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-   -   Minimal set of the strings for primes with at least two digits (https://www.mersenneforum.org/showthread.php?t=24972)

sweety439 2022-02-12 20:20

1 Attachment(s)
[QUOTE=sweety439;597067]* Case (2,1):

** [B]21[/B] is prime, and thus the only minimal prime in this family.

* Case (2,2):

** Since 21, 25, 12, 32, 52, [B]272[/B] are primes, we only need to consider the family 2{0,2,4,6,8}2 (since any digits 1, 3, 5, 7 between them will produce smaller primes)

*** All numbers of the form 2{0,2,4,6,8}2 are divisible by 2, thus cannot be prime

* Case (2,4):

** Since 21, 25, 14, 34, 74 are primes, we only need to consider the family 2{0,2,4,6,8}4 (since any digits 1, 3, 5, 7 between them will produce smaller primes)

*** All numbers of the form 2{0,2,4,6,8}4 are divisible by 2, thus cannot be prime

* Case (2,5):

** [B]25[/B] is prime, and thus the only minimal prime in this family.

* Case (2,7):

** Since 21, 25, 47, 67, 87 are primes, we only need to consider the family 2{0,2,3,7}7 (since any digits 1, 4, 5, 6, 8 between them will produce smaller primes)

*** If there are at least two 3's in {}, then 337 will be a subsequence.

*** If there are exactly one 3's in {}, then there cannot be 7's in {}, otherwise, either 377 or 737 will be a subsequence, thus the family is 2{0,2}3{0,2}7

**** All numbers of the form 2{0,2}3{0,2}7 are divisible by 2, thus cannot be prime

*** If there are no 3's in {}, then the family will be 2{0,2,7}7

**** Since [B]2027[/B] and 272 are primes, we only need to consider the family 2{2}{0,7}7 (since any digits combo 02, 72 between them will produce smaller primes, thus let "d" be the rightmost digit 2 in {}, then all digits before "d" are 2 (cannot be 0 or 7, otherwise 02 or 72 will be in {}, and hence either [B]2027[/B] or 272 will be a subsequence); also, all digits after "d" are 0 or 7, since "d" is the rightmost digit 2, thus the family is 2{2}{0,7}7)

***** Since [B]22227[/B] is prime, we only need to consider the families 2{0,7}7, 22{0,7}7, 222{0,7}7

****** Since [B]2207[/B] is prime, we only need to consider the families 2{0,7}7, 22{7}7, 222{7}7

******* For the 2{0,7}7 family, since the digit sum of primes must be odd (otherwise the number will be divisible by 2, thus cannot be prime), there is an odd total number of 7

******** If there are only 1 7's, then the form is 2{0}7

********* The smallest prime of the form 2{0}7 is [B]2000000000007[/B]

******** If there are at least 3 7's, then there cannot be any 0 before the 3rd rightmost 7, or [B]20777[/B] will be a subsequence, thus the family is 2{7}7{0}7{0}7

********* Since [B]270707[/B] is prime, we only need to consider the families 2{7}7{0}77 and 2{7}77{0}7

********** All numbers of the form 2{7}7{0}77 are divisible by 2 (if the total number of 7's is even) or 5 (if the total number of 7's is odd), thus cannot be prime.

********** For the 2{7}77{0}7 family, since [B]2770007[/B] is prime, we only need to consider the families 2{7}777, 2{7}7707, 2{7}77007

*********** All numbers of the form 2{7}777 are divisible by 2 (if the total number of 7's is even) or 5 (if the total number of 7's is odd), thus cannot be prime.

*********** The smallest prime of the form 2{7}7707 is [B]27777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777707[/B] with 687 7's, which can be written as 2(7^686)07 and equal the prime (23*9^688-511)/8 ([URL="http://factordb.com/cert.php?id=1100000002495467486"]primality certificate of this prime[/URL])

*********** All numbers of the form 2{7}77007 are divisible by 2 (if the total number of 7's is even) or 5 (if the total number of 7's is odd), thus cannot be prime.

