20211005, 12:07  #188 
"99(4^34019)99 palind"
Nov 2016
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110000101101_{2} Posts 
Quote:

20211006, 23:52  #189 
"99(4^34019)99 palind"
Nov 2016
(P^81993)SZ base 36
110000101101_{2} Posts 
New minimal prime (start with b+1) in base b is found for b=908: 8(0^243438)1, see post https://mersenneforum.org/showpost.p...&postcount=992
File https://docs.google.com/spreadsheets...RwmKME/pubhtml updated. 
20211016, 13:38  #190 
"99(4^34019)99 palind"
Nov 2016
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6055_{8} Posts 
Conjecture: If sequence (a*b^n+c)/gcd(a+c,b1) (a>=1 is integer, b>=2 is integer, c is (positive or negative) integer, c>=1, gcd(a,c) = 1, gcd(b,c) = 1) does not have covering set (full numerical covering set, full algebraic covering set, or partial algebraic/partial numerical covering set), then the sum of the reciprocals of the positive integers n such that (a*b^n+c)/gcd(a+c,b1) is prime is converge (i.e. not infinity) and transcendental number. (of course, this conjecture will imply that there are infinitely many such n)
For the examples of (a,b,c) triples (a>=1 is integer, b>=2 is integer, c is (positive or negative) integer, c>=1, gcd(a,c) = 1, gcd(b,c) = 1) such that (a*b^n+c)/gcd(a+c,b1) have covering set (full numerical covering set, full algebraic covering set, or partial algebraic/partial numerical covering set), see post https://mersenneforum.org/showpost.p...&postcount=678 
20211016, 13:45  #191 
"99(4^34019)99 palind"
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6055_{8} Posts 
Another conjecture (seems to already be proven, but I am not sure that): If all but finitely many primes p divide (a*b^n+c)/gcd(a+c,b1) (a>=1 is integer, b>=2 is integer, c is (positive or negative) integer, c>=1, gcd(a,c) = 1, gcd(b,c) = 1) for some n>=1, then a=1 and c=1, i.e. (a*b^n+c)/gcd(a+c,b1) is generalized repunit number (b^n1)/(b1)
The factor tables have many examples for the special case that b=10, e.g. {2}1 in base 10 is (a,b,c) = (2,10,11), the section Prime factors that appear periodically lists the primes that divide (a*b^n+c)/gcd(a+c,b1) = (2*10^n11)/9 for some n, and we note that the primes 2, 5, 11, 31, 37, 41, 43, 53, 71, 73, 79, 83, 101, 103, 107, 127, 137, 157, 173, 191, 199, 227, 239, 241, 251, 271, 281, 283, 307, 311, 317, 331, 347, 349, 353, 397, 409, 449, 523, 547, 563, 569, 599, 601, 613, 617, 631, 641, 643, 653, 661, 673, 691, 719, 733, 739, 751, 757, 761, 769, 773, 787, 797, 809, 827, 829, 839, 853, 859, 907, 911, 967, 991, 997, ..., divides no numbers of the form (a*b^n+c)/gcd(a+c,b1) = (2*10^n11)/9, and this sequence of primes seems to be infinite, another example is 1{3} in base 10 is (a,b,c) = (4,10,1), the section Prime factors that appear periodically lists the primes that divide (a*b^n+c)/gcd(a+c,b1) = (4*10^n1)/3 for some n, and we note that the primes 2, 3, 5, 11, 37, 41, 53, 73, 79, 101, 103, 137, 139, 173, 211, 239, 241, 271, 277, 281, 317, 331, 349, 353, 397, 421, 449, 463, 521, 547, 607, 613, 617, 661, 673, 733, 751, 757, 773, 797, 829, 853, 859, 907, 967, ..., divides no numbers of the form (a*b^n+c)/gcd(a+c,b1) = (4*10^n1)/3, and this sequence of primes seems to be infinite. Of course this conjecture also include for bases other than 10, i.e. for every (a*b^n+c)/gcd(a+c,b1) (a>=1 is integer, b>=2 is integer, c is (positive or negative) integer, c>=1, gcd(a,c) = 1, gcd(b,c) = 1) family other than generalized repunit family (b^n1)/(b1) (i.e. a=1 and c=1), there are infinitely many primes not dividing any number of the form (a*b^n+c)/gcd(a+c,b1), e.g. for the base 11 unsolved family 5{7} = (57*11^n7)/10, the primes dividing no numbers of the form (a*b^n+c)/gcd(a+c,b1) = (57*11^n7)/10 are 3, 7, 11, 