20191106, 13:46  #1 
Mar 2018
17·31 Posts 
binary form of the exponents 69660, 92020, 541456
pg(69660), pg(92020) and pg(541456) are probable primes with 69660, 92020 and 541456 multiple of 86
69660 in binary is 10001000000011100 92020 in binary is 10110011101110100 541456 in binary is 10000100001100010000 you can see that the number of the 1's is always a multiple of 5 a chance? 
20191218, 08:26  #2 
Mar 2018
17×31 Posts 
69660, 92020, 541456
69660, 92020 and 541456 are 6 mod 13 (and 10^m mod 41)
69660, 92020 and 541456 are multiple of 43 is there a reason why (696606)/26 is congruent to 13 mod 43 (920206)/26 is congruent to 13 mod 43 (5414566)/26 is congruent to 13 mod 43? 215, 69660, 92020, 541456 are multiple of 43 let be log the log base 10 int(x) let be the integer part of x so for example int(5.43)=5 A=10^2*log(215)215/41 int(A)=227=B 215 (which is odd) is congruent to B+1 mod 13 69660 which is even is congruent to B mod 13 92020 is congruent to B mod 13 and 541456 is also congruent to B mod 13 215 (odd) is congruent to 1215 mod 13 69660 (even) is congruent to 1215 mod 13 92020 (even) is congruent to 1215 mod 13 541456 (even) is congruent to 1215 mod 13 215, 69660, 92020, 541456 can be written as 13x+1763y+769 (54145676913*93)/1763=306 (6966076913*824)/1763=33 (9202076913*781)/1763=46 as you can see 306,33 and 46 are all 7 mod 13 769+13*824 769+13*781 769+13*93 are multiples of 43 (69660(10^3+215))/13=5265 which is 19 mod 43 (92020(10^3+215))/13=6985 which is 19 mod 43 (541456(10^3+215))/13=41557 which is 19 mod 43 10^3+215 is 6 mod 13 now 69660/13=5358,4615384... 92020/13=7078,4615384... 541456/13=41650,4615384... the repeating term 4615384 is the same...so that numbers must have some form 13s+k? 215 (odd) is congruent to 307*2^210^3 or equivalently to  (19*2^61) mod 13 69660 (even) is congruent to 307*2^21001 or equivalenly to (19*2^61) mod 13 92020 (even) the same 541456 (even) the same (19*2^6*(54145692020)/(13*43)+1)/13+10=75215 is a multiple of 307=(215*101)/7=(54145601)/17637 pg(51456) is another probable prime with 51456 congruent to 10^n mod 41 75215=(51456*19+1)/13+10=(19*2^6*(54145692020)/(13*43)+1)/13+10 215 is congruent to 1215=5*3^5 mod 13 69660 is congruent to 1215 mod 13 and so also 92020 and 541456 1215=(51456/22*13*10^243)/19 ...so summing up... 215 (odd) is congruent to 3*19*2^2 mod((41*43307)/(7*2^4)=13) where 41*43307 is congruent to 10^3+3*19*2^2 mod (3*19*2^2=228) 69660 (even) is congruent to (3*19*2^21) mod 13 ... the same for 92020 and 541456 another way is 215 (odd) is congruent to 15*81 mod ((41*43307)/(30715*13)) 69660 (even) is congruent to 15*81 mod ((41*43307)/(30715*13)) the same for 92020 and 541456 215 is congruent to (41*4330710^3)/2 mod ((41*43307)/(7*2^4)) 69660 is congruent to (41*4330710^3)/21 mod ((41*43307)/(7*2^4)) 92020 is congruent to (41*4330710^3)/21 mod ((41*43307)/(7*2^4)) 541456 is congruent to (41*4330710^3)/21 mod ((41*43307)/(7*2^4)) 215 is congruent to 3*19*2^2 mod ((41*43307)/(2*(19*31))) 69660 is congruent to 3*19*2^21 mod ((41*43307)/(2*(19*31))) and so 92020 and 541456 215 is congruent to 3*19*2^2 mod ((3^61)/(3*191)) 69660 is congruent to 3*19*2^21 mod((3^61)/(3*191)) and so 92020 and 541456 51456 (pg(51456) is probable prime and 51456 is 10^n mod 41) is congruent to 19*3*2^4 mod 13 Pg(2131) is probable prime 2131 is prime 227=2131307*426^2 So 215 is congruent also to 2131307*426^2 mod 13 And 69660 is congruent to 2131307*426^2+1 mod 13 and so also 92020 and 541456 69660 is congruent to 1763307*51 mod ((1763307)/112)=13) And so 92020 and 541456 215 which is odd is congruent to  1763+307*5+1 mod 13 215+1763307*51 is divisible by 17 and by 13 And also 5414561763+307*5+1 is divisible by 13 and 17 215 and 541456 have the same residue 10 mod 41 ((5414561763+307*5+1)/(13*17)+1) *200+51456=541456 69660 92020 541456 are congruent to 7*2^6 mod 26 7*2^6(1763307*51)=221=13*17 215+7*2^6 is a multiple of 221 5414567*2^6 is a multiple of 221 Last fiddled with by enzocreti on 20200126 at 17:58 