******* The smallest prime of the form 22{7}7 is [B]22777[/B]

******* All numbers of the form 222{7}7 are divisible by 2 (if the total number of 7's is even) or 5 (if the total number of 7's is odd), thus cannot be prime.

* Case (2,8):

** Since 21, 25, 18, 58, 78, [B]238[/B] are primes, we only need to consider the family 2{0,2,4,6,8}8 (since any digits 1, 3, 5, 7 between them will produce smaller primes)

*** All numbers of the form 2{0,2,4,6,8}8 are divisible by 2, thus cannot be prime[/QUOTE]

* Case (3,1):

** Since 32, 34, 21, 41, 81, [B]331[/B], [B]371[/B] are primes, we only need to consider the family 3{0,1,5,6}1 (since any digits 2, 3, 4, 7, 8 between them will produce smaller primes)

*** If there are at least two 5's in {}, then 355 will be a subsequence.

*** If there are exactly one 5's in {}, then the form is 3{0,1,6}5{0,1,6}1

**** Since 315, 65, [B]3501[/B], [B]3561[/B] are primes, we only need to consider the family 3{0}5{1}1 (since any digit 1, 6 between (3,5{0,1,6}1) will produce small primes, and any digit 0, 6 between (3{0,1,6}5,1) will produce small primes)

***** Since [B]305111[/B] is prime, we only need to consider the families 35{1}1, 3{0}51, 3{0}511

****** The smallest prime of the form 35{1}1 is [B]351111111[/B]

****** The smallest prime of the form 3{0}51 is [B]30000000000000000000051[/B]

****** All numbers of the form 3{0}511 are divisible by 2, thus cannot be prime

*** If there are no 5's in {}, then the form is 3{0,1,6}1

**** If there are no 1's in {}, then the form is 3{0,6}1

***** All numbers of the form 3{0,6}1 are divisible by 2, thus cannot be prime

**** If there are 0's and 1's and 6's in {}, since 3101 and 3611 are primes, thus the 0's must before the 1's and the 1's must before the 6's

***** We have the prime [B]30161[/B]

**** If there no 6's in {}, then the form is 3{0,1}1

***** Since [B]3101[/B] is prime, we only need to consider the family 3{0}{1}1

****** Since [B]301111[/B] is prime, we only need to consider the families 3{1}1, 3{0}1, 3{0}11, 3{0}111

******* All numbers of the form 3{1}1 factored as (27*10^n-1)/8 = (5*3^n-1)/2 * (5*3^n+1)/4 (if n is odd) or (5*3^n-1)/4 * (5*3^n+1)/2 (if n is even), thus cannot be prime

******* All numbers of the form 3{0}1 are divisible by 2, thus cannot be prime

******* The smallest prime of the form 3{0}11 is [B]300000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000011[/B] with 1158 0's, which can be written as 3(0^1158)11 and equal the prime 3*9^1160+10 ([URL="http://factordb.com/cert.php?id=1100000002376318423"]primality certificate of this prime[/URL])

******* All numbers of the form 3{0}111 are divisible by 2, thus cannot be prime

**** If there no 0's in {}, then the form is 3{1,6}1

***** Since [B]3611[/B] is prime, we only need to consider the family 3{1}{6}1

****** Since 661 is prime, we only need to consider the families 3{1}1 and 3{1}61

******* All numbers of the form 3{1}1 factored as (27*10^n-1)/8 = (5*3^n-1)/2 * (5*3^n+1)/4 (if n is odd) or (5*3^n-1)/4 * (5*3^n+1)/2 (if n is even), thus cannot be prime

******* The smallest prime of the form 3{1}61 is [B]311111111161[/B]

sweety439 2022-02-12 23:20

* Case (3,2):