19, 37, 43, 61, 83, 89, 107, 131, 137, 157, 191, 193, 199, 211, 229, 241, 257, 269, 307, 311, 313, 317, 379, 389, 397, 421, 431, 439, 449, 457, 479, 503, 509, 521, 523, 541, 547, 571, 577, 607, 617, 631, 641, 653, 659, 661, 691, 727, 739, 743, 751, 757, 773, 787, 797, 811, 827, 829, 907, 911, 919, 967, ..., and for the base 17 unsolved family F1{9} = (4105*17^n9)/16, the primes dividing no numbers of the form (a*b^n+c)/gcd(a+c,b1) = (4105*17^n9)/16 are 3, 5, 17, 29, 43, 59, 67, 71, 79, 101, 103, 137, 151, 157, 163, 179, 181, 191, 199, 223, 229, 239, 241, 257, 263, 281, 293, 307, 331, 337, 353, 359, 373, 383, 389, 409, 433, 443, 457, 461, 463, 491, 509, 541, 563, 587, 601, 619, 631, 647, 659, 661, 727, 733, 739, 757, 761, 769, 773, 797, 811, 821, 829, 859, 863, 877, 883, 919, 937, 947, 953, 967, 977, 991, ... (it is surprising that many primes do not divide (4105*17^n9)/16 for any n, since (4105*17^n9)/16 is a lowweight form, i.e. (4105*17^n9)/16 is divisible by a small prime for most n, note that (4105*17^n9)/16 has no algebraic factorization for any n, since 4105 is not perfect power) Last fiddled with by sweety439 on 20211018 at 14:46 
20211016, 13:54  #192 
"99(4^34019)99 palind"
Nov 2016
(P^81993)SZ base 36
3×1,039 Posts 
The smallest generalized nearrepdigit primes (i.e. of the form x{y} or {x}y) base b is always minimal primes (start with b+1) in base b unless the repeating digit (i.e. y for x{y}, or x for {x}y) is 1, since the generalized repunit numbers base b may be prime unless b is 9, 25, 32, 49, 64, 81, 121, 125, 144, ... (A096059) bases without any generalized repunit primes, and for a repdigit to be prime, it must be a repunit (i.e. the repeating digit is 1) and have a prime number of digits in its base (except trivial singledigit numbers), since, for example, the repdigit 77777 is divisible by 7, in any base > 7, thus (i.e. of the form x{y} or {x}y) base b is always minimal primes (start with b+1) in base b if the repeating digit (i.e. y for x{y}, or x for {x}y) is not 1, thus, the families A{1} in base 22 and 8{1} in base 33 and 4{1} in base 40 are not unsolved families in this problem (i.e. finding all minimal primes (start with b+1) in base b) although all they are nearrepdigit families and all they have no known primes or PRPs and none of them can be ruled out as only contain composites (only count numbers > base), since their repeating digit are 1, and the prime F(1^957) in base 24 (its value is (346*24^9571)/23) is not minimal prime (start with b+1) in base b=24, since its repeating digit is 1
Last fiddled with by sweety439 on 20211017 at 12:55 
20211017, 17:00  #193 
"99(4^34019)99 palind"
Nov 2016
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3×1,039 Posts 
The algebra form ((a*b^n+c)/d) of the unsolved families are:
Code:
base unsolved family algebra form 11 5(7^n) (57*11^n7)/10 13 9(5^n) (113*13^n5)/12 13 A(3^n)A (41*13^(n+1)+27)/4 16 (3^n)AF (16^(n+2)+619)/5 16 (4^n)DD (4*16^(n+2)+2291)/15 
20211018, 18:04  #194 
"99(4^34019)99 palind"
Nov 2016
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3·1,039 Posts 
Families which can be ruled out as contain no primes (only count numbers > base) by reasons other than trivial 1cover are:
Code:
Base 5: {1}3 (covering set {2,3}) (not produce minimal primes (start with b+1) since 111 is prime) {1}4 (covering set {2,3}) (not produce minimal primes (start with b+1) since 111 is prime) 3{1} (covering set {2,3}) (not produce minimal primes (start with b+1) since 111 is prime) 4{1} (covering set {2,3}) (not produce minimal primes (start with b+1) since 111 is prime) Base 8: 1{0}1 (sum of cubes) 6{4}7 (covering set {3,5,13}) (not produce minimal primes (start with b+1) since 44444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444447 is prime) Base 9: {1} (difference of squares) {1}5 (covering set {2,5}) 2{7} (covering set {2,5}) 3{1} (difference of squares) {3}5 (covering set {2,5}) 3{8} (difference of squares) {3}8 (covering set {2,5}) 5{1} (covering set {2,5}) 5{7} (covering set {2,5}) 6{1} (covering set {2,5}) {7}2 (covering set {2,5}) {7}5 (covering set {2,5}) {8}5 (difference of squares) Base 11: {1}3 (covering set {2,3}) (not produce minimal primes (start with b+1) since 11111111111111111 is prime) {1}4 (covering set {2,3}) (not produce minimal primes (start with b+1) since 11111111111111111 is prime) {1}9 (covering set {2,3}) (not produce minimal primes (start with b+1) since 11111111111111111 is prime) {1}A (covering set {2,3}) (not produce minimal primes (start with b+1) since 11111111111111111 is prime) 2{5} (covering set {2,3}) 3{1} (covering set {2,3}) (not produce minimal primes (start with b+1) since 11111111111111111 is prime) 3{5} (covering set {2,3}) 3{7} (covering set {2,3}) 4{1} (covering set {2,3}) (not produce minimal primes (start with b+1) since 11111111111111111 is prime) 4{7} (covering set {2,3}) {5}2 (covering set {2,3}) {5}3 (covering set {2,3}) {5}8 (covering set {2,3}) {5}9 (covering set {2,3}) {7}4 (covering set {2,3}) {7}9 (covering set {2,3}) {7}A (covering set {2,3}) 8{5} (covering set {2,3}) 9{1} (covering set {2,3}) (not produce minimal primes (start with b+1) since 11111111111111111 is prime) 9{5} (covering set {2,3}) 9{7} (covering set {2,3}) A{1} (covering set {2,3}) (not produce minimal primes (start with b+1) since 11111111111111111 is prime) A{7} (covering set {2,3}) Base 12: {B}9B (combined with covering set {13} and difference of squares) Last fiddled with by sweety439 on 20211026 at 20:56 
20211019, 11:04  #195 
"99(4^34019)99 palind"
Nov 2016
(P^81993)SZ base 36
3·1,039 Posts 
Like http://www.wiskundemeisjes.nl/wpcon...02/primes2.pdf, if you write down a prime > 10, then I can always strike out 0 or more digits to get a prime on this set: {11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97, 227, 251, 257, 277, 281, 349, 409, 449, 499, 521, 557, 577, 587, 727, 757, 787, 821, 827, 857, 877, 881, 887, 991, 2087, 2221, 5051, 5081, 5501, 5581, 5801, 5851, 6469, 6949, 8501, 9001, 9049, 9221, 9551, 9649, 9851, 9949, 20021, 20201, 50207, 60649, 80051, 666649, 946669, 5200007, 22000001, 60000049, 66000049, 66600049, 80555551, 555555555551, 5000000000000000000000000000027}
In fact, if a primes which do not contain any of the 1st to the 76th primes in the set, then this prime must be of the form 5{0}27 = 5*10^(n+2)+27, and if a primes which do not contain any of the 1st to the 75th primes in the set, then the prime must be of one of these forms: 5{0}27 = 5*10^(n+2)+27, {5}1 = (5*10^(n+1)41)/9, 8{5}1 = (77*10^(n+1)41)/9, by our theorem, such primes must contain either the 76th prime or the 77th prime as subsequence, however any such prime cannot contain both the 76th prime and the 77th prime as subsequences, since the 76th prime contain "1" and the 77th prime contain "7", any prime (in fact, any number, need not to be prime) containing both the 76th prime and the 77th prime as subsequences will contain both "1" and "7", and hence contain either "17" or "71" (or both) as subsequence, but both 17 and 71 are primes > 10 Last fiddled with by sweety439 on 20211019 at 12:47 
20211019, 13:04  #196 
"99(4^34019)99 palind"
Nov 2016
(P^81993)SZ base 36
6055_{8} Posts 
A minimal prime (start with b+1) base b is called infiniteminimal prime iff there are infinitely many primes which have this minimal prime (start with b+1) as subsequence (when written as base b string) but not have any other minimal prime (start with b+1) base b as subsequence (when written as base b string).
