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 2020-06-21, 22:01 #5 enzocreti   Mar 2018 17×31 Posts Pg(69660) pg(92020) and pg(541456) pg(k) primes with k multiple of 86 This identity involves pg(k) primes with k multiple of 86 k multiple of 86 with pg(k) primes satisfy this identity 541456-(69660*16/10+1000)=1000*sqrt(92020*2+1) pg(366770) is prime and pg(331259) is prime 366770 has the same residue 66 mod 344 and mod 559 331259 has the same residue 331 mod 344 and mod 559 pg(75894) is prime 75894 has the same residue 214 mod 215 and mod 344 215=559-344 also pg(39699) is prime and 39699 Leaves the same residue 139 mod 215 and mod 344 I thinkk that there is something related to this equation: 1720x+214=139y using Wolphram Alpha you get integers solutions are: x=139n+44 and y=1720n+546 so there are pg(k) primes with k multiple of 546... pg(75894) and pg(56238)..for example... i think that investigating further this will shed a light in these primes this could explain why there are pg(k) primes with k multiple of 546 and pg(k) primes with k multiple of 139 as pg(3336) for example and other pg(k) primes with k multiple of 215 as pg(215), pg(69660) and pg(92020) 69660 and 92020 leave the same residue 860 mod 1720 69660=(546+1720*40+314) 92020=(546+1720*53+314) 546+314=860 Curious the 314 which is the first three digits of pi So another conjecture could be there are infinitely many s such that ec(s) is prime and s has the form 860+1720k. Because the multiple of 43 are congruent to 344 mod 559 then k must have the form 13r+1, for some integer r The numbers 1720 and 546 must be involved in some misterious way in these primes Pg(2131) is prime Pg(2131*9) is prime Pg(331259) is prob prime 331259-1720*139-546=2131*43 I note that (2131*43+546)+1720*139=331259 2131*43+546 is prime and is equal to 92179 which is a number ending with 179 as 19179=9*2131 I note also that pg(92020) is prime and 92020=92179+159, maybe it is not chance So 2131*x+546+1720y 0r something like that... If x and y are 0, then you have pg(k) primes with k multiple of 546 If x and y are not 0, then you have maybe other infinite pg(k) primes If I am not wrong I am ready for the conjecture: There are infinitely many pg(k) primes with k of the form 2131*x+546*y+1720*z for some nonnegative integers x y and z I think that it is more general to restate the conjecture There are infinitely many pg(k) primes with k of the form 2131*x+546*y+860*z pg(2131) Pg(19179) Pg(92020) Pg(69660) Pg(331259) Pg(56238) Pg(75894) Should be pg primes with this form of k (1456+331*2-1000) /2=559 92020 is congruent to 2 mod 331 and mod 139 I notice that pg(75894) and pg(56238) are primes with 75894 and 56238 multiple of 546 75894 and 56238 leave the same residue 182 mod (3^6-1)=728 728*2=1456 (1390+728-1000) /2=559 182=728-546 Seem to exist primes pg(k) with k of the form 2131*x+182 as pg(541456) and pg primes with k of the form 728x+182 as pg(75894) and pg(56238) 541456 is congruent to 6 mod 182 75894 and 56238 are congruent to 0 mod 182 Maybe if k is congruent to 6 mod 182 then k is always of the form 2131x+182 and if k is 0 mod 182 then k is 728x+182 541456 and 92020 are of the form 8643s+5590. But 541456 is also of the form 860s+516 69660 is of the form 8643s+516 541456,92020,69660 are multiple of 86 Pg(56238) and pg(75894) are primes 56238 and 75894 are both congruent to 2184 mod (17^3+1) 2184=3*(3^6-1) 56238 and 75894 are 0 mod 546 Consider now 215, 69660, 92020, 541456 and 331259 Pg(215) pg(69660), pg(92020) pg(331259) pg(541456) are primes 215, 69660, 92020, 541456 plus or minus 2021 are divisible by 559 which is a sum of two positive cubes 6^3+7^3 331259-2021 is divisible by the sum of two positive cubes 1+17^3 I note that the sum of 1+17^3 and 6^3+7^3 are both multiple of 13 I notice that 215 69660 92020 541456 and 331259 are the k such that pg(k) is prime and k is congruent to plus or minus 6 mod 13 I also notice that 559 and 1+17^3 leave the same residue 23 mod 67 