20200124, 15:31  #3 
Mar 2018
17×31 Posts 
215 , 51456, 69660, 92020, 541456
51456, 69660, 92020, 541456 are even and congruent to 10^n mod 41
pg(51456), pg(69660), pg(92020) and pg(541456) are prp 51456, 69660, 92020, 541456 are all congruent to 7*2^q1 mod 13 with q a nonnegative integer 215 is odd and pg(215) is prp 215 is congruent to 7*2^q mod 13 
20200330, 11:52  #4 
Mar 2018
1000001111_{2} Posts 
69660 and 92020
69660 and 92020 are multiple of 215 and congruent to 344 mod 559
92020=lcm(215,344,559)+69660  denotes concatenation in base 10 2^696601  2^695591 is prime 2^920201  2^920191 is prime!!! Last fiddled with by enzocreti on 20200330 at 11:53 
20200330, 12:19  #5 
Romulan Interpreter
Jun 2011
Thailand
10001011100100_{2} Posts 
please show us a proof that they are prime

20200330, 12:50  #6 
Mar 2018
1017_{8} Posts 
...
Well...
actually they are only probable primes... maybe in future they will be proven primes Last fiddled with by enzocreti on 20200330 at 12:52 
20200330, 17:39  #7 
Mar 2018
20F_{16} Posts 
http://factordb.com/index.php?id=1100000001110801143 http://factordb.com/index.php?query=...%2B2%5E920191 Last fiddled with by enzocreti on 20200330 at 17:55 
20200330, 23:06  #8 
Mar 2018
17·31 Posts 
... I note also...
69660 I note also that
(lcm(215,344,559))^2=4999*10^5+6966060 I notice that lcm(215,344,559)=22360 22360/(18*18)=69.01234567... curious I notice also that 541456 (multiple of 43),is  215 mod (18*181) and 69660 (multiple of 43) is 215 mod (18*181) I note that the polynomial X^2X*429^2+7967780460=0 has the solution x=69660 If you see the discriminant of such polynomial you can see interesting things about pg primes with exponent multiple of 43 I note that 429^2 is congruent to 1 mod 215 and to 1 mod 344. I note that 92020*2+1=429^2 The discriminant of the polynomial is 429^44*7967780460 which is a perfect square and lcm(215,344,559) divides 429^44*79677804601 Pg(331259) is prime and pg(92020) is prime. 92020+(92020/2151)*559+546=331259 Again magic numbers 559 and 546 strike! So 331259 is a number of the form 215*(13s1)+(13*b2)*559+546 For some s, b positive integers So there are pg primes pg(75894) and pg(56238) with 75894 and 56238 multiple of 546 and pg(331259) with 331259 of the form 215m+559g+546 for some positive m and g. Pg(69660) is prime. 69660=(3067*8546)*311# where # is the primorial and 3067 is a prime of the form 787+456s notice that lcm(215,344,559)=22360 22360/(18*18)=69.01234567... curious I notice also that 541456 (multiple of 43),is  215 mod (18*181) and 69660 (multiple of 43) is 215 mod (18*181) So we have pg(215) is prime pg(69660) is prime pg(92020) is prime With 215 69660 and 92020 multiple of 43 69660=215+(18*181)*215 92020=69660+18*18*69+4 215 is congruent to 108 mod (18*181=323) 541456 is congruent to 108 mod (18*181) 92020 is congruent to (17*171) mod (18*181) 69660 is congruent to 215 mod (18*181) 108=6^318^2 17*171+36216=17*171+6^26^3=108 215=6^31 To make it easier 215, 69660, 541456 are congruent to plus or minus 215 mod 323 92020 is congruent to (17*171) mod (18*181) curious that 289/215 is about 1.