20211120, 10:39  #210  
"99(4^34019)99 palind"
Nov 2016
(P^81993)SZ base 36
C84_{16} Posts 
Quote:
An interesting minimal (probable) prime (start with b'+1) in base b'=b^2 is b=34, it is (34^294277+1)/35, see references http://www.primenumbers.net/prptop/s...&action=Search and http://www.fermatquotient.com/PrimSerien/GenRepuP.txt, this (probable) prime is (1122^147137):1123 in base b=1156, and this (probable) prime is minimal (probable) prime (start with b+1) in this base, also, for some bases, like 4, 8, 27, 32, such primes do not exist because of algebra factors, thus the families {12}:13 in base 16, {56}:57 in base 64, {702}:703 in base 729, {992}:993 in base 1024, etc. are ruled out as only contain composite numbers (only consider the numbers > base), for the smallest such prime (always minimal prime (start with b'+1) in base b' = b^2) (its formula is (#^n)$, where # = b^2b and $ = b^2b+1), see this list (for the first 100 bases b' = b^2, list the value n of the minimal prime (#^n)$): Code:
b'=b^2 the n of (#^n)$ 4,1 9,1 16,algebra 25,1 36,4 49,7 64,algebra 81,28 100,1 121,1 144,1 169,4 196,2 225,2 256,1 289,2 324,2 361,7 400,1 441,1 484,1 529,4 576,2 625,2 676,4 729,algebra 784,8 841,2 900,68 961,53 1024,algebra 1089,1 1156,147137 1225,4 1296,14 1369,1 1444,1 1521,5 1600,25 1681,7 1764,353 1849,1 1936,2 2025,50 2116,2 2209,1 2304,1 2401,2 2500,575 2601,73 2704,2 2809,10970 2916,2 3025,1 3136,17 3249,25 3364,7 3481,7 3600,467 3721,2 3844,4 3969,17 4096,algebra 4225,8 4356,2 4489,1172 4624,377 4761,4 4900,29 5041,1 5184,2 5329,2 5476,5 5625,1 5776,1 5929,17 6084,2 6241,52 6400,1 6561,1 6724,145 6889,8 7056,2 7225,82 7396,2 7569,2 7744,353 7921,5 8100,22 8281,4 8464,17 8649,43 8836,34 9025,20 9216,17 9409,>249998 9604,8 9801,2 10000,145 10201,2 Last fiddled with by sweety439 on 20211124 at 09:43 

20211120, 10:57  #211 
"99(4^34019)99 palind"
Nov 2016
(P^81993)SZ base 36
2^{2}·3^{2}·89 Posts 
This problem is better than the original minimal prime problem since this problem is regardless whether 1 is considered as prime or not, i.e. no matter 1 is considered as prime or not prime (in the beginning of the 20th century, 1 is regarded as prime), the sets M(Lb) in this problem are the same, while the sets M(Lb) in the original minimal prime problem are different, e.g. in base 10, if 1 is considered as prime, then the set M(Lb) in the original minimal prime problem is {1, 2, 3, 5, 7, 89, 409, 449, 499, 6469, 6949, 9049, 9649, 9949, 60649, 666649, 946669, 60000049, 66000049, 66600049}, while if 1 is not considered as prime, then the set M(Lb) in the original minimal prime problem is {2, 3, 5, 7, 11, 19, 41, 61, 89, 409, 449, 499, 881, 991, 6469, 6949, 9001, 9049, 9649, 9949, 60649, 666649, 946669, 60000049, 66000049, 66600049}, however, in base 10, the set M(Lb) in this problem is always {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}, no matter 1 is considered as prime or not prime.