** [B]32[/B] is prime, and thus the only minimal prime in this family.

* Case (3,4):

** [B]34[/B] is prime, and thus the only minimal prime in this family.

* Case (3,5):

** Since 32, 34, 25, 45, 65, [B]315[/B], [B]355[/B], [B]375[/B] are primes, we only need to consider the family 3{0,3,8}5 (since any digits 1, 2, 4, 5, 6, 7 between them will produce smaller primes)

*** If there is no 8 in {}, then the form is 3{0,3}5

**** If there is no 0 in {}, then the form is 3{3}5

***** All numbers of the form 3{3}5 are divisible by 2 (if the total number of 3's is odd) or 5 (if the total number of 3's is even), thus cannot be prime

**** If there is at least one 0 in {}, then there must be either <= 1 3's in {} before the rightmost 0 in {} or no 3's in {} after the rightmost 0 in {}, otherwise [B]333035[/B] will be a subsequence (in fact, not only for the rightmost 0 in {}, this is true for any 0 in {})

***** If there is no 3's in {} after the rightmost 0, then the form is 3{0,3}05

******

***** If there is no 3's in {} before the rightmost 0, then the form is 3{0}{3}5

****** Since [B]30333335[/B] is prime, we only need to consider the families 3{3}5, 3{0}35, 3{0}335, 3{0}3335, 3{0}33335

******* All numbers of the form 3{3}5 are divisible by 2 (if the total number of 3's is odd) or 5 (if the total number of 3's is even), thus cannot be prime

******* The smallest prime of the form 3{0}35 is [B]300000000035[/B]

******* All numbers of the form 3{0}335 are divisible by 2, thus cannot be prime

******* The smallest prime of the form 3{0}3335 is 30000000003335 (not minimal prime, since 300000000035 is prime)

******* All numbers of the form 3{0}33335 are divisible by 2, thus cannot be prime

***** If there is exactly one 3's in {} before the rightmost 0, then the form is 3{0}{3}05

****** Since 30333335 is prime, we only need to consider the families 3{3}05, 3{0}305, 3{0}3305, 3{0}33305, 3{0}333305

******* All numbers of the form 3{3}05 are divisible by 2 (if the total number of 3's is odd) or 5 (if the total number of 3's is even), thus cannot be prime

******* The smallest prime of the form 3{0}305 is 3000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000305 (not minimal prime, since 300000000035 is prime)

******* All numbers of the form 3{0}3305 are divisible by 2, thus cannot be prime

******* The smallest prime of the form 3{0}33305 is [B]300033305[/B]

******* All numbers of the form 3{0}333305 are divisible by 2, thus cannot be prime

sweety439 2022-02-14 14:44

3x5 base 9

x={0,3}

minimal elements of x in {0,3} such that 3x5 (base 9) is prime:

[CODE]
3303
03033
030000
033333
0003000
0003330
0000000003
[/CODE]

x must contain an odd number of 3's

if x is of these forms, then 3x5 (base 9) cannot be prime:

[CODE]
{3}
{3}{0}
30{3}
[/CODE]

Note that 3x cannot be prime since it is always divisible by 3, and x5 is a subset of the original numbers (i.e. 3x5), since it must be begin with 3 (numbers cannot be [URL="https://en.wikipedia.org/wiki/Leading_zero"]begin with 0[/URL])

Edit: Found 3033300005 (base 9) is prime, but this is not minimal prime (start with b+1) since 30300005 (base 9) is prime.

Since the numbers of the form {3}{0}5 can be excluded as they are always divisible by either 2 or 5, "03" must be subsequence of x, thus ....

