mersenneforum.org I think I discovered new largest prime
 Register FAQ Search Today's Posts Mark Forums Read

 2020-05-29, 18:11 #1 harrik   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
 2020-05-29, 18:15 #2 Uncwilly 6809 > 6502     """"""""""""""""""" Aug 2003 101×103 Posts 22×72×47 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 2020-05-29 at 19:16
2020-05-29, 18:57   #3
VBCurtis

"Curtis"
Feb 2005
Riverside, CA

10010000100112 Posts

Quote:
 Originally Posted by harrik 2^283.243.137 − 1 I dont use any computer. I use my new formule pls check it. Thnk you for everything
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!

 2020-06-28, 14:34 #4 kriesel     "TF79LL86GIMPS96gpu17" Mar 2017 US midwest 10010111100012 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 2283243137-1 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)
2020-07-10, 08:29   #5
Zcyyu

Jun 2020

23 Posts

Quote:
 Originally Posted by harrik 2^283.243.137 − 1 I dont use any computer. I use my new formule pls check it. Thnk you for everything

 2020-07-14, 07:50 #6 JeppeSN     "Jeppe" Jan 2016 Denmark 22×41 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
 2020-07-14, 15:51 #7 kriesel     "TF79LL86GIMPS96gpu17" Mar 2017 US midwest 12F116 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^n-1 where n= a x b is composite, is composite, and is a repdigit (number with repeating digits), with factors 2^a-1 and 2^b-1 easily visible when expressed in base 2^a and 2^b respectively. Consider 2^8-1 = 255 = 2^(2*4)-1 2^2-1 = 3 = 255/85. 2^4-1 = 15 = 255/17. 2^4-1 = 15 = 2^2-1 * 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^a-1 2^b-1 2^c-1 2^d-1 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^a-1) / (2^b-1) / (2^c-1) / (2^d-1) is. Which may be prime or composite.
2020-07-14, 19:33   #8
sweety439

Nov 2016

22·691 Posts

Quote:
 Originally Posted by kriesel 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^n-1 where n= a x b is composite, is composite, and is a repdigit (number with repeating digits), with factors 2^a-1 and 2^b-1 easily visible when expressed in base 2^a and 2^b respectively. Consider 2^8-1 = 255 = 2^(2*4)-1 2^2-1 = 3 = 255/85. 2^4-1 = 15 = 255/17. 2^4-1 = 15 = 2^2-1 * 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^a-1 2^b-1 2^c-1 2^d-1 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^a-1) / (2^b-1) / (2^c-1) / (2^d-1) is. Which may be prime or composite.
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, ...

Last fiddled with by sweety439 on 2020-07-14 at 19:34

 2020-07-14, 20:06 #9 Uncwilly 6809 > 6502     """"""""""""""""""" Aug 2003 101×103 Posts 22·72·47 Posts OP never replied. Let's stop beating this horse. It has passed on. Closing thread.

 Similar Threads Thread Thread Starter Forum Replies Last Post dabaichi News 571 2020-10-26 11:02 kgr Miscellaneous Math 3 2013-03-04 09:20 Frodo42 Lounge 1 2005-06-15 19:44 gbvalor Math 1 2002-12-10 04:11 wfgarnett3 Lounge 7 2002-11-25 06:34

All times are UTC. The time now is 06:51.

Tue Jan 26 06:51:57 UTC 2021 up 54 days, 3:03, 0 users, load averages: 2.57, 2.80, 2.76

This forum has received and complied with 0 (zero) government requests for information.

Permission is granted to copy, distribute and/or modify this document under the terms of the GNU Free Documentation License, Version 1.2 or any later version published by the Free Software Foundation.
A copy of the license is included in the FAQ.