(344)... and 541456 92020 69660 215 are congruent to plus or minus 344 mod 559 215, 69660, 541456 are congruent to plus or minus 6^31 mod 323 92020 is congruent to (12/9)*6^3 mod 323 92020*9/12 is congruent to 6^3 mod 323 92020 is congruent to (17^21) mod 323 and to  (6^21) mod 323 92020 is a number of the form 8686+13889s 13889=(6^3+1)*64 215 69660 92020 541456 are + or  344 mod 559 lcm(215,344,559)86*(10^2+1)+6^31=(6^3+1)*2^6+1 92020=69660+lcm(215,344,559) so you can substitute 92020=69660+86*(10^2+1)6^3+1+(6^3+1)*2^6+1 86*(10^2+1) mod 323 is 17^21 215, 69660, 541456 are multiple of 43 and congruent to 10 and 1 mod 41 They are congruent to plus or minus 215 mod 323 92020 is congruent to 2^4 (not a power of 10) mod 41 92020 is congruent to (2^4+1)^21 mod 323 288 is 17^21 288 in base 16 is 120 120=11^21 also 323=18^21 in base 16 is 143=12^21 344*((14444561763*2^9) /3441)=541456 1444456=lcm(13,323,344) 541456=lcm(13,323,344)344*(41*2^6+1) 215 69660 92020 541456 are congruent to plus or minus (3^a*2^b) mod 323 215 is congruent to  108 mod 323 541456 is congruent to 108 mod 323 69660 is congruent to  108 mod 323 92020 is congruent to 288 mod 323 108 and 288 are numbers of the form 3^a*2^b So exponents multiple of 43 are congruent to plus or minus 344 mod 559 and to plus or minus a 3smooth number mod 323 108 and 288 are both divisible by 36 Last fiddled with by enzocreti on 20200818 at 21:00 Reason: notice that lcm(215,344,559)=22360 22360/(18*18)=69.01234567... curious I notice also that 541456 (multiple of 43),is  21 
20200819, 05:48  #9  
Mar 2018
20F_{16} Posts 
Q77I
Quote:
215 69660 92020 541456 are congruent to plus or minus (6^k1) mod 323 for k=3,2 69660=(2^5*3^7)(2^3*3^4) so it is the difference of two 3 smooth numbers (2^a*3^b)(2^(a3)*3^(b3)) 69660 is multiple of 3 and congruent to 0 mod (6^21) 215, 92020, 541456 are not multiple of 3 and multiple of 43 and are congruent to plus or minus 2^k mod 36 for some k 1763*323(6^3+1)*(2^7+1)=541456 or (42^21)*(18^21)(6^3+1)*(2^7+1)=541456 I also note that 69660=(2^7+1)*(6^3+1)+(2^7+1)*(18^21) And by the way 215=(42^21)*4^2(2^7+1)*(6^3+1) 27993=(217*129)=(2^7+1)*(6^3+1) I notice that  541456 mod 27993=9202*2 92020=10*9202 And 9202*2*10+1=429^2 27993=3/5*(6^61) 27993 in base 6 is 333333 27993 has also the representation: (42^21)*4^2(6^31)=27993=2*43*(18^21)+(6^31) x/5+(42^21)*(18^21)3/5*(6^61)=20*3/5*(6^61) The solution of this equation x=92020 this identity: (42^21)*(18^21)=(10^3+18^2)*430+43*3 Maybe it is not a chance that pg(10^3+18^21=1323) is prime pg(1323), pg(215), pg(69660), pg(92020), pg(541456) are primes 1323, 215, 69660, 92020, 541456 are congruent to plus or minus (2^a*3^b1) mod 323 where 2^a*3^b is a 3 smooth number with 2^a*3^b<323 or 1323, 215, 69660, 92020, 541456 are congruent to plus or minus (p1) mod 323 where p is a perfect power 92020 has the factorization 2*(6^31)*(6^32) lcm(215,344,559)=(6^31)*(2*(6^32)18^2)=(2*(6^32)18^2)*(2*(6^32)18^2+111) 69660=92020lcm(215,344,559) 215=2*(6^32)+2*(6^3+1)2*18^2+1 92020=(2*(6^32)+2*(6^3+1)2*18^2+1)*(2*(6^32)+2*(6^3+1)2*18^2)*2 so 215=2*(6^32)+2*(6^3+1)2*18^2+1 69660==(2*(6^32)+2*(6^3+1)2*18^2+1)*(2*(6^32)+2*(6^3+1)2*18^2)*2((6^31)*(2*(6^32)18^2)) 92020=(2*(6^32)+2*(6^3+1)2*18^2+1)*(2*(6^32)+2*(6^3+1)2*18^2)*2 541456=(42^21)*(18^21)(7^36^3+2)*(6^3+1) 69660=1111115*11111+8*(42^21) 344=7^3+1=2*(1111115*11111)/(18^21)=A so lcm(215,344,559)=lcm(215, A, A+215) 541456=5*111112(42^21)*8 (12^21) divides (1111115*111116^3+1) 559=(1111115*111116^3+1)/(12^21)+(1111115*11111)/(18^21) Quite clear that 215, 69660, 92020,541456 multiple of 43 are congruent to plus or minus (18^26^k) mod (18^21) where k is 2 or 3 and 3 indeed