* The maximum number of 0's in x is 11, since
** There must be at least one 3 in x, or the number is of the form 3{0}5 and divisible by 2, thus let the form be 3y3z5 (i.e. let x be y3z), we can let the "3" in y3z be the rightmost 3 in x, i.e. there is no 3 in z
*** There are at most 8 0's in y, or 300000000035 will be a subsequence
*** There are at most 3 0's in z, or 30300005 will be a subsequence (note that there must be at least one 0 in y, or the form will be {3}{0}5 (note that we already assume that there is no 3 in z), and this form is already ruled out (as all numbers of this form are divisible by either 2 or 5))

* The maximum number of 3's in x is 5, since
** There must be at least one 0 in x, or the number is of the form 3{3}5 and divisible by either 2 or 5, thus let the form be 3y0z5 (i.e. let x be y0z), we can let the "0" in y0z be the leftmost 0 in x, i.e. there is no 0 in y
*** There are at most 1 3's in y, or 333035 will be a subsequence (note that there must be at least one 3 in z, or the form will be {3}{0}5 (note that we already assume that there is no 0 in y), and this form is already ruled out (as all numbers of this form are divisible by either 2 or 5))
*** There are at most 4 3's in z, or 30333335 will be a subsequence

Thus, the length of x is at most 16, and thus the length of the minimal prime (start with b+1) in base b=9, starting with 3 and ending with 5, is at most 18

sweety439 2022-02-25 11:42

4 Attachment(s)
update PRIMO certifate files of some primes in this project (B(0^1765)999B (base b=12) is not minimal prime (start with b+1), since B(0^27)9B (base b=12) is prime, but B(0^1765)999B (base b=12) still appear in the proof, currently the only unproven PRP appearing in the proof is 3(5^9234)4 (base b=7))

sweety439 2022-03-01 18:59

Proving that the set of the minimal primes (start with b+1) in base b is S is equivalent to:

* Proving that all strings in S are primes
* Proving that all strings in S are pairwise incomparable (for the subsequence ordering)
* Proving that all strings > b not contain any string in S as subsequence are composite

The first part needs to use [URL="http://www.ellipsa.eu/public/primo/primo.html"]Primo[/URL] to prove, for small primes it is very easy to prove, also for numbers < 10^1000 using [URL="https://www.numberempire.com/primenumbers.php"]this website[/URL] to prove or disprove (maybe they are [URL="https://primes.utm.edu/glossary/xpage/Pseudoprime.html"]pseudoprimes[/URL]), also for numbers whose N-1 and/or N+1 can be >=33.3333% factored, use [URL="https://primes.utm.edu/prove/prove3_1.html"]N-1 test[/URL] or [URL="https://primes.utm.edu/prove/prove3_2.html"]N+1 test[/URL] to prove

The second part is just obvious, or if not obvious, you can use the "subsequence checker"

The third part needs to use number theory theorems to show that some families contain no primes (only count the numbers > b), and if S contains large primes, you should use [URL="https://en.wikipedia.org/wiki/Trial_division"]trial division[/URL] and [URL="https://en.wikipedia.org/wiki/Fermat_primality_test"]Fermat primality test[/URL] to show that these primes are the first prime (only count the numbers > b) in the corresponding families (by using the sieving program [URL="https://www.rieselprime.de/ziki/Srsieve"]srsieve[/URL] and the primality testing program [URL="https://www.rieselprime.de/ziki/PFGW"]PFGW[/URL] or [URL="https://www.rieselprime.de/ziki/LLR"]LLR[/URL])

sweety439 2022-03-01 19:19

e.g.

proving S(2) = {11}

* 11 (decimal 3) is prime (trivial, just use the [URL="https://en.wikipedia.org/wiki/Sieve_of_Eratosthenes"]sieve of Eratosthenes[/URL])
* all strings in {11} are pairwise incomparable (trivial, since the set {11} has only one element: 11)
* all strings > 10 (decimal 2) not contain any string in {11} as subsequence are composite (very easy to prove, since such string must contain at most one 1, and since if it contains no 1 then its value will be 0 and not > 10, it must contain exactly one 1, and since it is > 10, it must contain at least two 0's in the right of the 1, thus the number is divisible by 100 and hence composite)