Besides, this problem is better than the original minimal prime problem since this problem (if we extend the problem to any larger and larger base) will cover "is there a prime of the form (a*b^n+c)/gcd(a+c,b1) (a>=1, b>=2, c != 0, gcd(a,c) = 1, gcd(b,c) = 1)?" for any given (a,b,c) triple satisfying that a>=1, b>=2, c != 0, gcd(a,c) = 1, gcd(b,c) = 1 (see this post), we can extend this problem to a large base which is power of b, while the original minimal prime problem will not, e.g. consider the form 67607*2^n+1, since if 67607*2^n+1 is prime then n == 3, 11 (mod 24) (for all other n, 67607*2^n+1 is divisible by at least one of {3, 5, 13, 17}) we can extend this problem to base b = 2^24 = 16777216, since 2^24 > 67607, thus the prime "67607" will be singledigit prime in base b = 2^24, which is excluded in this problem, and now we can consider the unsolved family (540856):{0}:1 (where 540856 = 67607 * 2^3) in base b = 16777216, whether there is a prime of this form is equivalent to whether there is a prime of the form 67607*2^n+1 with n == 3 (mod 24) 
20211121, 15:27  #212 
"99(4^34019)99 palind"
Nov 2016
(P^81993)SZ base 36
2^{2}×3^{2}×89 Posts 
For the forms in https://docs.google.com/spreadsheets...RwmKME/pubhtml, it is conjectured that every base has infinitely many primes of these forms:
1{0}z 1{z} y{z} z{0}1 {z}1 However .... * For the form 2{0}1, all bases b == 1 mod 3 can be ruled out as only contain composite numbers (by trivial 1cover {3}), the smallest base b != 1 mod 3 such that family 2{0}1 has no primes is conjectured to be 201446503145165177 (with covering set {3, 5, 17, 257, 641, 65537, 6700417}), but this conjecture is very hard to prove (however, since the smallest prime of the form 2{0}1 (if exists) is always minimal prime (start with b+1), the "minimal primes (start with b+1) problem" in all bases b<201446503145165177 covers the conjecture that 201446503145165177 is the smallest base b != 1 mod 3 such that there is no prime of the form 2{0}1 * Since in odd bases, 1{0}2 is the dual form of 2{0}1 (in even bases, form 1{0}2 can be ruled out as only contain composite numbers (by trivial 1cover {2})), 201446503145165177 is also conjectured to be the smallest base b == 3, 5 mod 6 (i.e. b is odd AND b != 1 mod 3) (with covering set {3, 5, 17, 257, 641, 65537, 6700417}), but again, this conjecture is also very hard to prove (however, since the smallest prime of the form 1{0}2 (if exists) is always minimal prime (start with b+1), the "minimal primes (start with b+1) problem" in all bases b<201446503145165177 covers the conjecture that 201446503145165177 is the smallest base b == 3, 5 mod 6 such that there is no prime of the form 1{0}2 * For the form {1}, all bases in A096059 (all are perfect powers) can be ruled out as only contain composite numbers (by differenceofrthpower factorization), but there is * For the form 4{0}1, all bases b == 1 mod 5 can be ruled out as only contain composite numbers (by trivial 1cover {5}), the smallest base b != 1 mod 5 such that family 4{0}1 has no primes is proven to be 14, for bases 5, 8, 12, the 3digit number 401 is prime, and for all other base 4<b<14 (b>4 is required since bases b<=4 have no digit "4" and hence no family "4{0}1") which not == 1 mod 5, the 2digit number 41 is prime * For the form 1{0}4, all bases b == 1 mod 5 and all even bases can be ruled out as only contain composite numbers (by trivial 1cover, {5} for bases b == 1 mod 5 and {2} for even bases), the smallest base which is neither == 1 mod 5 nor even base is proven to be 29, while the largest prime is 1000004 (with 7 digits) in base 23, which is equal to 23^6+4 
20211123, 15:45  #213 
