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Old 2022-07-01, 19:13   #353
sweety439
 
"99(4^34019)99 palind"
Nov 2016
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For a sub-problem 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.
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Old 2022-07-01, 19:24   #354
sweety439
 
"99(4^34019)99 palind"
Nov 2016
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Some minimal primes (start with b+1) in power-of-2 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^n-1)
7777777777771: smallest dual Riesel prime for k=7 (2^n-7)
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^n-223)

Base 16:

40000000000000000000085: a dual Proth prime for k=133 (2^n+133)
CAFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFF: a Riesel prime for k=203 (203*2^k-1)
52000000000000000000000000000000000000000000000000000000000000000000000001: a Proth prime for k=41 (41*2^n+1)
3FFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFF23: smallest dual Riesel prime for k=221 (2^n-221)
8888FFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFF: a Riesel prime for k=34953 (34953*2^n-1)
88FFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFF: a Riesel prime for k=137 (137*2^n-1)
2000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000321: a dual Proth prime for k=801 (2^n+801)

Last fiddled with by sweety439 on 2022-07-02 at 20:16
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Old 2022-07-02, 20:53   #355
sweety439
 
"99(4^34019)99 palind"
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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	base-b 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
("these three numbers are in fact composite" will only cause the unsolved families 5{7} in base 11, B{K}5 in base 22, I{0}D in base 30, respectively)

(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	base-b 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
(also, even when we find a prime in these three families, that number will be so big that proving its primality can cost weeks (or months or years ...), like the status of the large unproven probable primes in this project, since these three families are not of the form *{0}1 or *{b-1}, thus neither N-1 nor N+1 can be >= 25% factored, and instead ECPP primality proving such as PRIMO must be used)

Besides, there are unproven probable primes for bases 13 and 16:

Code:
b	base-b 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
The "number of minimal primes (start with b+1)" in base 13 is:

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, 17, 19, 21, 22, 30 data assume the primality of the strong probable primes)

Code:
b	number of minimal primes base b	base-b 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	10405~10428	G(7^32072)F	32074	(263×17^32073+121)/16
18	549	C(0^6268)5C	6271	12×18^6270+221
19	31400~31435	D17D(0^19750)1	19755	89674×19^19751+1
20	3314	G(0^6269)D	6271	16×20^6270+13
21	13373~13395	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 2022-08-13 at 13:27
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Old 2022-07-05, 12:36   #356
sweety439
 
"99(4^34019)99 palind"
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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,b-1) (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,b-1) (since if (a*b^n+c)/gcd(a+c,b-1) 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,b-1) 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 r-th powers of rational numbers, or a*b^n*c is perfect 4th power), then (a*b^n+c)/gcd(a+c,b-1) cannot have finite covering sets (covering congruence, algebraic factorization, or combine of them), i.e. if (a*b^n+c)/gcd(a+c,b-1) 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^n-11)/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 r-th 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^n-1

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^n-1, besides, there is no integer r>1 such that 5*14^0 and 1 are both r-th 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 r-th 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^n-1 does not apply this, since although 4*72^0-1 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^n-1 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^n-1)/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^n-1)/33

For n = -1, its value is 1, and 1 has no prime factors

And (133*100^n-1)/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^n-1)/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^n-7)/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^n-1 (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^n-1 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^n-1 and b^n-(2*b+1) are divisible by b+1), thus its Nash weight must * (1/2) and become very low (usually, low-weight (a*b^n+c)/gcd(a+c,b-1) (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,b-1) (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 (one-cover with b+1) and algebraic factorization (difference-of-two-squares factorization)

For the original form:

b = 46 is a classic example, its smallest prime is 93*46^24162-1

For the dual form:

b = 18 has smallest prime 18^768-37, 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^n-1 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 2022-07-13 at 06:25
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Old 2022-07-06, 13:28   #357
sweety439
 
"99(4^34019)99 palind"
Nov 2016
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Quote:
Originally Posted by sweety439 View Post
Records for the lengths of the numbers in these families in base b:

{1}: ((b^n-1)/(b-1), length n)

2 (2)
3 (3)
7 (5)
11 (17)
19 (19)
35 (313)
39 (349)

1{0}1: (b^n+1, length n+1)

2 (2)
14 (3)
34 (5)

1{0}2: (b^n+2, length n+1)

3 (2)
23 (12)
47 (114)
89 (256)

{z}y: (b^n-2, length n)

3 (2)
11 (4)
17 (6)
23 (24)
79 (38)
81 (130)
97 (747)

1{0}3: (b^n+3, length n+1)

4 (2)
22 (3)
32 (4)
46 (21)
292 (40)
382 (256)

{z}x: (b^n-3, length n)

4 (2)
16 (3)
22 (6)
28 (10)
50 (21)
52 (105)
94 (204)
152 (346)
154 (396)

1{0}4: (b^n+4, length n+1)

5 (3)
23 (7)

{z}w: (b^n-4, length n)

5 (5)
27 (7)
35 (13)
47 (65)
65 (175)
123 (299)
141 (395)

1{0}z: (b^n+(b-1), length n+1)

2 (2)
5 (3)
14 (17)
32 (109)
80 (195)

{z}1: (b^n-(b-1), length n)

2 (2)
5 (5)
8 (13)
20 (17)
29 (33)
37 (67)

1{0}11: (b^n+(b+1), length n+1)

2 (3)
9 (4)
11 (5)
23 (10)
35 (16)
63 (74)
68 (596)

{z}yz: (b^n-(b+1), length n)

2 (3)
13 (4)
19 (5)
33 (7)
37 (9)
43 (31)
52 (108)
99 (131)
190 (562)
213 (643)

2{0}1: (2*b^n+1, length n+1)

3 (2)
12 (4)
17 (48)
38 (2730)

1{z}: (2*b^n-1, length n+1)

2 (2)
5 (5)
20 (11)
29 (137)
67 (769)

3{0}1: (3*b^n+1, length n+1)

4 (2)
8 (3)
18 (4)
28 (8)
44 (10)
62 (13)
72 (15)
108 (271)
314 (281)

2{z}: (3*b^n-1, length n+1)

4 (2)
12 (3)
32 (12)
42 (2524)

4{0}1: (4*b^n+1, length n+1)

5 (3)
17 (7)
23 (343)

3{z}: (4*b^n-1, length n+1)

5 (2)
23 (6)
47 (1556)

z{0}1: ((b-1)*b^n+1, length n+1)

2 (2)
5 (3)
10 (4)
11 (11)
19 (30)
41 (81)
53 (961)

y{z}: ((b-1)*b^n-1, length n+1)

2 (3)
8 (4)
15 (15)
23 (56)
26 (134)

11{0}1: ((b+1)*b^n+1, length n+2)

2 (3)
9 (4)
18 (11)
51 (185)
63 (187)
108 (400)

10{z}: ((b+1)*b^n-1, length n+2)

2 (3)
12 (4)
17 (6)
23 (9)
42 (10)

{y}z: (((b-2)*b^n+1)/(b-1), length n)

3 (1)
5 (2)
13 (564)
83 (680)
An interesting thing:

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 2022-07-30 at 02:39
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Old 2022-07-07, 03:48   #358
sweety439
 
"99(4^34019)99 palind"
Nov 2016
(P^81993)SZ base 36

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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^n-1)/9 = ((28*10^n-1)/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 difference-of-cubes and two small prime factors (3 and 37))

Last fiddled with by sweety439 on 2022-07-14 at 00:35
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Old 2022-07-11, 07:48   #359
sweety439
 
"99(4^34019)99 palind"
Nov 2016
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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
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Old 2022-07-12, 02:16   #360
sweety439
 
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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
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Old 2022-07-14, 00:27   #361
sweety439
 
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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
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Old 2022-07-14, 03:17   #362
sweety439
 
