20200529, 18:11  #1 
May 2020
1 Posts 
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

20200529, 18:15  #2 
6809 > 6502
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Aug 2003
101×103 Posts
2·3^{2}·7·67 Posts 
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: https://www.mersenne.ca/exponent/283243137 Last fiddled with by Uncwilly on 20200529 at 19:16 
20200529, 18:57  #3 
"Curtis"
Feb 2005
Riverside, CA
2^{2}×1,061 Posts 
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!

20200628, 14:34  #4 
"TF79LL86GIMPS96gpu17"
Mar 2017
US midwest
10230_{8} Posts 
https://www.alpertron.com.ar/ECM.HTM: 283 243137 = 3 × 17 × 23 × 241469
Mersenne numbers with composite (factorable) exponents are never prime: https://www.mersenneforum.org/showpo...13&postcount=4 so we know that 2^{283243137}1 has several factors and is not prime. (OP may benefit from the beginning of the larger reference material at https://www.mersenneforum.org/showthread.php?t=24607) 
20200710, 08:29  #5 
Jun 2020
3×7 Posts 

20200714, 07:50  #6 
"Jeppe"
Jan 2016
Denmark
2·71 Posts 
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 
20200714, 15:51  #7 
"TF79LL86GIMPS96gpu17"
Mar 2017
US midwest
2^{3}×3^{2}×59 Posts 
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: https://www.alpertron.com.ar/ECM.HTM: 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. 
20200714, 19:33  #8  
Nov 2016
4173_{8} Posts 
Quote:
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, ... Last fiddled with by sweety439 on 20200714 at 19:34 

20200714, 20:06  #9 
6809 > 6502
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Aug 2003
101×103 Posts
2×3^{2}×7×67 Posts 
OP never replied. Let's stop beating this horse. It has passed on.
Closing thread. 
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