20220701, 19:13  #353 
"99(4^34019)99 palind"
Nov 2016
(P^81993)SZ base 36
7·509 Posts 
For a subproblem of this problem, finding the minimal (a,b,c) triple (i.e. there is no a' <= a, b' <= b, c' <= c, except the case a' = a and b' = b and c' = c) such that xxx...xxxyyy...yyyzzz...zzz (with a x's, b y's, c z's) is prime
e.g. in base 8, the family (7^a)(4^b)1 the minimal pairs of (a,b) are: (0,8) (1,7) (4,6) (12,0) they corresponding to minimal primes (start with base+1) 444444441, 744444441, 77774444441, 7777777777771, respectively. (for a = 2, the minimal b is infinity (since all numbers 77(4^b)1 are divisible by 5); for a = 3, the minimal b is 11; for a = 5, the minimal b is 143; for a = 6, the minimal b is infinity (since all numbers 777777(4^b)1 are divisible by 5); for a = 7, the minimal b is 17; for a = 8, the minimal b is 16; for a = 9, the minimal b is 15; for a = 10, the minimal b is infinity (since all numbers 7777777777(4^b)1 are divisible by 5); for a = 11, the minimal b is 97; for b = 1, the minimal a is 79; for b = 2, the minimal a is 84; for b = 3, the minimal a is 233; for b = 4, the minimal a is 56; for b = 5, the minimal a is infinity (since all numbers (7^a)444441 are divisible by 7); thus none of them can produce minimal primes (start with base+1)) e.g. in base 9, the family (8^a)(3^b)5 the minimal pairs of (a,b) are: (1,8) (9,4) (19,2) they corresponding to minimal primes (start with base+1) 8333333335, 88888888833335, 8888888888888888888335, respectively. 
20220701, 19:24  #354 
"99(4^34019)99 palind"
Nov 2016
(P^81993)SZ base 36
7·509 Posts 
Some minimal primes (start with b+1) in powerof2 bases related to Sierpinski problem, Riesel problem, dual Sierpinski problem, dual Riesel problem (see https://oeis.org/A046067, https://oeis.org/A046069, https://oeis.org/A067760, https://oeis.org/A096502, also see http://www.prothsearch.com/):
Base 8: 47777: a Riesel prime for k=5 (5*2^n1) 7777777777771: smallest dual Riesel prime for k=7 (2^n7) 777777777777777777777777777777777777777777777777777777777777777777777777777777777777441 (although not minimal prime (start with b+1) in base b = 8, but still appear in the proof of minimal prime (start with b+1) problem in base b = 8): smallest dual Riesel prime for k=223 (2^n223) Base 16: 40000000000000000000085: a dual Proth prime for k=133 (2^n+133) CAFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFF: a Riesel prime for k=203 (203*2^k1) 52000000000000000000000000000000000000000000000000000000000000000000000001: a Proth prime for k=41 (41*2^n+1) 3FFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFF23: smallest dual Riesel prime for k=221 (2^n221) 8888FFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFF: a Riesel prime for k=34953 (34953*2^n1) 88FFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFF: a Riesel prime for k=137 (137*2^n1) 2000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000321: a dual Proth prime for k=801 (2^n+801) Last fiddled with by sweety439 on 20220702 at 20:16 
20220702, 20:53  #355 
"99(4^34019)99 palind"
Nov 2016
(P^81993)SZ base 36
7·509 Posts 
We have completely solved the "minimal prime > base problem" in bases 2, 3, 4, 5, 6, 7, 8, 9, 10, 12, 14, 15, 18, 20, 24
Also, we have completely solved the "minimal prime > base problem" in the weaker case that "a number > 20000 decimal digits passes the strong primality test to all primes bases <= 61 (see https://oeis.org/A014233 and https://primes.utm.edu/glossary/xpage/StrongPRP.html) and passes the strong Lucas primality test with parameters (P, Q) defined by Selfridge's Method A (see https://oeis.org/A217255 and http://ntheory.org/pseudoprimes.html) and trial factored to 10^11 (see https://primes.utm.edu/glossary/xpag...lDivision.html)" can be regarded as primes, in bases 11, 22, 30, these unproven probable primes for bases 11, 22, 30 are: Code:
b baseb form of PRP algebraic form of PRP 11 5(7^62668) (57×11^62668−7)/10 22 B(K^22001)5 (251×22^22002−335)/21 30 I(0^24608)D 18×30^24609+13 (however, see https://primes.utm.edu/notes/prp_prob.html, bases 11, 22, 30 are in fact 99.999999999999...% (with over 10000 9's) solved, although not 100% solved, bases 2, 3, 4, 5, 6, 7, 8, 9, 10, 12, 14, 15, 18, 20, 24 are 100% solved (thus, bases 2, 3, 4, 5, 6, 7, 8, 9, 10, 12, 14, 15, 18, 20, 24 have "minimal prime > base theorem"), thus for example, we cannot definitely say that base 11 has 1068 minimal primes (start with b+1) (although this is very likely, and we can definitely say that base 11 has either 1067 or 1068 minimal primes (start with b+1), and base 11 has 1067 minimal primes (start with b+1) if and only if 5(7^62668) is in fact composite and there is no prime of the form 5{7} in base 11, but both are very impossible), and we cannot definitely say that the largest minimal primes (start with b+1) in base 11 has length 62669 (although this is very likely, and we can definitely say that the largest minimal primes (start with b+1) in base 11 has length either 1013 or >=62669, it is 1013 if and only if 5(7^62668) is in fact composite and there is no prime of the form 5{7} in base 11, and it is n (n>62669) if and only if 5(7^62668) is in fact composite and there is a larger prime of the form 5{7} in base 11 and this prime has length n), however, we can definitely say that base 24 has 3409 minimal primes (start with b+1), and we can definitely say that the largest minimal primes (start with b+1) in base 24 has length 8134, since all these primes are proven primes. Besides, for bases 13 and 16, the "minimal prime > base problem" is completely solved with the exception of these 3 families of the form x{y}z (again, in the weaker case that "a number > 20000 decimal digits passes the strong primality test to all primes bases <= 61 (see https://oeis.org/A014233 and https://primes.utm.edu/glossary/xpage/StrongPRP.html) and passes the strong Lucas primality test with parameters (P, Q) defined by Selfridge's Method A (see https://oeis.org/A217255 and http://ntheory.org/pseudoprimes.html) and trial factored to 10^11 (see https://primes.utm.edu/glossary/xpag...lDivision.html)" can be regarded as primes): Code:
b baseb form of unsolved family base b algebraic form of unsolved family base b 13 9{5} (113×13^n−5)/12 13 A{3}A (41×13^(n+1)+27)/4 16 {3}AF (16^(n+2)+619)/5 Besides, there are unproven probable primes for bases 13 and 16: Code:
b baseb form of PRP algebraic form of PRP 13 C(5^23755)C (149×13^23756+79)/12 13 8(0^32017)111 8×13^32020+183 16 D(B^32234) (206×16^32234−11)/15 16 (4^72785)DD (4×16^72787+2291)/15 3197 (likely) 3195~3197 (if strong PRP can be regarded as primes, and this is 99.999999999999...% (with over 10000 9's)) 3193~3197 (definitely say, since C(5^23755)C is in fact composite will only cause an unsolved family C{5}C, and 8(0^32017)111 is in fact composite will only cause an unsolved family 8{0}111, so there definitely cannot be more than 3197 minimal primes (start with b+1) in base 13) The "number of minimal primes (start with b+1): in base 16 is: 2347 (likely) 2346~2347 (if strong PRP can be regarded as primes, and this is 99.999999999999...% (with over 10000 9's)) 2343~2347 (definitely say, since D(B^32234) is in fact composite will only cause an unsolved family D{B}, and (4^72785)DD is in fact composite will only cause an unsolved family {4}DD, so there definitely cannot be more than 2347 minimal primes (start with b+1) in base 16) Condensed table: (bases 11, 13, 16, 22, 30 data assume the primality of the strong probable primes) Code:
b number of minimal primes base b baseb form of largest known minimal prime base b length of largest known minimal prime base b algebraic ((a×b^n+c)/d) form of largest known minimal prime base b 2 1 11 2 3 3 3 111 3 13 4 5 221 3 41 5 22 1(0^93)13 96 5^95+8 6 11 40041 5 5209 7 71 (3^16)1 17 (7^17−5)/2 8 75 (4^220)7 221 (4×8^221+17)/7 9 151 3(0^1158)11 1161 3×9^1160+10 10 77 5(0^28)27 31 5×10^30+27 11 1068 5(7^62668) 62669 (57×11^62668−7)/10 12 106 4(0^39)77 42 4×12^41+91 13 3195~3197 8(0^32017)111 32021 8×13^32020+183 14 650 4(D^19698) 19699 5×14^19698−1 15 1284 (7^155)97 157 (15^157+59)/2 16 2346~2347 (4^72785)DD 72787 (4×16^72787+2291)/15 17 10403~10428 34(7^16074) 16076 (887×17^16074−7)/16 18 549 C(0^6268)5C 6271 12×18^6270+221 20 3314 G(0^6269)D 6271 16×20^6270+13 21 13373~13396 5D(0^19848)1 19851 118×21^19849+1 22 8003 B(K^22001)5 22003 (251×22^22002−335)/21 24 3409 N00(N^8129)LN 8134 13249×24^8131−49 30 2619 O(T^34205) 34206 25×30^34205−1 36 35256~35263 (J^10117)LJ 10119 (19×36^10119+2501)/35 Last fiddled with by sweety439 on 20220730 at 01:15 
20220705, 12:36  #356 
"99(4^34019)99 palind"
Nov 2016
(P^81993)SZ base 36
7·509 Posts 
Some families cannot have covering sets (covering congruence, algebraic factorization, or combine of them) and thus there must be a prime of this form.
For the standard notation of family: (a*b^n+c)/gcd(a+c,b1) (a >= 1, b >= 2 (b is the base), c != 0, gcd(a,c) = 1, gcd(b,c) = 1), for this minimal prime (start with b+1) problem, we require a lower bound of n (i.e. n>=n_0 for a given n_0), however, in this research we can let n < n_0, even let n = 0 or n < 0, i.e. n can be any (positive or negative or 0) integer, and take the numerator of the absolute value of (a*b^n+c)/gcd(a+c,b1) (since if (a*b^n+c)/gcd(a+c,b1) have covering sets (covering congruence, algebraic factorization, or combine of them), then this covering sets must have a period of n (i.e. depending on (n mod N) for an integer N, where N is the period), even include n = 0 or n < 0, if there is a (positive or negative or 0) integer n such that the numerator of the absolute value of (a*b^n+c)/gcd(a+c,b1) has no prime factor p not dividing b, nor has algebraic factorization (it has algebraic factorization if and only if there is an integer r>1 such that a*b^n and c are both rth powers of rational numbers, or a*b^n*c is perfect 4th power), then (a*b^n+c)/gcd(a+c,b1) cannot have finite covering sets (covering congruence, algebraic factorization, or combine of them), i.e. if (a*b^n+c)/gcd(a+c,b1) has covering sets, then the covering sets must be infinite, and thus there must be a prime of this form e.g. the form D{B} in base 16, its formula is (206*16^n11)/15, and we have (take the numerator of the absolute value of the numbers): (although for this minimal prime (start with b+1) problem, n must be >= 1, but in this research of covering sets, we extend to all (positive or negative or 0) integer n n = 2: 5*19*37 n = 1: 3 * 73 n = 0: 13 n = 1: 1 n = 2: 3*29 Since for n = 1, its value is 1, and 1 has no prime factors, besides, there is no integer r>1 such that 206*16^(1) and 11 are both rth powers of rational numbers, and 206*16^(1)*(11) is not 4th power of rational number, thus the form D{B} in base 16 cannot have any kinds of covering sets, and there must be a prime of this form. The