"99(4^34019)99 palind"
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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
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Old 2022-07-14, 03:40   #363
sweety439
 
"99(4^34019)99 palind"
Nov 2016
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Quote:
Originally Posted by sweety439 View Post

Code:
b=2, d=0: 0
b=2, d=1: 2 (the prime 11)
b=3, d=0: 0
b=3, d=1: 3 (the prime 111)
b=3, d=2: 1 (the primes 12 and 21)
b=4, d=0: 0
b=4, d=1: 2 (the prime 11)
b=4, d=2: 2 (the prime 221)
b=4, d=3: 1 (the primes 13, 23, 31)
b=5, d=0: 93 (the prime 109313)
b=5, d=1: 3 (the prime 111)
b=5, d=2: 1 (the primes 12, 21, 23, 32)
b=5, d=3: 4 (the prime 33331)
b=5, d=4: 4 (the primes 14444 and 44441)
b=6, d=0: 2 (the prime 40041)
b=6, d=1: 2 (the prime 11)
b=6, d=2: 1 (the primes 21 and 25)
b=6, d=3: 1 (the primes 31 and 35)
b=6, d=4: 3 (the prime 4441)
b=6, d=5: 1 (the primes 15, 25, 35, 45, 51)
b=7, d=0: 7 (the prime 5100000001)
b=7, d=1: 5 (the prime 11111)
b=7, d=2: 3 (the prime 1222)
b=7, d=3: 16 (the prime 3161)
b=7, d=4: 2 (the primes 344, 445, 544, 4504, 40054)
b=7, d=5: 4 (the prime 35555)
b=7, d=6: 2 (the prime 6634)
b=8, d=0: 3 (the prime 500025)
b=8, d=1: 3 (the prime 111)
b=8, d=2: 2 (the prime 225)
b=8, d=3: 3 (the prime 3331)
b=8, d=4: 220 (the prime 42207)
b=8, d=5: 14 (the prime 51325)
b=8, d=6: 2 (the primes 661 and 667)
b=8, d=7: 12 (the prime 7121)
b=9, d=0: 1158 (the prime 30115811)
b=9, d=1: 36 (the prime 56136)
b=9, d=2: 4 (the prime 22227)
b=9, d=3: 8 (the prime 8333333335)
b=9, d=4: 11 (the prime 5411)
b=9, d=5: 4 (the prime 55551)
b=9, d=6: 329 (the prime 763292)
b=9, d=7: 687 (the prime 2768607)
b=9, d=8: 19 (the prime 819335)
b=10, d=0: 28 (the prime 502827)
b=10, d=1: 2 (the prime 11)
b=10, d=2: 3 (the prime 2221)
b=10, d=3: 1 (the primes 13, 23, 31, 37, 43, 53, 73, 83, 349)
b=10, d=4: 2 (the prime 449)
b=10, d=5: 11 (the prime 5111)
b=10, d=6: 4 (the prime 666649)
b=10, d=7: 2 (the primes 277, 577, 727, 757, 787, 877)
b=10, d=8: 2 (the prime 881)
b=10, d=9: 3 (the prime 9949)
b=11, d=0: 126 (the prime 5012657)
b=11, d=1: 17 (the prime 117)
b=11, d=2: 6 (the prime 5222222)
b=11, d=3: 10 (the prime 3107)
b=11, d=4: 44 (the prime 4441)
b=11, d=5: 221 (the prime 8522005]
b=11, d=6: 124 (the prime 326124)
b=11, d=7: 62668 (the prime 5762668)
b=11, d=8: 17 (the prime 8173)
b=11, d=9: 32 (the prime 9321)
b=11, d=A: 713 (the prime A71358)
b=12, d=0: 39 (the prime 403977)
b=12, d=1: 2 (the prime 11)
b=12, d=2: 3 (the prime 222B)
b=12, d=3: 1 (the primes 31, 35, 37, 3B)
b=12, d=4: 3 (the prime 4441)
b=12, d=5: 2 (the primes 565 and 655)
b=12, d=6: 2 (the prime 665)
b=12, d=7: 3 (the primes 4777 and 9777)
