
Your sets for 2 and 3 are complete, but the others are not.
For example, all Fermat numbers that may be prime are of the form "1000.....1" in base 4, and additionally, you will have and endless number of combinations of patterns of 0 and 1 that may be prime, and yet not appear in your list, so your base 4 may look like: Code:
List(["2", "3", "11", "101", "10001", "100000001", "100001001001001", "100100100001001"]) Code:
List(["2", "3", "10", "111", "401", "414", "1404", "4041", "14004", "14404", "14411", "14444", "40041", "40441", "41141", "44001 ", "44441", "144404", "404001", "404441", "1440044", "1444114", "4000001", "4000144", "4001411", "4114001", "11440001", "1400000 4", "14000114", "14000404", "14000411", "40400001", "40400014", "40400041", "44000441", "44400014", "44440001", "140000404", "14 0000411", "140000444", "140011411", "140014001", "141144004", "144000044", "144000114", "144000444", "400040014", "400114001", " 400140004", "400144004", "404000041", "404004001", "404040001", "404400001", "404400041", "440040001", "1400040004", "1400044114 ", "1411400011", "1440000404", "1440004114", "1444000114", "1444004404", "4000000041", "4000400041", "4000400441", "4004400041", "4040000041", "4040440001", "4044000041", "4044004001", "4400000001", "4400011441", "4404000441", "4444400041", "11400000001", "11400000041", "11400114441", "11440000144", "11444000014", "14000011444", "14000014441", "14000040011", "14000040044", "1400004 4004", "14000044114", "14000140001", "14001440001", "14114000011", "14114000444", "14400000004", "14400000011", "14400001441", " 14400004114", "14400040044", "14400400141", "14400400444", "14400404044", "14440000004", "14440004044", "14440040004", "14440040 444", "40000000041", "40000001411", "40000004441", "40000040014", "40000141144", "40000400441", "40001400004", "40004001141", "4 0004004441", "40011400001", "40014000044", "40014001144", "40014440044", "40040000001", "40040040001", "40044004001", "400440044 41", "40044400001", "40044411441", "40400004441", "40440400441", "40444000041", "40444004441", "41144400001", "44000400014", "44 004440001", "44040004441", "44044040001", "44114000441", "44114400001", "44400001444", "44400004001", "44400004441", "4444000000 1", "44440000144", "44444000041"]) Last fiddled with by LaurV on 20191126 at 05:55 
20191126, 06:05  #3  
Jan 2017
2^{3}×11 Posts 
Quote:
Quote:


20191126, 06:10  #4 
Romulan Interpreter
Jun 2011
Thailand
9369_{10} Posts 
Ok, scratch the former post, I am stupid.
It looked odd, and I could not believe those paper were right, then I went and read them (skimmed) and realized the sequence does not need to be consecutive. In fact, you stated as much as that, and your example with 251664 shows very clearly what you mean. Sorry for being stupid (in fact, my excuse is that you post so much rubbish here that I don't read your post more than fast skimming mode ) Give me 5 minutes... (look how I am spending my lunch break!) (edit @uau: crosspost, I was talking to sweety) Last fiddled with by LaurV on 20191126 at 06:12 
20191126, 06:46  #5 
Romulan Interpreter
Jun 2011
Thailand
3^{3}·347 Posts 
Ok, here is the code, and I searched much higher, there are no new numbers.
Code:
\\finds text substrings in text strings, or smallsubvecteors in large smallvectors :P \\(kind of efficient, but it still could be optimized...) find(subtxt,txt,startfrom=1)= { my(i,j,s=Vecsmall(subtxt), t=Vecsmall(txt)); i=startfrom1; while(i<=#t#s, j=1; while(j<=#s, if(s[j]!=t[i+j], break ); j++ ); if(j>#s, return(i+1) ); i++ ); return(0); } \\similar, but the strings do not need to be consecutive findweak(subtxt,txt,startfrom=1)= { my(i,j,s=Vecsmall(subtxt), t=Vecsmall(txt)); i=startfrom; j=1; while(i<=#t && j<=#s, if(s[j]==t[i], j++ ); i++ ); if(j>#s, return(1), return(0) ); } \\base from 2 to 36, no validation! bprint(n,base=10)= { if(base==10, return(Str(n)) ); my(v=digits(n,base)); for(i=1,#v, if(v[i]<10, v[i]+=48, v[i]+=55 ) ); return(Strchr(v)); } minprime(startfrom=2, base=10)= { my(n,k,x,cnt); lstminprimes=List([]); n=nextprime(startfrom); listput(lstminprimes,bprint(n,base)); print(lstminprimes); while(1, \\stop it with ctrl+c when fedup with it, etc n=nextprime(n+1); x=bprint(n,base); k=1; while(k<=#lstminprimes, \\ if(find(lstminprimes[k],x), if(findweak(lstminprimes[k],x), break ); k++ ); if(k>#lstminprimes, listput(lstminprimes,x); print(lstminprimes) ); cnt++; if(cnt%10000==0, printf("...%d...%c",n,13) ) ); } 
20191126, 11:21  #6 
Romulan Interpreter
Jun 2011
Thailand
22231_{8} Posts 
I let that script running in background for 8, during I was doing some real jobrelated work, and it still produced 77774444441.
Meantime I realized that it is terrible inefficient, and it can be done much cleverer. So. factory stopped for now, I may revisit it in the future (probably weekend). I also have an idea how 7 and 8 (starting from 2 digits) can be "proved" formally. I go home now, 18:20 here. Last fiddled with by LaurV on 20191126 at 11:23 
20191128, 11:29  #7 
Romulan Interpreter
Jun 2011
Thailand
22231_{8} Posts 
I found an easy way to generate those sets, and to prove that they are complete.
For the "starting from two digits" version, neither one of the exposed sets for 7 and 8 are complete. Some larger primes are still lurking in the dark there. I have the complete sets for both 8, and 7 for the both cases when the base itself is included in the set or not*, but I don't want to spoil the puzzle, this is an interesting little problem... hehe... Hint: Code:
gp > a=(7^175)/2 %1 = 116315256993601 gp > isprime(a) %2 = 1 gp > digits(a,7) %3 = [3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 1] gp > *when the base is prime, like for 5 and 7, the sets are different; including the base results in automatic elimination of all possible extension numbers with "0 after 1" from the set, which is quite restrictive, so I also calculated the lists for the "base is not included" version, i.e. base5 starting from 6, and base7 starting from 8; in this case, for example, base5 will include numbers like 104 and 10103 which are prime, and base7 list will include 1022, 1051, 1202, .... 1100021 ... etc, they are "enriched" compared with the case when the first "10" is included. So I have the complete list for 8, and the complete two lists for 7, the normal one, and the "enriched" one. Base5 is easy, in any case. Last fiddled with by LaurV on 20191128 at 11:33 
20191128, 22:16  #8  
Nov 2016
101100000100_{2} Posts 
Quote:


20191128, 22:40  #9  
Nov 2016
101100000100_{2} Posts 
Quote:
Code:
a(n,k,b)=v=[];for(r=1,length(digits(n,b)),if(r+length(digits(k,2))length(digits(n,b))>0 && digits(k,2)[r+length(digits(k,2))length(digits(n,b))]==1,v=concat(v,digits(n,b)[r])));fromdigits(v,b) g(n)=if(n<10,n+48,if(n<36,n+55,if(n<62,n+61,if(n<77,n29,if(n<84,n19,if(n<90,n+7,if(n<94,n+33,n+67))))))) f(n,b)=for(k=1,length(digits(n,b)),print1(Strchr(g(digits(n,b)[k])))) is(n,b)=for(k=1,2^length(digits(n,b))2,if(ispseudoprime(a(n,k,b)) && a(n,k,b)>=b,return(0)));1 c(b)=forprime(p=b,2^32,if(is(p,b),f(p,b);print1(", "))) Code:
a(n,k,b)=v=[];for(r=1,length(digits(n,b)),if(r+length(digits(k,2))length(digits(n,b))>0 && digits(k,2)[r+length(digits(k,2))length(digits(n,b))]==1,v=concat(v,digits(n,b)[r])));fromdigits(v,b) g(n)=if(n<10,n+48,if(n<36,n+55,if(n<62,n+61,if(n<77,n29,if(n<84,n19,if(n<90,n+7,if(n<94,n+33,n+67))))))) f(n,b)=for(k=1,length(digits(n,b)),print1(Strchr(g(digits(n,b)[k])))) is(n,b)=for(k=1,2^length(digits(n,b))2,if(ispseudoprime(a(n,k,b)),return(0)));1 c(b)=forprime(p=2,2^32,if(is(p,b),f(p,b);print1(", "))) Last fiddled with by sweety439 on 20191128 at 22:40 

20191129, 02:17  #10 
Romulan Interpreter
Jun 2011
Thailand
2499_{16} Posts 
I didn't look yet how good is your code, but my former one is lousy, so there are chances that yours is better. I mean, not the code, but my method itself was lousy, to look at all primes one by one. The authors of that paper you linked describe a method which is much better and somehow similar to what I am doing now.
Right now, I split the problem in two steps, first I let the zero apart, and solve the problem with "digits" from 1 to b1, by starting from the end with all possible cases in a set. Starting from the end or from the beginning makes no difference, but in the case the base is even, I only have n/2 elements in the initial set (because numbers ending in 2, 4, 6, etc, can never be prime), so the search dimension is reduced in half. Then, for all elements in set, I check what digit I can add in front of them and still avoiding conflicts. If any of the resulting numbers is prime, I add it to the set. Here is where the algorithm "strikes", because I can do this in about linear time, by creating a matrix with the possible candidates, and then eliminating them from the matrix, by different criteria (like, it produces conflict, it is a prime and I add it to the list, or it is always composite regardless of how you extend it, etc), and sometimes full rows and columns can be eliminated. This gives me the complete set, excluding the numbers that contain zero, in just minutes. The second part comes from the realization that the numbers that contain zero and have to be in the set, if we delete zeroes from them, the new created are (1) still not in the set, and (2) can not be covered with numbers in the set, and (3) are the same magnitude as the numbers in the set except maybe the first digit, that can repeat indefinitely till the first prime is found. The (3) is very important (and it can be proved) so the second part of the algorithm is to create a list with all such numbers (like 56 digit numbers in our case) and see which one becomes a prime when it is "stuffed" with zeroes, which is piece of cake. Mind that the zeros have to be "between" the digits, as "leading zeros do not count"^{(TM)} and numbers ending in zero in any base are not prime. Last fiddled with by LaurV on 20191129 at 02:25 
20191129, 06:50  #11  
Nov 2016
2^{2}×3×5×47 Posts 
Quote:
There is C code and C++ code (given by others, not me), but I do not know how to run https://github.com/curtisbright/mepndata/ https://github.com/RaymondDevillers/primes Last fiddled with by sweety439 on 20191129 at 06:51 

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