"99(4^34019)99 palind"
Nov 2016
(P^81993)SZ base 36
2^{2}·3^{2}·89 Posts 
Every base b family is the dual (for the definition, see http://www.kurims.kyotou.ac.jp/EMIS...rs/i61/i61.pdf and https://oeis.org/A076336/a076336c.html and https://mersenneforum.org/showthread.php?t=10761 and https://mersenneforum.org/showthread.php?t=6545 and https://mersenneforum.org/showthread.php?t=26328 and https://mersenneforum.org/showthread.php?t=21954, also see Five or Bust forum, the dual form has same Nash weight (or difficulty), the difficulty page has many forms shown as having same difficulty and next to each other (e.g. (107*10^n71)/9 and (71*10^n107)/9, 4*10^n+3 and 75*10^n+1, (53*10^n+37)/9 and (37*10^n+53)/9, (5*10^n437)/9 and (874*10^n1)/9), they are exactly the dual forms each other), for the forms in base b, these forms are dual forms each other:
1{0}1 (b^n+1) is the dual of itself (if b is even) 1{0}2 (b^n+2) and 2{0}1 (2*b^n+1) (if b is odd) #{0}1 (# = b/2) ((b/2)*b^n+1) and 2{0}1 (2*b^n+1) (if b is even) 1{0}3 (b^n+3) and 3{0}1 (3*b^n+1) (if b == 2, 4 mod 6) #{0}1 (# = b/3) ((b/3)*b^n+1) and 3{0}1 (3*b^n+1) (if b == 0 mod 6) 1{0}z (b^n+(b1)) and z{0}1 ((b1)*b^n+1) 1{0}11 (b^n+(b+1)) and 11{0}1 ((b+1)*b^n+1) {1} ((b^n1)/(b1)) is the dual of itself 1{z} (2*b^n1) and {z}y (b^n2) (if b is odd) 1{z} (2*b^n1) and #{z} (# = b/21) ((b/2)*b^n1) (if b is even) 2{z} (3*b^n1) and {z}x (b^n3) (if b == 2, 4 mod 6) 2{z} (3*b^n1) and #{z} (# = b/31) ((b/3)*b^n1) (if b == 0 mod 6) 2{1} (((2*b1)*b^n1)/(b1)) and {1}0z ((b^n(2*b1))/(b1)) {y}z (((b2)*b^n+1)/(b1)) and {1}2 ((b^n+(b2))/(b1)) (if b is odd) y{z} ((b1)*b^n1) and {z}1 (b^n(b1)) 10{z} ((b+1)*b^n1) and {z}yz (b^n(b+1)) also .... for any digits X and Y: X{0}Y and Y{0}X (if gcd(X,b) = 1, gcd(Y,b) = 1) X{0}1 and Y{0}1 (if X*Y = b) X{z} and Y{z} (if (X+1)*(Y+1) = b) Last fiddled with by sweety439 on 20211219 at 21:26 
20211123, 18:13  #214 
"99(4^34019)99 palind"
Nov 2016
(P^81993)SZ base 36
2^{2}·3^{2}·89 Posts 
See https://stdkmd.net/nrr/prime/primesize.txt, sequences of the form k*10^n+1 includes Generalized Cullen prime numbers, and sequences of the form k*10^n1 includes Generalized Woodall prime numbers, thus, sequences of the form k*b^n+1 (of the form *{0}1 in base b) includes Generalized Cullen prime numbers base b (numbers of the form n*b^n+1, i.e. the special case that k=n), and sequences of the form k*b^n1 (of the form *{z} in base b) includes Generalized Woodall prime numbers base b (numbers of the form n*b^n1, i.e. the special case that k=n), for such numbers, see:
Generalized Cullen primes: http://www.prothsearch.net/cullen.html (broken link: from wayback machine cached copy) (b=2) http://guenter.loeh.name/gc/status.html (3<=b<=100) https://harvey563.tripod.com/GClist.txt (101<=b<=10000) http://www.primzahlenarchiv.de/index.html (broken link: from wayback machine cached copy) (101<=b<=200) http://www.primzahlenarchiv.de/news.html (broken link: from wayback machine cached copy) (101<=b<=200) http://www.primzahlenarchiv.de/status.html (broken link: from wayback machine cached copy) (101<=b<=200) https://www.rieselprime.de/ziki/Gen_Cullen_prime_table (all available bases b) http://www.primegrid.com/forum_thread.php?id=7073 (primegrid search page) http://www.primegrid.com/stats_cullen_llr.php (primegrid status page, b=2) http://www.primegrid.com/stats_gcw_llr.php (primegrid status page, b>=3) https://primes.utm.edu/top20/page.php?id=6 (top 20 page, b=2) https://primes.utm.edu/top20/page.php?id=42 (top 20 page, b>=3) https://oeis.org/A240234 (smallest n for given base b) https://oeis.org/A327660 (smallest n>b2 for given base b) Generalized