proving S(3) = {12, 21, 111}

* 12 (decimal 5), 21 (decimal 7), 111 (decimal 13) are all primes (trivial, just use the [URL="https://en.wikipedia.org/wiki/Sieve_of_Eratosthenes"]sieve of Eratosthenes[/URL])
* all strings in {12, 21, 111} are pairwise incomparable (just obvious, none of 12, 21, 111 contains another of 12, 21, 111 as subsequence)
* all strings > 10 (decimal 3) not contain any string in {12, 21, 111} as subsequence are composite (since such numbers cannot contain 1 and 2 simultaneously (or either 12 or 21 will be a subsequence), and if it does not contain 1 then it contains only 0 and 2 (or only 0, or only 2), and hence its value must be even, and since it is > 10, it must be composite, besides, if it does not contain 2 then it contains only 0 and 1 (or only 0, or only 1), but it can contain at most two 1's (or 111 will be a subsequence), however, if it contain no 1 or exactly two 1, its value must be even, and since it is > 10, it must be composite, and if it contain exactly one 1, since it is > 10, it must contain at least two 0's in the right of the 1, thus the number is divisible by 100 and hence composite)

proving S(10) = {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}

* 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 are all primes (you can use [URL="https://oeis.org/A000040/b000040_1.txt"]this online list[/URL] to check all primes in S up to 5200007, and for large primes you can either [URL="http://www.ellipsa.eu/public/primo/primo.html"]Primo[/URL] to prove or use [URL="https://www.numberempire.com/primenumbers.php"]this website[/URL] to check whether they are prime or not)
* all strings in {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} are pairwise incomparable (just obvious, if not obvious for you, you can use the "subsequence checker" to check if they contain another string in S as subsequence or not)
* all strings > 10 not contain any string in {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} as subsequence are composite (first, you need to use the divisibility rule of 2, 3, 5 to remove the numbers in some families, then prove that all numbers of the form 28{0}7 or 4{6}9 (as well as the numbers containing only 0 and 7) are divisible by 7, for some non-obvious numbers, like 6049 and 50027, you need to use trial division or Fermat primality test to show that they are composite)

sweety439 2022-03-05 21:17

1 Attachment(s)
Upload newest pdf file about this research.

sweety439 2022-03-11 21:41

Like the sense of [URL="https://github.com/curtisbright/mepn-data/commit/7acfa0656d3c6b759f95a031f475a30f7664a122"]https://github.com/curtisbright/mepn-data/commit/7acfa0656d3c6b759f95a031f475a30f7664a122[/URL] and [URL="https://github.com/curtisbright/mepn-data/commit/e6b2b806f341e9dc5b96662edba2caf3220c98b7"]https://github.com/curtisbright/mepn-data/commit/e6b2b806f341e9dc5b96662edba2caf3220c98b7[/URL], for the minimal prime (start with b+1) problem base b:

* Bases 2, 3, 4 have no minimal primes (start with b+1) with >3 digits (see [URL="https://cs.uwaterloo.ca/~cbright/reports/mepn.pdf"]this article[/URL]), thus the "unsolved family" list for these bases will be empty.

* For base 5, "unsolved family" list should include the non-simple families 1{0}1{0}3 and 3{0}3{0}1, and with primes 10103, 30301, 33001, these families become 1{0}13, 3{0}31 (note: 11{0}3 can be removed, since all numbers in this family are divisible by 3)

* For base 5, "unsolved family" list should include families 1{4}, {3}1, {4}1

* For base 10, "unsolved family" list should include the non-simple families 2{0}2{0}1 and 5{0}2{0}7 and 6{0,6}49, and with primes 20201, 20021, 50207, 60649, these families become 22{0}1, 52{0}7, 5{0}27, {6}{0}49, and with the prime 666649, the family {6}{0}49 becomes 6{0}49, 66{0}49, 666{0}49

* For base 10, "unsolved family" list should include families {5}1, 8{5}1, 80{5}1, and with the prime 80555551, the family 8{5}1 become unneeded and removed, and only leave the family {5}1

sweety439 2022-03-25 19:12

[URL="https://oeis.org/A034388"]https://oeis.org/A034388[/URL]: Smallest prime containing at least n consecutive identical digits.