form 4{D} in base 14, its formula is 5*14^n1 n = 2: 11 * 89 n = 1: 3 * 23 n = 0: 2^2 n = 1: 3^2 n = 2: 191 Since for n = 0, its value is 4, and the only prime factor of 4 is 2, but since 2 divides 14 and then 2 cannot appear in the covering set of 5*14^n1, besides, there is no integer r>1 such that 5*14^0 and 1 are both rth powers of rational numbers, and 5*14^0*(1) is not 4th power of rational number, thus the form 4{D} in base 14 cannot have any kinds of covering sets, and there must be a prime of this form. The form 9{6}M in base 25, its formula is (37*25^n+63)/4 n = 2: 11 * 17 * 31 n = 1: 13 * 19 n = 0: 5^2 n = 1: 13 * 31 n = 2: 59 * 167 Since for n = 0, its value is 25, and the only prime factor of 25 is 5, but since 5 divides 25 and then 5 cannot appear in the covering set of (37*25^n+63)/4, besides, there is no integer r>1 such that 37*25^0 and 63 are both rth powers of rational numbers, and 37*25^0*63 is not 4th power of rational number, thus the form 9{6}M in base 25 cannot have any kinds of covering sets, and there must be a prime of this form. Note: 4*72^n1 does not apply this, since although 4*72^01 has no prime factor p not dividing 72, but 4*72^0 and 1 are both squares, thus it has algebraic factorization of difference of squares, thus we cannot show that 4*72^n1 has no covering set. An interesting case, there is two such k but still no easy prime: Base 312, form C{0}1, its formula is 12*312^n+1: n = 2: 229 * 5101 n = 1: 5 * 7 * 107 n = 0: 13 n = 1: 3^3 n = 2: 7 * 19 * 61 For n = 0, its value is 13, and the only prime factor of 13 is 13, but since 13 divides 312 and 13 cannot appear in the coveting set of 12*312^n+1 For n = 1, its value is 27, and the only prime factor of 27 is 3, but since 3 divides 312 and 3 cannot appear in the coveting set of 12*312^n+1 And 12*312^n+1 has no algebraic factorization for any n (since the difference of the exponents of 3 and the exponents of 13 is 1, and thus there cannot be r>1 dividing both of them, thus 12*312^n is never perfect power nor of the form 4*m^4), however, 12*312^n+1 has no easy prime, the first prime is n=21162 Base 100, family 4{3}, its formula is (133*100^n1)/33 n = 2: 41 * 983 n = 1: 13 * 31 n = 0: 2^2 n = 1: 1 n = 2: 13 * 23 For n = 0, its value is 4, and the only prime factor of 4 is 2, but since 2 divides 100 and 2 cannot appear in the coveting set of (133*100^n1)/33 For n = 1, its value is 1, and 1 has no prime factors And (133*100^n1)/33 has no algebraic factorization for any n (since the exponents of 7 and 19 are both 1, thus 133*100^n is never perfect power nor of the form 4*m^4), however, (133*100^n1)/33 has no easy prime, the first prime is n=5496 However, family 5{7} in base 11 does not apply this (there is no (positive or negative or 0) integer n such that (57*11^n7)/10 has no prime factor p not dividing 11), thus we cannot show that 5{7} in base 11 has no covering set. An interesting case is (2*b+1)*b^n1 (for even b), and its dual form b^n(2*b+1) (for even b), it applies this (by choose n = 0 for both (2*b+1)*b^n1 and b^n(2*b+1), both will get 2*b, and since b is even, it cannot have prime factors not dividing b, but since all prime factors of b+1 divides 1/2 of numbers in these two sequences (for all odd b, both (2*b+1)*b^n1 and b^n(2*b+1) are divisible by b+1), thus its Nash weight must * (1/2) and become very low (usually, lowweight (a*b^n+c)/gcd(a+c,b1) (a >= 1, b >= 2 (b is the base), c != 0, gcd(a,c) = 1, gcd(b,c) = 1) forms have a prime factor p dividing (a*b^n+c)/gcd(a+c,b1) (a >= 1, b >= 2 (b is the base), c != 0, gcd(a,c) = 1, gcd(b,c) = 1) for either all even