Woodall primes: http://www.prothsearch.net/woodall.html (broken link: from wayback machine cached copy) (b=2) https://harvey563.tripod.com/GWlist.txt (3<=b<=10000) https://harvey563.tripod.com/General...dallPrimes.txt (3<=b<=10000) http://science.kennesaw.edu/~jdemaio/generali.htm (broken link: from wayback machine cached copy) (3<=b<=100) https://www.rieselprime.de/ziki/Gen_Woodall_prime_table (all available bases b) http://www.primegrid.com/forum_thread.php?id=7073 (primegrid search page) http://www.primegrid.com/stats_woodall_llr.php (primegrid status page, b=2) http://www.primegrid.com/stats_gcw_llr.php (primegrid status page, b>=3) https://primes.utm.edu/top20/page.php?id=7 (top 20 page, b=2) https://primes.utm.edu/top20/page.php?id=45 (top 20 page, b>=3) https://oeis.org/A240235 (smallest n for given base b) https://oeis.org/A327661 (smallest n>b2 for given base b) Also, since the dual for n*b^n+1 and n*b^n1 are b^n+n and b^nn, respectively (we assume that gcd(n,b) = 1) (since the dual of k*b^n+1 and k*b^n1 are b^n+k and b^nk, respectively), and n*b^n+1 is including in *{0}1, n*b^n1 is including in *{z}, b^n+n is including in 1{0}*, b^nn is including in z{*}, thus the OEIS sequence: https://oeis.org/A093324 (b^n+n) https://oeis.org/A084743 (b^n+n, corresponding primes) https://oeis.org/A084746 (b^nn) https://oeis.org/A084745 (b^nn, corresponding primes) Last fiddled with by sweety439 on 20211123 at 18:31 
20211124, 18:45  #215  
"99(4^34019)99 palind"
Nov 2016
(P^81993)SZ base 36
2^{2}·3^{2}·89 Posts 
Quote:
e.g. A014233(12) = A014233(13) = 318665857834031151167461 = 399165290221 * 798330580441 = (399165290220+1) * (2*399165290220+1), and 8*318665857834031151167461+1 = 1596661160883^2 is a square, and for the currently only "probable primes that failed primality proof or had a factor" with order 21 in factordb, 1396981702787004809899378463251 = 835757651112751 * 1671515302225501 = (835757651112750+1) * (2*835757651112750+1), and 8*1396981702787004809899378463251+1 = 3343030604451003^2 is a square. (both of them are triangular numbers) Factorization of A014233(n): Code:
1: 2047 = 23 * 89 (form (m+1) * (4*m+1) with m+1 and 4*m+1 both primes) 2: 1373653 = 829 * 1657 (form (m+1) * (2*m+1) with m+1 and 2*m+1 both primes) 3: 25326001 = 2251 * 11251 (form (m+1) * (5*m+1) with m+1 and 5*m+1 both primes) 4: 3215031751 = 151 * 751 * 28351 (can also be written as 28351 * 113401, is still form (m+1) * (4*m+1), although 4*m+1 is not prime) 5: 2152302898747 = 6763 * 10627 * 29947 (form (7*m+1) * (11*m+1) * (31*m+1) with 7*m+1, 11*m+1 and 31*m+1 all primes) 6: 3474749660383 = 1303 * 16927 * 157543 (form (m+1) * (13*m+1) * (121*m+1) with m+1, 13*m+1 and 121*m+1 all primes) 7,8: 341550071728321 = 10670053 * 32010157 (form (m+1) * (3*m+1) with m+1 and 3*m+1 both primes) 9,10,11: 3825123056546413051 = 149491 * 747451 * 34233211 (form (m+1) * (5*m+1) * (229*m+1) with m+1, 5*m+1 and 229*m+1 all primes) 12,13: 318665857834031151167461 = 399165290221 * 798330580441 (form (m+1) * (2*m+1) with m+1 and 2*m+1 both primes) In fact, the hexagonal number (m+1)*(2*m+1) (if both m+1 and 2*m+1 are primes) are Fermat pseudoprime to all bases which are coprime to this triangular number and quadratic residue mod 2*m+1 (reference for a special example that base = 4, which is quadratic residue mod all numbers since it is square number) Since the Fermat primality test is b^(n1) == 1 mod n, composite number n is Fermat pseudoprime base b if and only if znorder(Mod(b,n)) divides n1, thus, composite number n is Fermat pseudoprime base b if and only if these three conditions are all satisfied: * gcd(b,n) = 1 * For every prime p