This sequence is related the project in this forum, since for example, the largest minimal prime (start with b+1) in base b=10 is 5(0^28)27, it makes that [URL="https://oeis.org/A034388"]A034388[/URL](28) <= 5(0^28)27

Similarly, for base b=16, D(B^32234) is minimal prime (start with b+1), assuming its primality, thus for the analog sequence of [URL="https://oeis.org/A034388"]https://oeis.org/A034388[/URL] in base 16, a(32234) <= D(B^32234), also for base b=14, 4(D^19698) is minimal prime (start with b+1), thus for the analog sequence of [URL="https://oeis.org/A034388"]https://oeis.org/A034388[/URL] in base 14, a(19698) <= 4(D^19698), and for base b=13, 8(0^32017)111 is minimal prime (start with b+1), assuming its primality, thus for the analog sequence of [URL="https://oeis.org/A034388"]https://oeis.org/A034388[/URL] in base 13, a(32017) <= 8(0^32017)111

sweety439 2022-04-02 20:00

[QUOTE=sweety439;599662]In base b=8 such prime does not exist, since the family 1{0}1 has infinite subset whose elements are pairwise coprime but the family 1{0}1 can be ruled out as only contain composite numbers, but if we only consider the numbers not in the family 1{0}1, such prime will be the smallest prime factor of (4*8^217+17)/7 (if it is > 7885303569123738614221) or 7885303569123738614221, which is needed to remove the composites (4^216)7 and (4^116)7, respectively, see [URL="http://factordb.com/index.php?query=%284*8%5E%28n%2B1%29%2B17%29%2F7&use=n&n=1&VP=on&VC=on&EV=on&OD=on&PR=on&FF=on&PRP=on&CF=on&U=on&C=on&perpage=200&format=1&sent=Show"]factorization of {4}7 in base 8[/URL][/QUOTE]

(4*8^217+17)/7 has been factored by me (using [URL="http://www.alpertron.com.ar/ECM.HTM"]ECM Integer factorization calculator[/URL]), one prime factor 46096827569809626166519 was found, and the remaining number is a 174-digit composite.

Since 46096827569809626166519 (23 digits) > 7885303569123738614221 (22 digits), if we only consider the numbers not in the family 1{0}1, such prime is 46096827569809626166519 if [URL="http://factordb.com/index.php?id=1100000002999513593"]the 174-digit composite[/URL] has no prime factor < 46096827569809626166519, I just use the ECM factorization calculator to find this 23-digit factor, this factor may not be the smallest prime factor of (4*8^217+17)/7, like the status of the Mersenne number [URL="https://www.mersenne.org/report_exponent/?exp_lo=1237&full=1"]M1237[/URL], it is unknown whether the only known 70-digit prime factor is the smallest prime factor of M1237, see [URL="https://oeis.org/A016047"]https://oeis.org/A016047[/URL], if there is a smaller prime factor of (4*8^217+17)/7, then....