n or all odd n) and may have no easy prime. If 2*b+1 is perfect square, then these two forms have a covering set with combine of covering congruence (onecover with b+1) and algebraic factorization (differenceoftwosquares factorization) For the original form: b = 46 is a classic example, its smallest prime is 93*46^241621 For the dual form: b = 18 has smallest prime 18^76837, and this prime is a minimal prime (start with b+1) in this base (the smallest prime of the dual form (if exists) must be a minimal prime (start with b+1) in this base if b is divisible by 6) A similar example is (b+2)*b^n1 with even b, it applies this (by choose n = 1), but since all prime factors of b+1 divides 1/2 of numbers, it have no easy prime for b = 352, 430, and many bases b, but it cannot produce minimal primes (start with b+1) Last fiddled with by sweety439 on 20220713 at 06:25 
20220706, 13:28  #357  
"99(4^34019)99 palind"
Nov 2016
(P^81993)SZ base 36
7·509 Posts 
Quote:
b = 20 and b = 23, family z{0}1 both have length 15 b = 20 and b = 23, family {z}1 both have length 17 Last fiddled with by sweety439 on 20220730 at 02:39 

20220707, 03:48  #358 
"99(4^34019)99 palind"
Nov 2016
(P^81993)SZ base 36
3563_{10} Posts 
This minimal prime (start with b+1) problem in base b = 100:
* Including 12:11^68 and 45:11^386 and 56:11^9235 and 67:11^105 (see https://oeis.org/A069568) * Including 2:22^9576:99 (see https://oeis.org/A111056) * Including 64:0^529396:1 and 75:0^16391:1 (see http://www.noprimeleftbehind.net/cru...tures.htm#S100) * Including 73:99^44709 (see http://www.noprimeleftbehind.net/cru...tures.htm#R100) * Including 4:3^5496 (see https://docs.google.com/document/d/e...FgpcOr1XfA/pub) * Including 52:11^4451 (see https://stdkmd.net/nrr/prime/primecount.txt) * The unsolved families 3:{43} and 7:{17} (see http://www.worldofnumbers.com/undulat.htm) But not including 7:11^5452:17 (see https://stdkmd.net/nrr/7/71117.htm), since both 7:{11} and {11}:17 have small primes. But not including 1:11^356:33 (see https://stdkmd.net/nrr/prime/primecount.txt), since this prime is covered by 1:11^9 Note: 87:{11} is not unsolved family, since it can be ruled out as only contain composites, (784*100^n1)/9 = ((28*10^n1)/9) * (28*10^n+1) Note: 38:{11} is not unsolved family, since it can be ruled out as only contain composites (the proof is more complex, it is combine of differenceofcubes and two small prime factors (3 and 37)) Last fiddled with by sweety439 on 20220714 at 00:35 
20220711, 07:48  #359 
"99(4^34019)99 palind"
Nov 2016
(P^81993)SZ base 36
110111101011_{2} Posts 
Two unusual things:
For the number of minimal primes: (This situation will not occur in larger bases) * This new problem (i.e. prime > base is needed) Base 9 (151) > Base 12 (106) > Base 10 (77) > Base 8 (75) * Original problem (i.e. prime > base is not needed) Base 10 (26) > Base 12 (17) > Base 8 (15) > Base 9 (12) The bases which are completely solved but with large minimal primes (start with b+1) in some classic simple families: * Base 7: {3}1 (indeed corresponding to the largest minimal prime (start with b+1)) * Base 8: {z}1 * Base 11: {1}, z{0}1 * Base 14: 1{0}z, 4{z} (indeed, family 4{z} corresponding to the largest minimal prime (start with b+1)) * Base 15: {3}1, y{z} * Base 20: 1{0}7, 1{z}, z{0}1, {z}1 * Base 24: 5{0}1, y{z} For unsolved bases such as: * Base 13: {y}z * Base 17: 2{0}1 * Base 19: z{0}1 * Base 28: {z}x The most unusual is for the original minimal prime problem (i.e. prime > base is not needed) in base 23, which is already solved (if probable primes are allowed), but these families all have large