I think I discovered new largest prime
2^283.243.137 − 1 I dont use any computer. I use my new formule pls check it. Thnk you for everything

That is not a prime number. For it to be a Mersenne Prime, the exponent mus be prime. Your's is not.
Edit to insert link: [url]https://www.mersenne.ca/exponent/283243137[/url] 
[QUOTE=harrik;546751]2^283.243.137 − 1 I dont use any computer. I use my new formule pls check it. Thnk you for everything[/QUOTE]
Next time you think your formula found a prime, check it yourself the software is found at mersenne.org. That way you won't have to share credit with anyone else! 
[URL]https://www.alpertron.com.ar/ECM.HTM:[/URL] 283 243137 = 3 × 17 × 23 × 241469
Mersenne numbers with composite (factorable) exponents are never prime: [URL]https://www.mersenneforum.org/showpost.php?p=512813&postcount=4[/URL] so we know that 2[SUP]283243137[/SUP]1 has several factors and is not prime. (OP may benefit from the beginning of the larger reference material at [URL]https://www.mersenneforum.org/showthread.php?t=24607[/URL]) 
[QUOTE=harrik;546751]2^283.243.137 − 1 I dont use any computer. I use my new formule pls check it. Thnk you for everything[/QUOTE]
283,243,137=3*17*23*241469,so Your conclusion is wrong. 
To make what everybody else said already, more explicit:
Your exponent 283243137 is divisible by 3. So 283243137 = 3*N. Then the number you propose, namely 2^283243137  1, is equal to 2^(3*N)  1 = (2^3)^N  1. And that will be divisible by 2^3  1 = 7. So the number you suggest, is divisible by 7 (because your exponent is divisible by 3). /JeppeSN 
What formula?
The alleged prime 2^283243137  1 can easily be shown to be composite without using a computer or calculator. Sum the decimal digits of the exponent: 33. (Sometimes called "casting out nines") The digit sum is obviously divisible by 3. Any number 2^n1 where n= a x b is composite, is composite, and is a repdigit (number with repeating digits), with factors 2^a1 and 2^b1 easily visible when expressed in base 2^a and 2^b respectively. Consider 2^81 = 255 = 2^(2*4)1 2^21 = 3 = 255/85. 2^41 = 15 = 255/17. 2^41 = 15 = 2^21 * cofactor 5. For an exponent with 4 distinct prime factors, for example from the OP, 283243137: [url]https://www.alpertron.com.ar/ECM.HTM:[/url] 283243137 = 3 × 17 × 23 × 241469 a=3 (repdigit 2^3  1 = 7's in base 2^3 = 8) b=17 (repdigit 2^17  1 = 131071's in base 2^17 = 131072) c=23 (repdigit 2^23  1 = 8388607's in base 2^23 = 8388608) d=241469 (repdigit 2^241469  1 in base 2^241469) The number has numerous factors (at least 14, as shown below), each of which corresponds to being able to express the number as a repdigit in some base 2^B where 2^B=factor+1. For an exponent with four distinct prime factors, a, b, c, d, there are unique factors as follows prime factors 2^a1 2^b1 2^c1 2^d1 composite factors 2^(ab)1 2^(ac)1 2^(ad)1 2^(bc)1 2^(bd)1 2^(cd)1 2^(abc)1 2^(abd)1 2^(bcd)1 There's also a cofactor, whatever 2^(abcd)1 / (2^a1) / (2^b1) / (2^c1) / (2^d1) is. Which may be prime or composite. 
[QUOTE=kriesel;550583]What formula?
The alleged prime 2^283243137  1 can easily be shown to be composite without using a computer or calculator. Sum the decimal digits of the exponent: 33. (Sometimes called "casting out nines") The digit sum is obviously divisible by 3. Any number 2^n1 where n= a x b is composite, is composite, and is a repdigit (number with repeating digits), with factors 2^a1 and 2^b1 easily visible when expressed in base 2^a and 2^b respectively. Consider 2^81 = 255 = 2^(2*4)1 2^21 = 3 = 255/85. 2^41 = 15 = 255/17. 2^41 = 15 = 2^21 * cofactor 5. For an exponent with 4 distinct prime factors, for example from the OP, 283243137: [url]https://www.alpertron.com.ar/ECM.HTM:[/url] 283243137 = 3 × 17 × 23 × 241469 a=3 (repdigit 2^3  1 = 7's in base 2^3 = 8) b=17 (repdigit 2^17  1 = 131071's in base 2^17 = 131072) c=23 (repdigit 2^23  1 = 8388607's in base 2^23 = 8388608) d=241469 (repdigit 2^241469  1 in base 2^241469) The number has numerous factors (at least 14, as shown below), each of which corresponds to being able to express the number as a repdigit in some base 2^B where 2^B=factor+1. For an exponent with four distinct prime factors, a, b, c, d, there are unique factors as follows prime factors 2^a1 2^b1 2^c1 2^d1 composite factors 2^(ab)1 2^(ac)1 2^(ad)1 2^(bc)1 2^(bd)1 2^(cd)1 2^(abc)1 2^(abd)1 2^(bcd)1 There's also a cofactor, whatever 2^(abcd)1 / (2^a1) / (2^b1) / (2^c1) / (2^d1) is. Which may be prime or composite.[/QUOTE] The general number is Phi_n(2), where Phi is the cyclotomic polynomial, which may be prime or composite, the value of Phi_n(2) for n = 1, 2, 3, ... are 1, 3, 7, 5, 31, 3, 127, 17, 73, 11, 2047, 13, 8191, 43, 151, 257, 131071, 57, 524287, 205, 2359, 683, 8388607, 241, 1082401, 2731, 262657, 3277, 536870911, 331, 2147483647, 65537, 599479, 43691, 8727391, 4033, 137438953471, 174763, 9588151, 61681, 2199023255551, 5419, 8796093022207, 838861, 14709241, 2796203, 140737488355327, 65281, 4432676798593, 1016801, ... Phi_n(2) is prime for n = 2, 3, 4, 5, 6, 7, 8, 9, 10, 12, 13, 14, 15, 16, 17, 19, 22, 24, 26, 27, 30, 31, 32, 33, 34, 38, 40, 42, 46, 49, 56, 61, 62, 65, 69, 77, 78, 80, 85, 86, 89, 90, 93, 98, 107, 120, 122, 126, 127, 129, 133, 145, 150, 158, 165, 170, 174, 184, 192, 195, 202, 208, 234, 254, 261, 280, 296, 312, 322, 334, 345, 366, 374, 382, 398, 410, 414, 425, 447, 471, 507, 521, 550, 567, 579, 590, 600, 607, 626, 690, 694, 712, 745, 795, 816, 897, 909, 954, 990, 1106, 1192, 1224, 1230, 1279, 1384, 1386, 1402, 1464, 1512, 1554, 1562, 1600, 1670, 1683, 1727, 1781, 1834, 1904, 1990, 1992, 2008, 2037, 2203, 2281, 2298, 2353, 2406, 2456, 2499, 2536, ... 
OP never replied. Let's stop beating this horse. It has passed on.
:deadhorse: Closing thread. 
All times are UTC. The time now is 14:11. 
Powered by vBulletin® Version 3.8.11
Copyright ©2000  2021, Jelsoft Enterprises Ltd.