is the maximum exponent such that 18^26^k i 1s positive (54145618^2+6^3)/323=41^25=1676 (9202018^2+6^2)/323=17^25=284 (69660+18^26^3)/323=6^3 (215+18^26^3)/323=1 There is clearly a pattern!!! I notice also that (1+5)=6 is a semiprime (6^3+5)=13*17 is a semiprime (1676+5)=41*41 is a semiprime (284+5)=17*17 is a semiprime (6,221,1681,289) are either squares of primes or product of two consecutive primes so when pg(k) is prime and k is a multiple of 43, then k can be expressed in this way: (p*q5)*(18^21) plus or minus (18^26^k) with 6^k<18^2 p and q are primes when pg(k) is prime and k is a multiple of 546, then k is congruent to 78 mod (18^26^3) as in the cases k=56238 and k=75894 Curio of the curios: pg(331259) is probable prime and 331259 is prime magic: 71*6^6331*(10^43*331)=331259 71*6^6 and 331259 have in common the first five digits 33125=182^2+1 If you consider 71*6^n for n>3 For n even 71*6^n is congruent to (11^21) mod (13^21) and for n odd Is congruent to (7^21) mod (13^21) For n=4 71*6^4=92016 for n=6 71*6^6=3312576 As you can see Pg(92020) is prime and pg(331259) is prime 92020 has the last two digits 20 different from 92016 The same for 3312576 and 331259 Moreover 331259 mod (71*6^3) =9203 Both 331259 and 92020 are 5 mod 239 Is there something connected with the fact ord (71*6^k)=4 I mean the smallest value k for which 71*6^k is congruent to 1 mod 239 is k=4??? 71 and 6 are both quadratic residues mod 239 92020 and 331259 are congruent to 71*3^3 mod (239*13) I wonder if this concept could be generalized pg(51456), pg(92020), pg(331259) are probable primes 51456 is congruent to 71 mod (239*(6^31)) 92020 is congruent to 71*3^3 mod (239*13) 331259 is congruent to 71*3^3 mod (239*13) 92020 and 331259 are congruent to 6 mod 13 51456 and 331259 are congruent to 2^3 mod 109 71 and 71*3^3 are both congruent to 6 mod 13 so 51456, 92020, 331259 are congruent to 13*(5+71*k)+6 either mod (239*215) or mod (239*13) are these primes infinite? the odd thing is that 51456, 92020, 331259 are either congruent to 10^m mod 41 or prime (331259 infact is prime) So I think it is no chance that 51456, 92020, 331259 are either congruent to 2^j mod 71 or to 13*2^i mod 71 No chance at all! There is a file Rouge! Pg(541456) is probable prime as pg(51456) And 541456 mod (239*215)=9202*3, that is 3*( 92020/10 )and pg(92020) is prime 541456 is congruent to (3/20)*(429^21) mod (239*215) 541456 is congruent to 14*71*3^3+3*2^8 (mod (239*215)) Maybe there is some connection to the fact that 215, 69660, 92020 and 541456 are plus or minus 344 mod 559 lcm(344,559)=4472=71*(2^61)1 I suspect that something in field F(239) is in action! 331259^(1) mod (215*239)=49999=(10^52)/2 ((71*(2^61))1)/2=lcm(344,559)=92026966 Multiplying both sides by 10 you have 92020=69660+lcm(215,344,559) I would suggest to study these exponents in field F(51385=239*215) (239*10*215+3*(429^21)/10541456) /3=9202 So 10* (239*10*215+3*(429^21)/10541456) /3=92020 In this equation we have 215, 92020, 541456 multiple of 43 and not of 3 The other multiple of 43 is 69660 which is multiple of 3 And 92020=69660+lcm(344,215,559) 239*10*215+3*(429^21)/10 is a multiple of 559 239*10*215+3*(429^21)/10=(42^21)*(18^22)+43*2^5 This identity (239*10*215+(3/10)*(429^21)541456+92020)/559=92020/(215*2) One can play around with this expression containing 215 and 559 And substitute for example 92020 with (429^21)/2 Pg(331259) is prime and also 331259 is prime The inverse modulo (215*239) of 331259 is the prime 5*10^41=49999 (331259*499991)/(215*239)=7*(18^21)*(12^21)10^3+1 So 331259=((((18^21)*(10^3+1)10^3+1)*239*215)+1)/(5*10^41) Using