dividing n, b is ((p1)/gcd(p1,n1))th power residue (quadratic residue for 2, cubic residue for 3, quartic residue for 4, octic residue for 8, ...) mod p * For every prime power p^r (r>1) dividing n, p is Wieferich prime base b with order >= r1 (i.e. b^(p1) == 1 mod p^r) (for Carmichael numbers n, all bases b such that gcd(b,n) = 1 satisfy these three conditions, thus Carmichael numbers n are Fermat pseudoprimes to all bases b coprime to n) (if (p1)/gcd(p1,n1) = 1, i.e. p1 divides n1, then every b is ((p1)/gcd(p1,n1))th power residue mod p, thus this part of conditions (e.g. p=7 for 91, which 71 divides 911) are not listed below, since every b satisfies this condition) e.g. * 15 is Fermat pseudoprime base b if and only if gcd(b,15) = 1 and b is 2nd power residue (quadratic residue) mod 5 * 21 is Fermat pseudoprime base b if and only if gcd(b,21) = 1 and b is 3rd power residue (cubic residue) mod 7 * 22 is Fermat pseudoprime base b if and only if gcd(b,22) = 1 and b is 10th power residue mod 11 * 24 is Fermat pseudoprime base b if and only if gcd(b,24) = 1 and b is 2nd power residue (quadratic residue) mod 3 and 2 is Wieferich prime with order >=2 base b * 25 is Fermat pseudoprime base b if and only if gcd(b,25) = 1 and 5 is Wieferich prime base b * 27 is Fermat pseudoprime base b if and only if gcd(b,27) = 1 and 3 is Wieferich prime with order >=2 base b * 28 is Fermat pseudoprime base b if and only if gcd(b,28) = 1 and b is 2nd power residue (quadratic residue) mod 7 and 2 is Wieferich prime base b * 35 is Fermat pseudoprime base b if and only if gcd(b,35) = 1 and b is 2nd power residue (quadratic residue) mod 5 and b is 3rd power residue (quadratic residue) mod 7 * 65 is Fermat pseudoprime base b if and only if gcd(b,65) = 1 and b is 3rd power residue (cubic residue) mod 13 * 87 is Fermat pseudoprime base b if and only if gcd(b,87) = 1 and b is 14th power residue mod 29 * 91 is Fermat pseudoprime base b if and only if gcd(b,91) = 1 and b is 2nd power residue (quadratic residue) mod 13 * 117 is Fermat pseudoprime base b if and only if gcd(b,117) = 1 and b is 3rd power residue (cubic residue) mod 13 and 3 is Wieferich prime base b * 148 is Fermat pseudoprime base b if and only if gcd(b,148) = 1 and b is 12th power residue mod 37 and 2 is Wieferich prime base b * 169 is Fermat pseudoprime base b if and only if gcd(b,169) = 1 and 13 is Wieferich prime base b * 205 is Fermat pseudoprime base b if and only if gcd(b,205) = 1 and b is 10th power residue mod 41 * 231 is Fermat pseudoprime base b if and only if gcd(b,231) = 1 and b is 3rd power residue (cubic residue) mod 7 * 285 is Fermat pseudoprime base b if and only if gcd(b,285) = 1 and b is 9th power residue mod 19 * 325 is Fermat pseudoprime base b if and only if gcd(b,325) = 1 and 5 is Wieferich prime base b * 341 is Fermat pseudoprime base b if and only if gcd(b,341) = 1 and b is 3rd power residue (cubic residue) mod 31 * 493 is Fermat pseudoprime base b if and only if gcd(b,493) = 1 and b is 4th power residue (quartic residue) mod 17 and b is 7th power residue (septic residue) mod 29 (for more informations, see attached text files and https://de.wikibooks.org/wiki/Pseudo...eudoprimzahlen and https://de.wikibooks.org/wiki/Pseudo...2815__4999%29) Thus, if n is hexagonal number (m+1)*(2*m+1) with m+1 and 2*m+1 both primes, then n is Fermat pseudoprime base b for all bases b which is coprime to n and 2nd power residue (quadratic residue) mod 2*m+1, which has m*m = m^2 bases in all n (= (m+1)*(2*m+1)) modular classes of n, which has about a half modular classes of n, thus these number are easily to be Fermat pseudoprime (if the base b is random