So 215 69660 92020 541456 and 331259 are congruent to plus or minus 6 mod 13 and congruent to plus or minus 2021 mod the sum of two positive cubes either 559=6^3+7^3 or 1+17^3 ...with 559 and 1+17^3 congruent to 23 mod 67 2021=559*4-215=43*47 product of two consecutive primes but 2021 is also of the form 559s+344 so 331259 (not multiple of 43 and congruent to 6 mod 13) is congruent to 559s+344 mod (1+17^3) whereas 215, 69660, 92020, 541456 (multiple of 43 and congruent to + or - 6 mod 13) are congruent to + or - 344 mod(6^3+7^3) It is quite clear that 92020, 69660 are congruent to 0 mod 860 541456 is congruent to (344+172) mod 860 331259 is congruent to (331-172) mod 860 I notice that 92020 541456 69660 are multiple of 86 331259 is congruent to - 13 mod 86 So mod 860 215 69660 92020 are congruent to (344+172) mod 860 331259 which is not multiple of 86 but - 13 mod 86 is congruent to (344-13-172) mod 860 Where naturally 172=344/2 215, 92020, 69660, 541456 are 10^m mod 41 and so they have the form 1763s+r where r is a residue (215,344,903,1677)...anyway also 331259 can be written as 1763s+r and the residue r seems to be not random...infact I think that r=344+1234 quite curious 1234-6 is a multiple of 307 and 331259-6 is a multiple of 307...this number 307 seems to be involved...215*10-1 and 541456*10-1 is multiple of 307 331259 is congruent to 1234 mod (307*215) 331259 is congruent to (1234-215*5=159) MOD 860 26^2-1+559=1234 so you can sobstitute (331259-19)/(3^6-1)=2*(3^6-1)-1001 331259 Leaves the same reminder 19 mod (3^6-1) and mod (26^2) so 1234 and 331259 leave the same residue 159 mod 215 and the same residue 6 mod 307. 1763-1234=23^2 maybe this is useful Infact 23^2+26^2=42^2-559 so if pg(k) is prime and k is congruent to + or - 6 mod 13, then k is either or the form 215s or 215s+73+13 or of the form 215s+2*73+13 this could be coincidence : 331259=215s+159 541456=215s+86 for some s s in both cases has the form (163r+73) 92020, 69660, 541456,331259 are congruent to (331-73x-13y) mod 344 for some x and y 215, 69660, 92020, 541456 are multiple of 86 and they are multiple either of 215=6^3-1 or 344=7^3+1 215, 69660, 92020 divided by 215 give a number that is congruent to plus or minus 1 mod 13 541456/344=1574 gives a number congruent to 1 mod 13 92020, 541456 and 331259 are congruent to 6 mod 13 and not multiple of 3 92020-929=91091 which is divisible by 1001 541456-(929-13)=540540 which is divisible by 1001 but also by 13# where #is primorial 331259-929=330330 which is divisible by 1001 but also by 13# Consider pg(75894) and pg(56238) they are primes 75894 and 56238 are multiple of 546 75894 and 56238 are congruent to 1638 mod 2184 546, 1638, 2184 in base 16 are repdigits numbers 75894 and 56248 are congruent to 546 mod 1092 1092 is a repdigit in base 16 A curious fact: 546 in base 16 is 222 546-222=18^2 pg(215*18^2) is prime note that 324 is a square in base 10 but also in base 16 541456=(215+1456)*324+52 92020=324*284+4 69660=324*215 0, 4, 52 which are the residues mod 324 are numbers of the form n^3*(5n+3)/2 pg(331259) is prime 331259=(429^2-1)/2+427*559+546 pg((429^2-1)/2) is prime pg(69660) pg(92020) and pg(331259) are probable primes 69660, 92020, 331259 are 6 mod 13 92020=69660+lcm(344,215,559) 331259=13*(429^2-11)/10+(429^2-1)/2 where (429^2-1)/2=92020 pg(75894) and pg(56238) are primes with 75894 and 56238 multiple of 546 75894+56238=132132 curious! pg(6231) and pg(51456) are primes with 6231 and 51456 multiple of 67 (67 in base 16 is 43) I would hazard that when pg(k) is prime and k is a multiple of 67, then k is congruent to plus or minus 10^m mod 41 pg(215), pg(69660), pg(92020), pg(541456) are probable primes with 215, 69660, 92020, 541456 multiple of 43. 215, 69660, 92020, 541456 are congruent to plus or minus (2580-559k) (mod 2795) with k=0 in the cases 215, 69660, 92020 (multiple of 215) and k=1 in the last case 541456 (multiple of 344) Last fiddled with by enzocreti on 2020-08-07 at 07:44 Reason: O