* If this prime factor is < 7885303569123738614221, then such prime in base 8 is 7885303569123738614221
* If this prime factor is between 7885303569123738614221 and 46096827569809626166519, then such prime in base 8 is this prime

See [URL="http://factordb.com/index.php?query=%284*8%5E%28n%2B1%29%2B17%29%2F7&use=n&n=1&VP=on&VC=on&EV=on&PR=on&FF=on&PRP=on&CF=on&U=on&C=on&perpage=200&format=1&sent=Show"]factorization of (4*8^(n+1)+17)/7 in factordb[/URL], since for odd n this number is divisible by 3 (and 3 must be the smallest prime factor of this number), we only list even n

sweety439 2022-04-05 07:59

The smallest prime of the form y{z} in base b for b = 2, 3, 4, ..., 36 are (always minimal prime (start with b+1) base b):

3, 5, 11, 19, 29, 41, 3583, 71, 89, 109, 131, 2027, 181, 408700964355468749, 239, 271, 5507, 846825857, 379, 419, 461, 17276416353328819798072137388863592892072278184923153720493777138850572564953, 839967991029301247, 599, 3885038158778096269468893991882380063764065770433606110283149695964997245520484669311748838825973451239771955518933348332721403496018696846203290707966794803507099534240007184258836096614399, 701, 2368778164222232774191928573951, 811, 26099, 929, 991, 34847, 3095263992211830248865791, 1457749, 1259

The smallest prime of the form z{0}1 in base b for b = 2, 3, 4, ..., 36 are (always minimal prime (start with b+1) base b):

3, 7, 13, 101, 31, 43, 449, 73, 9001, 259374246011, 19009, 157, 2549, 211, 241, 1336337, 307, 218336795902605993201009018384568383223, 31129600000000000001, 421, 463, 255042399139852495799, 13249, 601, 16901, 13817467, 757, 23549, 23490001, 858874531, 35740566642812256257, 34849, 1123, 41651, 45361

The smallest prime of the form {z}1 in base b for b = 2, 3, 4, ..., 36 are (always minimal prime (start with b+1) base b):

3, 7, 13, 3121, 31, 43, 549755813881, 73, 991, 1321, 248821, 157, 2731, 211, 241, 34271896307617, 307, 6841, 13107199999999999999981, 421, 463, 141050039560662968926081, 331753, 601, 17551, 7625597484961, 757, 1816075630094014572464024421543167816955354437761, 21869999971, 29761, 34359738337, 1185889, 1123, 42841, 60466141

The smallest prime of the form 1{0}z in base b for b = 2, 3, 4, ..., 36 are (always minimal prime (start with b+1) base b):

3, 5, 7, 29, 11, 13, 71, 17, 19, 131, 23, 181, 2177953337809371149, 29, 31, 83537, 5849, 37, 419, 41, 43, 279863, 47, 15649, 701, 53, 811, 420707233300229, 59, 61, 3599131035634557106248430806148785487095757694641533306480604458089470064537190296255232548883112685719936728506816716098566612844395439751206812144692131084107807, 35969, 67, 1259, 71

The smallest prime of the form {y}z in base b for b = 3, 4, 5, ..., 36 are (always minimal prime (start with b+1) base b): (this form is not interpretable in base 2)

5, 11, 19, 29, 41, 439, 71, 89, 109, 131, 16836900297891418080414469547118518955584357920776290786511507224819852347973193037600665289070901330976115445902783343792856149076064327963454445124840887022352433623214149015015943271257627167012185236811023315748308075343126054090560004563875124190448995227748073744916159908957819701603274854998000296763254125672206384758348891742961717040363229489213108521955314350073857925001010097317113705164622416602981584525394558649693204742511309000575073486313783914987497483013408328355077527202814535784777000148396721007194688339582681878366906510944731328876064735814127172451578146421749559114747412555063799277435883965467381, 181, 535461077009, 239, 271, 98801, 754617425461612781, 379, 419, 461, 74751395041, 317351, 599, 11406121, 701, 21139, 811, 703862069, 929, 991, 44846087994920604803621301910895025159221288572184381044859909289485641865275583407994915558605570950946734680988452965394713688106491174400948294112101434213220953385460836118066048540254897963702856420719776463509047941852241566428193327643900875942259940601970070393196751457213209881319085660701362499099654878864165624569792466096820413058979402445330988995938988157868541715781, 38113, 7385772222129586256251615636489, 1259


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