chosen), and if n is octagonal number (m+1)*(3*m+1) with m+1 and 3*m+1 both primes, then n is Fermat pseudoprime base b for all bases b which is coprime to n and 3rd power residue (cubic residue) mod 3*m+1, which has m*m = m^2 bases in all n (= (m+1)*(3*m+1)) modular classes of n, which has about one third modular classes of n. Note that no polygonal numbers can be primes (except the trivial case, i.e. every n is the second ngonal number), see this reference, for generalized polygonal numbers, only index 2 (n3), 3 (n), and 4 (3*n8) can be primes, no generalized polygonal numbers with index > 4 can be primes. Thus, for a number to test primality, suggest that if this number passes the strong test to first few prime bases (strong pseudoprimes are always Fermat pseudoprimes to the same bases), test if 3*n+1, 8*n+1, 24*n+1 are squares (this is very easily to test), if at least one of them are squares and n>7, then n must be composite because of the algebraic factorization of difference of two square numbers, if none of these three numbers are squares, then n is very likely to be prime. Last fiddled with by sweety439 on 20211230 at 19:02 

20211124, 18:59  #216 
"99(4^34019)99 palind"
Nov 2016
(P^81993)SZ base 36
2^{2}×3^{2}×89 Posts 
Since triangular number * 9 + 1 are always triangular numbers, and no triangular numbers > 3 are primes, thus these families in base 9 contain no primes (only count numbers > base), since they can be factored as difference of squares divided by 8:
1{1} (={1}), 3{1}, 6{1}, 11{1} (={1}), 16{1}, 23{1}, 31{1} (=3{1}), 40{1}, ... (all numbers in these families are triangular numbers, and thus all such numbers except 3 are composite, but the prime 3 is not allowed since prime must be > base) they are the families (triangular number){1}, in fact, families 6{1} and 16{1} are already ruled out as only contain composite by covering set {2,5} Since generalized octagonal numbers * 4 + 1 are always generalized octagonal numbers, and no generalized octagonal numbers > 5 are primes, thus these families in base 4 contain no primes (only count numbers > base), since they can be factored as difference of squares divided by 3: 1{1} (={1}) (except the number 11 = 5 in decimal), 11{1} (={1}), 20{1}, 100{1}, 111{1} (={1}), 201{1} (=20{1}), 220{1}, ... (all numbers in these families are generalized octagonal numbers, and thus all such numbers except 11 in base 4 (= 5 in decimal) are composite, the prime 5 is allowed in base 4, but not allowed in bases > 4 which are powers of 4 since prime must be > base) they are the families (generalized octagonal number){1}, and since base b families can be converted to base b^n families for n>1 (the repeating digit (i.e. the digit in {}) will be multiple of Rn(b), where Rn(b) is the base b repunit with length n), these base 4 families can be converted to base 16 (=4^2) families, {1} in base 4 = 1{5} in base 16, 20{1} in base 4 = 8{5} in base 16, 100{1} in base 4 = 10{5} in base 16, 220{1} in base 4 = A1{5} in base 16, etc. (one base b^n digit is equivalent to n base b digits, and thus minimal prime (start with b+1) base b is always minimal prime (start with b'+1) base b' = b^n if this prime is > b' (references: the case base 3 families *{1} converted to base 9=3^2 families *{4} the case n base 10 digits is equivalent to one base 10^n digit, in research of truncatable primes (instead of minimal primes) to other bases the case of padic numbers, one base p^n digit is equivalent to n base p digits, also see the automorphic numbers to base b and base b^n, they are equivalent) Since generalized pentagonal numbers * 25 + 1 are always generalized pentagonal numbers, and