minimal primes: 4{0}1, y{z}, z{0}1, {z}1, {z}y (in fact, also 1{0}2, but 1{0}2 produces minimal prime only in this new problem (i.e. prime > base is needed) Finally, base 7 is much easier in base 9 (and all bases > 7), for both “number of minimal primes (start with b+1)” and “the largest minimal prime (start with b+1)”, base 7 is both much less than base 9, the largest minimal prime (start with b+1) in base 7 has only length 17, less than all bases except 2, 3, 4, 6, but for both the smallest GFN prime and the smallest GRU prime, base 7 sets records: 3334 and 11111, with lengths 4 and 5, larger than all other bases <= 10 
20220712, 02:16  #360 
"99(4^34019)99 palind"
Nov 2016
(P^81993)SZ base 36
3563_{10} Posts 
Family x{y} always produce minimal primes (start with b+1), unless y=1
Family {x}y always produce minimal primes (start with b+1), unless x=1 Family xy{x} always produce minimal primes (start with b+1) if there is no possible prime of the form y{x}, unless x=1 Family {x}yx always produce minimal primes (start with b+1) if there is no possible prime of the form {x}y, unless x=1 All minimal prime (start with b’+1) in base b’=b^n with integer n is always minimal prime (start with b+1) in base b, if this prime is > b Family A{1} in base 22 cannot produce minimal prime (start with b+1) in base b=22, since its repeating digits is indeed 1, however, it can produce minimal prime (start with b+1) in base b=484=22^2, since we can separate this family to: A:{11} in base 484 A1:{11} in base 484 And both of them can produce minimal prime (start with b+1) in base b=484, and both of them have prime candidates Family 19{1} in base 11 cannot produce minimal prime (start with b+1) in base b=11, neither can in base b=121=11^2, since: 1:91:{11} in base 121 ——> covered by the prime 1:11^8, thus cannot produce minimal prime (start with b+1) 19:{11} in base 121 ——> always divisible by 2 and cannot be prime However, it can produce minimal prime (start with b+1) in base b=1331=11^3, since we can separate this family to: 1:911:{111} in base 1331 ——> covered by the prime 1:111^6, thus cannot produce minimal prime (start with b+1), but this situation does not exist since it is always divisible by 19 and cannot be prime 19:{111} in base 1331 191:{111} in base 1331 And both the second and third of them can produce minimal prime (start with b+1) in base b=1331, and both the second and third of them have prime candidates 
20220714, 00:27  #361 
"99(4^34019)99 palind"
Nov 2016
(P^81993)SZ base 36
7×509 Posts 
the base 17 data
we use the sense of post #218 for largest minimal primes (start with b+1): B(0^^n)901: B(0^^n)91 is always divisible by 11 and cannot be prime (5^^n)2F: (5^^n)2 is divisible by 2 if the number of 5's is even and divisible by 3 if the number of 5's is odd, thus cannot be prime 2(5^^n)8: 2(5^^n) is divisible by 2 if the number of 5's is even and divisible by 3 if the number of 5's is odd, thus cannot be prime; (5^^n)8 is divisible by 2 if the number of 5's is even and divisible by 3 if the number of 5's is odd, thus cannot be prime (B^^n)E8: (B^^n)E is divisible by 2 if the number of B's is even and divisible by 3 if the number of B's is odd, thus cannot be prime; (B^^n)8 is divisible by 2 if the number of B's is even and divisible by 3 if the number of B's is odd, thus cannot be prime 9(5^^n)09: 9(5^^n) is divisible by 3 if the number of 5's is even and divisible by 2 if the number of 5's is odd, thus cannot be prime; (5^^n)09 is divisible by 3 if the number of 5's is even and divisible by 2 if the number of 5's is odd, thus cannot be prime; but 9(5^^n)9 is indeed an unsolved family 109(0^^n)D: 19(0^^n)D is always divisible by 13 and cannot be prime F(0^^n)103: F(0^^n)13 is always divisible by 5 and cannot be prime 40(D^^n): 4(D^^n) is divisible by 2 if the number of D's is even and divisible by 3 if the number of D's is odd, thus cannot be prime EG(7^^n): G(7^^n) is divisible by 2 if the number of 7's is even and divisible by 3 if the number of 7's is odd, thus cannot be prime 