Wolphram Alpha i considered this equation: ((323*(10^x+1)10^x+1)*239*215+1)/(5*10^(x+1)1)=y wolphram say that the integer solution is x=3, y=331259 wolphram gives an alternate form: 84898302/(5*(2^(x+1)*5^(x+2)1))+1654597/5=y the number 1654597=69660+(30^21)*(42^21) so 69660 and 331259 (both 6 mod 13 and pg(69660) pg(331259) primes) are linked by this equation: 84898302/(5*(2^(3+1)*5^(3+2)1))+(69660+(30^21)*(42^21))/5=331259 541456 in field F(239*215) and in field F(239*323) 541456 mod (239*215)=9202*3 541456 mod (239*323)=359*3 Pg(359) is prime, pg(9202*10) is prime 9202359 is a multiple of 239 Pg(92020+239239=331259) is prime I note also 331259=71*6^6331*(10^43*331)=6^2*920213=92020+239239 239239 is congruent to 13 mod 107 and mod 43 22360=(10*(239239+13))/107 92020=69660+22360 92020=331259239239 92020 is a multiple of 107 lcm(215,344,559)*10=22360 multiple of 86, that is 69660, 92020 and 541456 are congruent to a square mod 428 so the muliple of 86=k for which pg(k) is prime are numbers for which there is a solution to this modular equation: y is congruent to 36*x^2 mod 428 infact 92020 is congruent to 36*0^2 mod 428 69660 is congruent to 36*3^2 mod 428 541456 is congruent to 36*1^2 mod 428 There are pg(k) primes with k multiple of 215 and k multiple of 546 probably that numbers 215 and 546 are not random at all look at this equality: 71*6^6182^23^2331*(10^43*331)10=546^2 546215=331=546(6^31) by the way 182^2+3^2+546^2 is prime 546^2 is congruent to 6^3 mod 331 541456((3*239) ^2239+(429^21)/20)=(429^21)/10 541456=(18/5)*(429^21)/2+13*((429^211)/102235) where 2235=lcm(215,344,559)1 541456*10=3*331259+239*(136^2+1) 92020 and 331259 are both 5 mod 239 71*6^k is congruent to 1 mod 239 for k=4 But 71*6^4 is 920... And 71*6^6 is 33125... 92020 mod 71 is 4 and mod 331 is 3 71*6^4 is congruent to 1 mod (385*239) 541456 is congruent to  239 mod 385 I notice that 92020 and 331259 are congruent to 5 mod 239 but also to  72 mod (1001) I notice that 92020 is congruent to 146 mod 71 541456 and 331259 are congruent to 146 mod 703 I notice that 215, 69660, 541456 are plus or minus 215 mod 323 92020 which is congruent to 16 mod 41, is congruent to 288 mod 323 and mod 71 288 and 92020 are both 4 mod 71 so 92020 is congruent to 4 mod 71 and is congruent to 4+284 (mod 323) where 284 is the residue mod 323 of 71*6^4 215, 69660, 541456 are congruent to plus or minus 215 mod 323 where 215=4+211 211 is the residue of 71*6^4 mod 215 in particular 92020 is congruent to 288 mod 323 and mod 284, infact 92020=71*6^4+4 I think that also 331259 has something to do with 71*6^6 so multiple of 43, that is 215, 69660, 92020, 541456 are either of the form 323k+108, or 323k108, or 323k288 I notice that (323108)=6^31 and (323288)=6^21 211 and 284 are also the residues of 71*6^4 and 71*6^6 mod 323 the 18th pg prime is pg(1323), the 36th is pg(360787) modulo (18^21=323), 1323 is 31 and 360787 is 319 the difference between 319 and 31 is again 288 1323 is congruent to 31 mod (6^44) 360787 is congruent to 6^2*2^3+31 mod (6^44) 319 is congruent to 31 mod 6^2 infact here: 215, 69660, 541456 are congruent to plus or minus 108 mod 323 92020 is congruent to (6^41008=288) mod 323 so multiple of 43 are either congruent to plus or minus (18+90) mod (18^21) or to 6^4(18+90+900) mod (18^21) 92020 is congruent to (11*6^2108) mod (18^21) and to  11*6^2 mod (304^2) so great fact 215, 69660, 92020, 541456 are either congruent to plus or minus (108*21) mod (108*31) or to plus/minus (108/31) mod (108*31) Last fiddled with by enzocreti on 20201026 at 14:20 Reason: 3 