no generalized pentagonal numbers > 7 are primes, thus these families in base 25 contain no primes (only count numbers > base), since they can be factored as difference of squares divided by 24: 1{1} (={1}), 2{1}, 5{1}, 7{1}, C{1}, F{1}, M{1}, 11{1} (={1}), 1A{1}, 1F{1}, 21{1} (=2{1}), 27{1}, 2K{1}, 32{1}, 3H{1}, 40{1}, ... (all numbers in these families are generalized pentagonal numbers, and thus all such numbers except 2, 5, 7 are composite, but the primes 2, 5, 7 are not allowed since prime must be > base) they are the families (generalized pentagonal number){1}, in fact, families 1F{1} and 3H{1} are already ruled out as only contain composite by covering set {2,13} Last fiddled with by sweety439 on 20220117 at 06:21 
20211125, 11:48  #217 
"99(4^34019)99 palind"
Nov 2016
(P^81993)SZ base 36
2^{2}·3^{2}·89 Posts 
Now I consider to add these families to the list, although they do not always provide minimal primes (start with b+1):
* 10{z} * {z}yz * 11{0}1 * 1{0}11 * {1}2 * {1}3 * {1}4 * {1}0z * 2{1} * 3{1} * 4{1} * {z0}z1 
20211126, 14:56  #218  
"99(4^34019)99 palind"
Nov 2016
(P^81993)SZ base 36
110010000100_{2} Posts 
Quote:
base 11 unsolved family 5{7}: {7}: divisible by 7 base 13 unsolved family A{3}A: A{3}: covering set {2,7} {3}A: covering set {2,7} {3}: divisible by 3 base 16 unsolved family {3}AF: {3}A: divisible by 2 {3}F: divisible by 3 {3}: divisible by 3 base 16 unsolved family {4}DD: {4}D: covering set {3,7,13} {4}: divisible by 2 base 17 unsolved family F1{9}: F{9}: divisible by 3 1{9}: odd length: difference of squares divided by 16; even length: divisible by 2 {9}: divisible by 9 base 19 unsolved family EE1{6}: EE{6}: divisible by 2 E1{6}: divisible by 3 E{6}: divisible by 2 1{6}: odd length: difference of squares divided by 3; even length: divisible by 5 {6}: divisible by 6 base 25 unsolved family CM{1}: C{1}: difference of squares divided by 24 M{1}: difference of squares divided by 24 {1}: difference of squares divided by 24 base 25 unsolved family E{1}E: E{1}: covering set {2,13} {1}E: covering set {2,13} {1}: difference of squares divided by 24 base 25 unsolved family F{1}F1: F{1}F: divisible by 5 F{1}1 = F{1}: difference of squares divided by 24 {1}F1: covering set {2,13} {1}F: divisible by 5 {1}1 = {1}: difference of squares divided by 24 base 25 family LO{L}8: LO{L}: divisible by 3 L{L}8 = {L}8: covering set {2,13} O{L}8: (another unsolved family) L{L} = {L}: divisible by 21 O{L}: divisible by 3 base 27 family 999{G}: 99{G}: divisible by 2 9{G}: difference of cubes divided by 13 {G}: divisible by 16 base 33 family FFF{W}: FF{W}: divisible by 2 F{W}: odd length: difference of squares; even length: divisible by 17 {W}: divisible by 32 base 47 family 8{0}FF1: 8{0}FF: divisible by 8 8{0}F1: divisible by 2 8{0}F: divisible by 23 8{0}1: covering set {3, 5, 13} (other subfamilies have either leading zeros or trailing zeros (or both)) Last fiddled with by sweety439 on 20211211 at 18:53 

20211126, 17:44  #219  
"99(4^34019)99 palind"
Nov 2016
(P^81993)SZ base 36
2^{2}·3^{2}·89 Posts 
Quote:
For 11{0}1 (the dual of 1{0}11), see https://www.rieselprime.de/ziki/Williams_prime_PP_least For 10{z} (the dual of {z}yz), see https://www.rieselprime.de/ziki/Williams_prime_PM_least Last fiddled with by sweety439 on 20211126 at 17:44 

20211126, 22:14  #220 
"99(4^34019)99 palind"
Nov 2016
(P^81993)SZ base 36
2^{2}×3^{2}×89 Posts 
For families X{Y}Z_{b} = (a*b^n+c)/gcd(a+c,b1) (a>=1, b>=2, c != 0, gcd(a,c) = 1, gcd(b,c) = 1), which can be ruled out as only containing composite, which are by covering congruence, and which are by algebra factors, and which are by combine of them?