20220714, 03:17  #362 
"99(4^34019)99 palind"
Nov 2016
(P^81993)SZ base 36
7·509 Posts 
If we write all minimal primes (start with b+1) in base b one time, then we will write these numbers of digits:
Code:
b=2: 0 0's 2 1's b=3: 0 0's 5 1's 2 2's b=4: 0 0's 5 1's 3 2's 3 3's b=5: 105 0's 24 1's 4 2's 22 3's 14 4's b=6: 3 0's 9 1's 2 2's 2 3's 8 4's 5 5's b=7: 43 0's 73 1's 21 2's 67 3's 24 4's 51 5's 9 6's b=8: 22 0's 49 1's 13 2's 24 3's 283 4's 48 5's 25 6's 59 7's b=9: 1350 0's 174 1's 32 2's 108 3's 24 4's 107 5's 357 6's 794 7's 58 8's b=10: 69 0's 33 1's 27 2's 9 3's 19 4's 45 5's 21 6's 31 7's 22 8's 34 9's b=11: 2666 0's 523 1's 227 2's 250 3's 722 4's 1514 5's (unless 5(7^62668) is in fact composite and there is no prime of the form 5{7}, in this case there are 1513 5's, but this is very impossible) 251 6's 66917 7's (unless 5(7^62668) is in fact composite, in this case there are n+4249 (must be > 66917) 7's where n is the smallest number n such that 5(7^n) is prime (must be > 62668) if there is a prime of the form 5{7}, or there are 4249 7's if there is no prime of the form 5{7}, but 5(7^62668) is in fact composite is very impossible) 357 8's 592 9's 1395 A's b=12: 105 0's 42 1's 28 2's 4 3's 25 4's 23 5's 14 6's 43 7's 4 8's 38 9's 38 A's 69 B's 
20220714, 03:40  #363  
"99(4^34019)99 palind"
Nov 2016
(P^81993)SZ base 36
3563_{10} Posts 
Quote:
b=11, d=7: * If 5(7^62668) is in fact composite but there is a larger prime 5{7}, then it will be the smallest n such that 5(7^n) is prime, and must be > 62668 * If 5(7^62668) is in fact composite and there is no prime of the form 5{7}, then it is 1011 (the prime 557_{1011}) b=13, d=0: * If 8(0^32017)111 is in fact composite but there is a larger prime 8{0}111, then it will be the smallest n such that 8(0^n)111 is prime, and must be > 32017 * If 8(0^32017)111 is in fact composite and there is no prime of the form 8{0}111, then it is 6540 (the prime B0_{6540}BBA) b=13, d=3: * If there is no prime of the form A{3}A, then it is 178 (the prime 3_{178}5) b=13, d=5: * If there is no prime of the form 9{5} and C(5^23755)C is in fact prime, then it is 23755 * If there is no prime of the form 9{5} and C(5^23755)C is in fact composite but there is a larger prime C{5}C, then it will be the smallest n such that C(5^n)C is prime, and must be > 23755 * If there is no prime of the form 9{5} and C(5^23755)C is in fact composite and there is no prime of the form C{5}C, then it is 713 (the prime CC5_{713}) b=16, d=3: * If there is no prime of the form {3}AF, then it is 24 (the prime 3_{24}1) b=16, d=4: * If (4^72785)DD is in fact composite but there is a larger prime {4}DD, then it will be the smallest n such that (4^n)DD is prime, and must be > 72785 * If (4^72785)DD is in fact composite and there is no prime of the form {4}DD, then it is 263 (the prime D4_{263}D) b=16, d=B: * If D(B^32234) is in fact composite but there is a larger prime D{B}, then it will be the smallest n such that D(B^n) is prime, and must be > 32234 * If D(B^32234) is in fact composite and there is no prime of the form D{B}, then it is 17804 (the prime D0B_{17804}) 

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