b=12, d=8: 1 (the primes 81, 85, 87, 8B)
b=12, d=9: 4 (the prime 9999B)
b=12, d=A: 4 (the prime AAAA1)
b=12, d=B: 7 (the prime BBBBBB99B)
b=13, d=0: 32017 (the prime 8032017111)
b=13, d=1: 5 (the prime 11111)
b=13, d=2: 77 (the prime 7277)
b=13, d=3: >82000 (the prime A3nA)
b=13, d=4: 14 (the prime 9414)
b=13, d=5: >88000 (the prime 95n)
b=13, d=6: 137 (the prime 6137A3)
b=13, d=7: 1504 (the prime 715041)
b=13, d=8: 53 (the prime 8537)
b=13, d=9: 1362 (the prime 913625)
b=13, d=A: 95 (the prime C5A95)
b=13, d=B: 834 (the prime B83474)
b=13, d=C: 10631 (the prime C1063192)
b=14, d=0: 83 (the prime 408349)
b=14, d=1: 3 (the prime 111)
b=14, d=2: 3 (the prime B2225)
b=14, d=3: 5 (the prime A33333)
b=14, d=4: 63 (the prime 46309)
b=14, d=5: 36 (the prime 8536)
b=14, d=6: 10 (the prime 861099)
b=14, d=7: 2 (the primes 771, 77D)
b=14, d=8: 86 (the prime 886B)
b=14, d=9: 37 (the prime 93689)
b=14, d=A: 59 (the prime A593)
b=14, d=B: 78 (the prime 6B772B)
b=14, d=C: 79 (the prime 8C793)
b=14, d=D: 19698 (the prime 4D19698)
b=15, d=0: 33 (the prime 503317)
b=15, d=1: 3 (the prime 111)
b=15, d=2: 9 (the prime 2222222252)
b=15, d=3: 12 (the prime 3121)
b=15, d=4: 3 (the prime 4434)
b=15, d=5: 8 (the prime 555555557)
b=15, d=6: 104 (the prime 9610408)
b=15, d=7: 156 (the prime 715597)
b=15, d=8: 8 (the prime 8888888834)
b=15, d=9: 10 (the prime 9999999999D)
b=15, d=A: 4 (the prime AAAA52)
b=15, d=B: 31 (the prime EB31)
b=15, d=C: 10 (the prime DCCCCCCCCCC8)
b=15, d=D: 16 (the prime D16B)
b=15, d=E: 145 (the prime E145397)
b=16, d=0: 3542 (the prime 90354291)
b=16, d=1: 2 (the prime 11)
b=16, d=2: 32 (the prime 2327)
b=16, d=3: >76000 (the prime 3nAF)
b=16, d=4: 72785 (the prime 472785DD)
b=16, d=5: 70 (the prime A01570)
b=16, d=6: 87 (the prime 5687F)
b=16, d=7: 20 (the prime 71987)
b=16, d=8: 1517 (the prime F81517F)
b=16, d=9: 1052 (the prime D91052)
b=16, d=A: 305 (the prime DA3055)
b=16, d=B: 32234 (the prime DB32234)
b=16, d=C: 3700 (the prime 5BC3700D)
b=16, d=D: 39 (the prime 4D39)
b=16, d=E: 34 (the prime E34B)
b=16, d=F: 1961 (the prime 300F1960AF)
Some of them are conjectured numbers (they are 99.9999999999999... (> 10000 9's) correct, but not 100% correct), they are:

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 5571011)

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 B06540BBA)

b=13, d=3:

* If there is no prime of the form A{3}A, then it is 178 (the prime 31785)

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 CC5713)

b=16, d=3:

* If there is no prime of the form {3}AF, then it is 24 (the prime 3241)

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 D4263D)

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 D0B17804)
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