Hint: * If by covering congruence, then all X{Y}Z_{b} numbers must have small prime factor (say less than 10^6, if b is small), especially, if there is (positive or negative) integer n such that all prime factors of numerator(abs((a*b^n+c)/gcd(a+c,b1))) are prime factors of b (including the case that numerator(abs((a*b^n+c)/gcd(a+c,b1))) = 1), then (a*b^n+c)/gcd(a+c,b1) cannot have covering congruence. * If by (full or partial) algebra factors, then there must be integers n>=1 and r>1 such that a*b^n and c are both rth powers, or there must be integer n>=1 such that a*b^n*c is of the form 4*m^4 with integer m * If by full algebra factors, then there must be integer r>1 such that a, b, c are all rth powers, or a*c is of the form 4*m^4 with integer m and b is 4th power Thus .... * Base 25 family C{1} must be only algebra factors, not by covering congruence, since Code:
length factors 6 19 · 101 · 233 · 263 12 69173177 · 415039063 18 71 · 91337821853 · 1080830891927 36 277 · 383 · 4253 · 9157 · 305497 · 2649281 · 5391097 · 141581477 · 39961446977 72 337 · 499 · 1051 · 4598699002567 · 5169269868350933 · 132787369068406184845865339939 · 38714450336182811476979541935153363 84 103 · 139 · 197 · 9277 · 500921 · 13288909 · (94digit composite with no small factor) * Base 25 family E{1} must be only covering congruence, not by algebra factors, since E(1^n) = (337*25^n1)/24, and there is no n such that 337*25^n is perfect power (since the exponent of the prime 337 in prime factorization of 337*25^n is always 1), thus there is no n>=1 and r>1 such that 337*25^n and 1 are both rth powers thus, E{1} can only by covering congruence, indeed, E(1^n) is divisible by 2 if n is even, and divisible by 13 if n is odd For the base 9 families .... * {1} must be only algebra factors, not by covering congruence, since (9^n1)/8 is divisibility sequence, by Zsigmondy's theorem, (9^n1)/8 has primitive prime factor for all n>=1, and cannot have covering congruence thus {1} can only by algebra factors, indeed, (1^n) = (9^n1)/8 = (3^n1) * (3^n+1) / 8 * 3{1} must be only algebra factors, not by covering congruence, since all prime factors of (25*9^n1)/8 are prime factors of 9 when n=0, thus 3{1} cannot have covering congruence thus 3{1} can only by algebra factors, indeed, 3(1^n) = (25*9^n1)/8 = (5*3^n1) * (5*3^n+1) / 8 * 5{1} must be only covering congruence, not by algebra factors, since 5(1^n) = (41*9^n1)/8, and there is no n such that 41*9^n is perfect power (since the exponent of the prime 41 in prime factorization of 41*9^n is always 1), thus there is no n>=1 and r>1 such that 41*9^n and 1 are both rth powers thus, 5{1} can only by covering congruence, indeed, 5(1^n) is divisible by 5 if n is even, and divisible by 2 if n is odd * However, 6{1} by both covering congruence and algebra factors, both of them can prove that 6{1} contain no primes, since: ** 6(1^n) is divisible by 2 if n is even, and divisible by 5 if n is odd ** 6(1^n) = (49*9^n1)/8 = (7*3^n1) * (7*3^n+1) / 8 Last fiddled with by sweety439 on 20211211 at 15:14 
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