e.g. the largest minimal prime (start with b+1) 5000000000000000000000000000027 is likely infiniteminimal prime in base 10, since there are likely infinitely many primes of the form 5{0}27 in base 10 (reference: https://stdkmd.net/nrr/prime/primecount.txt, search: “50w27”), so is the secondlargest minimal prime (start with b+1) 555555555551, since there are likely infinitely many primes of the form {5}1 (search “5w1” in the reference page) or 8{5}1 (search “85w1” in the reference page), but the thirdlargest minimal prime (start with b+1) and the fourthlargest minimal prime (start with b+1), 80555551 and 66600049 are not infiniteminimal prime in base 10, since the only primes which have no minimal prime (start with b+1) base b=10 but 80555551 as subsequence are 80555551 and 8055555551, and the only prime which have no minimal prime (start with b+1) base b=10 but 66600049 as subsequence is 66600049. In base 2, of course, 11 is infiniteminimal prime, since all odd primes have 11 as subsequence (at least, the first digit and the last digit of all odd primes are both 1), and there are infinitely many such primes (reference: Euclid's Proof), and in base 3, all the three minimal primes (start with b+1) are very likely infiniteminimal primes, since for 12, all primes of the form 1{2} (A003307) or 1{0}2 (A051783) have no subsequence 21 or 111, and for 21, all primes of the form {2}1 (A014224) or 2{0}1 (A003306) have no subsequence 12 or 111, and for 111, all primes of the form {1} (A028491) or 1{0}11 (A058958) or 11{0}1 (A005537) have no subsequence 12 or 21, and all these seven sequences are conjectured to be infinite. Problem: In base 10, how many of the 77 minimal primes (start with b+1) are infiniteminimal primes? (assume the conjecture in this post) In base 10, 1235607889460606009419 the smallest prime containing all 26 original minimal primes (i.e. p>b not needed) as subsequences (see https://www.primepuzzles.net/puzzles/puzz_178.htm and https://primes.utm.edu/curios/page.p...606009419.html), problem: Find the smallest prime containing all 77 minimal primes (start with b+1) in base b=10, also this problem can be generalized to other bases, in base 2, this prime is clearly 11, and in base 3, this prime is 1121 (since this number is the smallest number containing both 21 and 111 as subsequences, and indeed this number is prime and containing 12 as subsequence). Last fiddled with by sweety439 on 20211111 at 03:14 
20211023, 20:29  #197 
"99(4^34019)99 palind"
Nov 2016
(P^81993)SZ base 36
110000101101_{2} Posts 
Base b=72 has largest known minimal prime (start with b+1) for all bases 2<=b<=1024: 3(71^1119849), its value is 4*72^11198491, it has 2079933 digits when written in decimal (this number is proven prime, for unproven probable prime, the largest known minimal prime (start with b+1) for all bases 2<=b<=1024 is base b=23, the PRP 9(14^800873), its value is (106*23^8008737)/11, it has 1090573 digits when written in decimal)

20211023, 20:39  #198  
"99(4^34019)99 palind"
Nov 2016
(P^81993)SZ base 36
6055_{8} Posts 
Quote:
* 1{0}2 (b^n+2): 12 digits [a record value for bases b] (minimal prime in my project, but not minimal prime in original project (i.e. the primes p<=base are also included), since this prime has "10" and "2" as subsequences) * 4{0}1 (4*b^n+1): 343 digits [a record value for bases b] * {4}1: 13 digits [not a record value for bases b, since base b=11 requires 45 digits] * 8{0}1 (8*b^n+1): 119216 digits [a record value for bases b] * {K}L (extended Sierpinski problem with k>base (k=10)): 3762 digits [a record value for base b for extended Sierpinski problem with k=10 (the prime (10*23^3762+1)/11, exponent is 3762), the previous record is b=17, the prime 10*17^1356+1, exponent is 1356] * y{z} ((b1)*b^n1, Williams prime of the 1st kind): 56 digits [a record value for bases b] * z{0}1 ((b1)*b^n+1, Williams prime of the 2nd kind): 15 digits [not a record value for bases b, since base b=19 requires 30 digits, also the same as base b=20, both require 15 digits] * {z}1 (b^n(b1), dual Williams prime of the 1st kind): 17 digits [a record value for bases b, the same as base b=20, both require 17 digits] * {z}y (b^n2): 24 digits [a record value for bases b] Last fiddled with by sweety439 on 20211026 at 22:08 

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